Two circles, C1 and C2, intersect at point A and B. Passes through the center O of circle C2. The point P lies on circle C2 so that the line PAT is target to circle Ci and point A. Let ∠APB =.

Your chances of winning a raffle with one ticket out of 325 sold is approximately 0.31% or 1 in 325.

The probability of winning a raffle is determined by dividing the number of tickets you have by the total number of tickets sold. In this case, since there are 325 tickets sold and you have only one ticket, your chances of winning are 1 in 325, which is equivalent to a probability of approximately 0.31%.

This means that you have a very low chance of winning, but it's not impossible. However, the more tickets you have, the greater your chances of winning will be. It's important to remember that winning a raffle is a matter of luck and chance, and not a guaranteed outcome.

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The statement that is best supported by the data is this: C. The range of data in Mr. Joe’s class is less than the range of data in Ms. Gambino’s class.

The true statement about the data is that the range of data in Mr. Joe's class is less than the range of data in Ms. Gambino's class.

The range of data in Mr. Joe's class spans from 4 to 10 hours while the range of data in Ms. Gambino's class spans from 4 to 12 hours. So, the data range for the latter class is higher than the former.

Complete Question:

The average number of hours of sleep of Ms. Gambino’s and Mr. Joe’s classes is shown below. Which of the following statements is best supported by the data?

The image shows a line graph:

Mr. Gambino's Class: Range 4 - 12

Hours: 4 = 0

5 = 1

6 = 1

7 = 3

8 = 5

9 = 3

10 = 2

11 = 1

12 = 1

Mr. Joe's class: Range 4 -12

4 = 0

5 = 1

6 = 5

7 = 3

8 = 1

9 = 2

10 = 5

11 = 0

12 = 0

The median number of hours slept in Ms. Gambino’s class is less than the median number of hours in Mr. Joe’s class.

The data for Ms. Gambino’s class is symmetrical, while the data for Mr. Joe’s class is skewed right.

The range of data in Mr. Joe’s class is less than the range of data is Ms. Gambino’s class.

The mode of the data in Ms. Gambino’s class was equal to the mode of the data in Mr. Joe’s class.

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Answer:

2√6 + 9

Step-by-step explanation:

√24 + √81

= √4√6 + 9

= 2√6 + 9

Answer: TJ has 17 quarters and Demi has 26 quarters.

Step-by-step explanation:

Let t be the number of quarters TJ has and d be the number of quarters Demi has.

First, we will write an equation based on the first detail given:

       d = t + 9

Next, we will write an equation based on the second detail given:

       2t + 3d = 112

Now, we will substitute the first equation into the second and solve for t.

       2t + 3d = 112

       2t + 3(t + 9) = 112

       2t + 3t + 27 = 112

       5t + 27 = 112

       5t = 85

       t = 17 quarters

Lastly, we know that Demi has 9 more quarters than TJ. We will add 9.

       17 quarters + 9 quarters = 26 quarters

Answer:

TJ has 17 quarters.

Step-by-step explanation:

Let d and t equal the numbers of quarters Demi and TJ have, respectively.

"Demi has nine more quarters than TJ."

d = t + 9

"if you double the number of quarters TJ has and triple the number of quarters then he has you would get 112 quarters in total"

I think that "then he" above really should read "Demi."

2t + 3d = 112

d = t + 9

2t + 3d = 112

2t + 3(t + 9) = 112

2t + 3t + 27 = 112

5t = 85

t = 17

Answer: TJ has 17 quarters.

Starting with a 1-by-1 square, a sequence of squares J1, J2, J3, ... is created by attaching squares half as wide as the previous squares to the outer edges of each successive square. The area of J∞, the limit of this sequence, is 4/3.

To find the area of J1, we simply calculate the area of the original 1-by-1 square, which is 1.

To find the area of J2, we need to attach squares half as wide (and half as tall) to the middle of each side of J1. The area of each attached square is (1/2)² = 1/4, so the total area added to J1 is 4(1/4) = 1. Thus, the area of J2 is 1 + 4(1/4) = 2.

To find the area of J3, we need to attach squares half as wide as the squares added in the previous step to every outer edge of J2. The area of each attached square is (1/4)² = 1/16, so the total area added to J2 is 4(1/16) = 1/4. Thus, the area of J3 is 2 + 4(1/4) = 3.

We can continue this process to find the areas of J4, J5, and so on. In general, the area of Jn is equal to the area of the previous square plus the area added by the attached squares, which is 4(1/2^(n-1))^2 = 1/2^(2n-2). Therefore, the area of Jn is 1 + 1/4 + 1/16 + ... + 1/4^(n-1) = (4/3)(1 - 1/4^n).

As n approaches infinity, the area of Jn approaches the limit of (4/3)(1 - 0) = 4/3. Therefore, the area of J∞, the limit of the sequence, is 4/3.

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Answer:

Step-by-step explanation:

The data set in order is: 11, 13, 15, 19, 19, 21, 22, 22, 25, 26, 34, 34, 41, 43, 45, 48, 49, 50, 55, 58, 60, 62.

The seventh percentile is 15.

The twenty fifth percentile is 22.

The seventy ninth percentile is 58.

The length of QR to the nearest tenth of a foot is approximately 92.3 feet.

To find the length of QR in AQRS, we can use the Law of Cosines. This formula relates the lengths of the sides of a triangle to the cosine of one of its angles. In this case, we want to find QR, which is opposite the known angle ZQ.

So, let's write the formula:

QR² = SQ² + QS² - 2(SQ)(QS)cos(ZQ)

Substituting in the given values, we get:

QR² = 94² + QS² - 2(94)(QS)cos(41°)

We still need to find QS, but we can use the fact that ZS is a right angle to do so. Since ZQ and ZS are complementary angles (add up to 90°), we know that:

cos(ZQ) = sin(ZS)

So, we can rewrite the Law of Cosines formula as:

QR² = 94² + QS² - 2(94)(QS)sin(ZS)

Now we need to use the sine ratio to find QS. Since ZS is opposite the side SQ, we can write:

sin(ZS) = QS / SQ

Rearranging this equation gives:

QS = SQ sin(ZS)

Substituting in the values we know:

QS = 94 sin(90°)

Since sin(90°) = 1, we can simplify to:

QS = 94

Plugging this into our Law of Cosines equation:

QR² = 94² + 94² - 2(94)(94)sin(ZS)

QR² = 2(94)² - 2(94)²cos(41°)

QR² = 2(94)²(1 - cos(41°))

QR ≈ 92.3 feet (rounded to the nearest tenth)

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Answer:

Sure, here is the solution for the equation G/B=M/n:

```

B = Gn/M

```

Here is the step-by-step solution:

1. Multiply both sides of the equation by B.

```

G/B * B = M/n * B

```

2. Simplify both sides of the equation.

```

G = Gn/n

```

3. Divide both sides of the equation by n.

```

G/n = Gn/n * 1/n

```

4. Simplify both sides of the equation.

```

B = Gn/M

```

Therefore, the solution for B is Gn/M.

Step-by-step explanation:

1)  All the solution are,

(x, y) = (√2, 2) , (- √2, 2), (√2, -2) , (- √2, -2), (√2i, 2) , (- √2i, 2), (√2i, -2) , (- √2i, -2),

2) Solutions are,

⇒ x = 3, 3, - √3 / 2, - 5

3) All the even integers which are divisible by 5 is,

⇒ 10, 20, 30, 40, 50, ....

Given that;

1) Expression is,

⇒ x² = y, and y² = 4

2) Expression is,

⇒ (x - 3)² (2x + √3) (x + 5) = 0

Now, We can simplify as;

⇒ x² = y,

⇒ x⁴ = y²

x⁴ = 4

x⁴ - 2² = 0

(x²)² - 2² = 0

(x² - 2) (x² + 2) = 0

This gives,

x² = 2

x = ± √2

x² = - 2

x = ±√2 i

Hence, We get;

y² = 4

y = ± 2

Thus, All the solution are,

(x, y) = (√2, 2) , (- √2, 2), (√2, -2) , (- √2, -2), (√2i, 2) , (- √2i, 2), (√2i, -2) , (- √2i, -2),

Since, 2) Expression is,

⇒ (x - 3)² (2x + √3) (x + 5) = 0

Simplify as;

⇒ (x - 3)² = 0

⇒ x = 3, 3

⇒ (2x + √3) = 0

⇒ x = - √3 / 2

⇒ (x + 5) = 0

⇒ x = - 5

3) All the even integers which are divisible by 5 is,

⇒ 10, 20, 30, 40, 50, ....

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The length of the sides AB, BC and AC are √13, √13 and √26, hence the triangle is a isosceles triangle.

a. Points (-2, 3), (1, 1), (-1, 2) in the triangle are vertex. The distance formula for any two points is,

AB = √[(d-b)²+(c-a)²]

= √[(1 - (-2))² + (1 - 3)²]

= √[3² + (-2)²]

= √13

BC = √[(d-b)²+(c-a)²]

= √[(-1-1)²+(-2-1)²]

= √[(-2)² + (-3)²]

= √13

AC = √[(d-b)²+(c-a)²]

= √[(-1 - (-2))² + (-2 - 3)²]

= √[1² + (-5)²]

= √26

b. The triangle is an isosceles triangle because AB = BC.

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In the surface area, the amount of tin needed to make a milk can is 87.92 [tex]cm^2[/tex].

What is surface area?

A three-dimensional object's surface area is the space it takes up when viewed from the outside.

Here we know that the tin is in the shape of cylinder.

Now to find the amount we need to determine the surface area of the cylinder.

Now Height h = 5 cm, Diameter = 4 cm then radius r = d/2 = 4/2 = 2 cm.

Now using formula then,

Surface  Area  = 2[tex]\pi\\[/tex]r(h+r) square unit.

=> Surface area = [tex]2\times3.14\times2(5+2)=2\times3.14\times2\times7[/tex] = 87.92 [tex]cm^2[/tex]

Hence the amount of tin needed to make a milk can is 87.92 [tex]cm^2[/tex].

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The segment lengths are given as follows:

6. AB = 15.

7. RS = 47.

The length of segment TU is obtained applying the trapezoid midsegment theorem, which states that the length of the midsegment of the trapezoid is equals to the mean of the length of the bases of the trapezoid.

For item 6, we have that the mean of AB = x and 29 is of 22, hence:

(x + 29)/2 = 22

x + 29 = 44

x = AB = 15.

The value of x in item 7 is obtained as follows:

3x + 5 = (2x + 15 + 6x - 37)/2

8x - 22 = 6x + 10

2x = 32

x = 16.

Hence the length of RS is given as follows:

RS = 2 x 16 + 15

RS = 47.

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The cost of a bag of chips is $0.75 and the cost of a cheeseburger is $3.

The definition of an equation is a mathematical statement that shows that two mathematical expressions are equal. For instance, 3x + 5 = 14 is an equation, in which 3x + 5 and 14 are two expressions separated by an 'equal' sign.

let the cost of a cheesburger be x.

let the cost of a bag of chips be y

therefore, it is given that x + y = 3.75                ...........(i)

it is also given that 2x + 3y = 8.25                     ............(ii)

multiplying the equation in( i) by 2 we get 2x + 2y = 7.50     ......(iii)

subtracting the equation in iii) from the equation in ii) we get y = $0.75

Therefore ,the cost of a bag of chips is $0.75

Substituting the value of y found in (ii) we get x = 3.

therefore ,the cost of a cheeseburger is $3.

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The complete question is ,

Nathan ordered 1 cheeseburger and 1 bag of chips for 3. 75 jack ordered 2 cheeseburgers and 3 bags of chips for 8. 25.find the value of one cheeseburger and one bag?

0.1859 is the relative frequency for male goats, female goats, adult goats, and baby goats

To find the relative frequency of marked mountain goats by gender and age group, we need to divide the number of marked goats in each group by the total number of marked goats.

The total number of marked goats is:

Total = Male + Female + Adult + Baby = 71 + 93 + 103 + 61 = 328

The relative frequency for male goats is:

Male/Total = 71/328 = 0.2165 or 433/2000 (simplified fraction)

The relative frequency for female goats is:

Female/Total = 93/328 = 0.2835 or 567/2000 (simplified fraction)

The relative frequency for adult goats is:

Adult/Total = 103/328 = 0.3140 or 157/500 (simplified fraction)

The relative frequency for baby goats is:

Baby/Total = 61/328 = 0.1859 or 93/500 (simplified fraction)

Therefore, the relative frequency for male goats is 433/2000, for female goats is 567/2000, for adult goats is 157/500, and for baby goats is 93/500, all expressed as simplified fractions.

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The expression to represent the number of circles at stage 20, assuming a starting circle, is 2²⁰.

To calculate the exponential growth of number of circles at stage 20, we need to consider the number of circles that appear at each stage of a process. Assuming that we start with one circle and that each subsequent stage doubles the number of circles from the previous stage, we can use the expression 2²⁰ to represent the number of circles at stage 20.

This expression is derived from the fact that at each stage, the number of circles is doubled from the previous stage. So, if we start with one circle, the number of circles at each stage is:

Stage 1: 1

Stage 2: 2 (doubled from stage 1)

Stage 3: 4 (doubled from stage 2)

Stage 4: 8 (doubled from stage 3)

...

Stage 20: 2²⁰

This expression gives us the number of circles at stage 20, which is an extremely large number. This shows how exponential growth can lead to very large numbers in a short period.

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The number of circles at stage 20 is 1141

The pattern of circles at each stage is as follows:

We can observe that the number of circles at each stage is equal to the sum of the number of circles in the previous stage, plus the number of circles in a new layer surrounding the previous layer.

Using this pattern, we can write a recursive expression to represent the number of circles at each stage:

C(n) = C(n-1) + 6(n-1)

where C(n) represents the number of circles at stage n.

Using this expression, we can find the number of circles at stage 20 as follows:

C(20) = C(19) + 6(19)

= C(18) + 6(18) + 6(19)

= C(17) + 6(17) + 6(18) + 6(19)

= ...

= C(1) + 6(1) + 6(2) + ... + 6(19)

Using the formula for the sum of an arithmetic series, we can simplify this expression to:

C(20) = C(1) + 6(1+2+...+19)

= 1 + 6(190)

= 1141

Therefore, the number of circles at stage 20 is 1141.

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Answer:

x = 65°

Step-by-step explanation:

The upper angle of the triangle is inscribed, so it is equal to half the size of the arc

Let's call this angle α:

[tex] \alpha = \frac{100°}{2} = 50°[/tex]

Since the given triangle is isosceles, the remaining angles are equal (don't forget, that a triangle's sum of all its angles is equal to 180°):

[tex]2x + 50° = 180°[/tex]

[tex]2x = 180° - 50°[/tex]

[tex]2x = 130°[/tex]

Divide both sides of the equation by 2 to make x the subject:

[tex]x = 65°[/tex]

Answer:47.25

Step-by-step explanation:

Answer:

48.5

Step-by-step explanation:

To find the first quartile (Q1) of the data set, we need to arrange the numbers in ascending order:

42, 46, 51, 53, 66, 70, 79, 90

Q1 is the median of the lower half of the data set. Since we have 8 data points, the lower half will be the first four numbers.

42, 46, 51, 53

To find the median of these numbers, we take the average of the two middle numbers:

(Q1) = (46 + 51) / 2 = 48.5

Therefore, the first quartile (Q1) of the data set is 48.5.

The simplified ratios are:  1:6  &  3:25

To simplify the first ratio, we need to convert the units so they are the same. We can either convert 500g to kilograms or 3kg to grams. Let's convert 3kg to grams since it will be easier to compare with 500g.

3 kg = 3000g

Now the ratio becomes:

500g : 3000g

We can simplify this ratio by dividing both sides by 500:

500g/500 = 1 and 3000g/500 = 6

So the simplified ratio is:

1 : 6

For the second ratio, we need to convert either 12cm to meters or 1m to centimeters. Let's convert 1m to centimeters since it will be easier to compare with 12cm.

1m = 100cm

Now the ratio becomes:

12cm : 100cm

We can simplify this ratio by dividing both sides by 4:

12cm/4 = 3 and 100cm/4 = 25

So the simplified ratio is:

3 : 25

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The quantity of heat which must be removed from the system in order to reduce the pressure from 2 bar to 1.6 bar is 4.23 kJ. The change in enthalpy per unit mass of the system is -123 kJ/kg, and the change in entropy per unit mass of the system is 0.134 kJ/kg-K.

To solve this problem, we need to use the steam tables to determine the properties of the steam at the initial and final conditions. We will assume that the system is undergoing a reversible, adiabatic process, so there is no heat transfer into or out of the system.

First, we determine the specific volume and enthalpy of the steam at the initial conditions of 2 bar and 0.9 dryness fraction. From the steam tables, we find that the specific volume is 0.4019 m^3/kg and the specific enthalpy is 2895.5 kJ/kg.

Next, we use the steam tables to find the specific volume and enthalpy of the steam at the final conditions of 1.6 bar. We find that the specific volume is 0.5059 m^3/kg and the specific enthalpy is 2772.5 kJ/kg.

The change in specific enthalpy per unit mass of the system is then given by:

Δh = h2 - h1 = 2772.5 - 2895.5 = -123 kJ/kg

The change in specific entropy per unit mass of the system is given by:

Δs = s2 - s1 = s2 - s1 = s2 - sf - x2*(sg - sf)

where sf and sg are the specific entropy of saturated liquid and saturated vapor at the final pressure of 1.6 bar, and x2 is the final dryness fraction. From the steam tables, we find that sf = 7.4332 kJ/kg-K, sg = 8.1248 kJ/kg-K, and x2 = 0.714.

Thus, we have:

Δs = s2 - s1 = s2 - sf - x2*(sg - sf) = (7.9757 - 7.4332) - 0.714*(8.1248 - 7.4332) = 0.134 kJ/kg-K

Finally, we can calculate the quantity of heat that must be removed from the system using the first law of thermodynamics:

Q = m*(h1 - h2) = m*Δh

where m is the mass of the steam in the system. To determine the mass of the steam, we use the specific volume at the initial conditions:

V = m/v1

where V is the volume of the system and v1 is the specific volume at the initial conditions. Substituting the given values, we have:

V = 85 L = 0.085 [tex]m^3[/tex]

m = Vv1 = 0.0850.4019 = 0.0344 kg

Substituting this value into the equation for Q, we obtain:

Q = mΔh = 0.0344(-123) = -4.23 kJ

Therefore, the quantity of heat which must be removed from the system in order to reduce the pressure from 2 bar to 1.6 bar is 4.23 kJ.

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To determine if the 275 fish are the right number for the cylindrical fish tank, we need to calculate the tank's capacity and compare it to the recommended average density of 16 fish per 100 gallons of water.

The volume of a cylinder is given by the formula V = πr^2h, where V is the volume, r is the radius, and h is the height of the cylinder.

Assuming the tank has a height of h and a radius of r, we can calculate its volume as follows:

[tex]V = πr^2h[/tex]

Since the tank has a vertical tube in the middle, we need to subtract the volume of the tube from the total volume of the tank. Let's assume the tube has a radius of 2 feet and a height of 8 feet. Then the volume of the tube is:

Vtube = π(2)^2(8) = 100.53 cubic feet

Thus, the volume of the tank without the tube is:

Vtank = πr^2h - Vtube

To find the value of r, we need to know the diameter of the tank. Let's assume the tank has a diameter of 10 feet, which means the radius is 5 feet.

Then the volume of the tank without the tube is:

Vtank = π(5)^2h - 100.53

We need to convert the volume of the tank from cubic feet to gallons, so we multiply by 7.48 (1 cubic foot = 7.48 gallons):

Vtank(gallons) = 7.48[π(5)^2h - 100.53]

Now we can calculate the recommended number of fish for the tank:

Recommended number of fish = 16 fish/100 gallons x Vtank(gallons)

Recommended number of fish = 16 fish/100 gallons x 7.48[π(5)^2h - 100.53]

Recommended number of fish = 1.175[π(5)^2h - 100.53]

So, if the number of fish available is 275, we can set up the following equation:

275 = 1.175[π(5)^2h - 100.53]

Solving for h, we get:

h = (275/1.175π(5)^2) + (100.53/π(5)^2)

h ≈ 8.3 feet

Therefore, the cylindrical fish tank with a height of 8.3 feet and a radius of 5 feet can hold 275 fish with an average density of 16 fish per 100 gallons of water. If the aquarium manager wants to add more fish, they should recalculate the volume of the tank and adjust the height accordingly to maintain the recommended density of 16 fish per 100 gallons of water. Conversely, if they want to remove fish, they can do so without changing the height of the tank.

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The balloon is rising at a rate of approximately 3.532 km/h at this moment.

We will use the tangent function in trigonometry and the concept of related rates.

Let x be the horizontal distance from the observer to the lift-off point (5 km) and y be the vertical distance from the ground to the balloon. The angle between the observer's line of sight and the horizontal is given as π/5.

We can use the tangent function:
tan(θ) = y/x

At the given moment, x = 5 km and θ = π/5, so:
tan(π/5) = y/5

Now, let's differentiate both sides with respect to time (t):
d(tan(θ))/dt = d(y)/dt / 5

We know that d(θ)/dt = 0.4 rad/h, so we can find d(tan(θ))/dt using the chain rule:
d(tan(θ))/dt = (sec^2(θ)) * d(θ)/dt
d(tan(θ))/dt = (sec^2(π/5)) * 0.4

Now, substitute this back into the first equation and solve for d(y)/dt:
(sec^2(π/5)) * 0.4 = d(y)/dt / 5
d(y)/dt = 5 * (sec^2(π/5)) * 0.4

After calculating the expression, you'll find that d(y)/dt ≈ 3.532 km/h. Thus, the balloon is rising at a rate of approximately 3.532 km/h at this moment.

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The new diet was not effective, the researchers may need to continue searching for other solutions.

In the trial to determine whether a new diet reduces cholesterol, a group of twenty volunteers with high cholesterol were selected. The trial likely involved splitting the volunteers randomly into two groups - a treatment group and a control group.

The treatment group would be given the new diet to follow, while the control group would continue with their normal diet. The participants' cholesterol levels would be measured at the beginning of the trial, and then again at regular intervals throughout the trial to track any changes.

After the trial has ended, the researchers would analyze the results to see if there was a significant difference in cholesterol levels between the treatment and control groups. If the new diet was effective in reducing cholesterol, the researchers may recommend it as a potential treatment option for people with high cholesterol. However, if the new diet was not effective, the researchers may need to continue searching for other solutions.

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Assuming that the muffins were actually ready at two minutes to five in the afternoon, we can determine that it took Robert approximately 38 minutes to prepare and bake the muffins.

To arrive at this conclusion, we can use the following logic:

Robert got home from school at 3:33 PM (twenty-seven minutes before 4:00 PM).

The muffins were ready at 4:58 PM (two minutes before 5:00 PM).

Therefore, the time between when Robert got home and when the muffins were ready is 85 minutes (58 minutes + 27 minutes).

Since Robert decided to bake the muffins immediately upon arriving home, it took him 85 minutes to prepare and bake them.

Of course, this assumes that Robert did not take any breaks or perform other activities during the time between getting home and the muffins being ready. In reality, the actual time it took to prepare and bake the muffins may have been longer or shorter depending on various factors, such as the recipe, equipment used, and Robert's baking experience.

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The percent increase in Patti's hourly wage is 23%.

Percent​ increase is aimply the amount of increase from the initial value to the new value in terms of 100 parts of the initial value.

It is expressed as:

percent increase = ((new value - old value) / old value) × 100%

Given that, the old hourly wage was $10.00 and the new hourly wage is $12.30.

Substituting the values into the formula, we get:

percent increase = ((new value - old value) / old value) × 100%

percent increase = (($12.30 - $10.00) / $10.00) × 100%

percent increase = ($2.30 / $10.00) × 100%

percent increase = 0.23 × 100%

percent increase = 23%

Therefore, the percent increase is 23%.

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The formula for the volume of a cone is:

V = (1/3)πr^2h

where r is the radius of the base of the cone and h is the height of the cone.

We are given that the radius of the cone is 5 cm and the slant height is 13 cm. We can use the Pythagorean theorem to find the height of the cone:

h^2 = l^2 - r^2

where l is the slant height of the cone. Substituting the given values, we get:

h^2 = 13^2 - 5^2

h^2 = 144

h = 12

Now we can substitute the values of r and h into the formula for the volume of the cone:

V = (1/3)πr^2h

V = (1/3)π(5^2)(12)

V = (1/3)π(25)(12)

V = (1/3)π(300)

V = 100π

Therefore, the volume of the cone is 100π cubic centimeters.

The Surface Area of the sphere is approximately 339.79 square inches.

The surface area of a sphere is given by the formula:

surface area= [tex]4\pi r^{2}[/tex]

where r is the radius of the sphere.

The diameter of the sphere measures 10.4 inches, hence the radius can be calculated as:

r=10.4/2=5.2inches

Hence, the surface area can be calculated as by substituting r=5.2 inches

Therefore, surface area of the sphere is:

Surface Area = [tex]4\pi (5.2)^{2}[/tex]=[tex]4\pi (27.04)[/tex]= 108.16[tex]\pi[/tex] square inches.

So, the Surface Area of the sphere is approximately 339.79 square inches(if we use [tex]\pi[/tex]=3.14 as an approximation)

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Answer:

Step-by-step explanation:

To find the experimental probability of a student in Homeroom 203 taking a certain number of minutes to get to school, we need to divide the number of students who take that amount of time by the total number of students in the homeroom.

The total number of students in the homeroom is:

3 + 5 + 10 + 7 + 2 + 3 = 30

The probability of a student taking less than 10 minutes to get to school is:

3/30 = 0.1 or 10%

The probability of a student taking 10-19 minutes to get to school is:

5/30 = 0.166 or 16.6%

The probability of a student taking 20-29 minutes to get to school is:

10/30 = 0.333 or 33.3%

The probability of a student taking 30-39 minutes to get to school is:

7/30 = 0.233 or 23.3%

The probability of a student taking 40-49 minutes to get to school is:

2/30 = 0.066 or 6.6%

The probability of a student taking 50 minutes or more to get to school is:

3/30 = 0.1 or 10%

To make a probability distribution, we can list the possible outcomes (in this case, the time it takes to get to school) and their corresponding probabilities:

Time (min) Probability

Less than 10 0.1

10-19 0.166

20-29 0.333

30-39 0.233

40-49 0.066

50 or more 0.1

Note that the probabilities add up to 1, which is what we expect for a probability distribution.

The value of 6x-5y = -31

The given equations are 2x-y=-7 -eq (1)

& 4x=-3y+16 -eq (2)

Multiplying eq (1) with 2 and rearranging to get equation as follows

(2x-y=-7)*2

4x-2y=-14 -eq (3)

Now, subtracting eq (3) from eq (2) to get y.

(4x+3y=16) - (4x-2y=-14)

5y=30

y=5

Substituting y=5 in eq (1) to get x.

2x-5=-7

x=-1

Now to determine 6x-5y substitute obtained values of x & y.

6*(-1)-5(5)=-31

Hence the value of 6x-5y=-31.

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The range of x values between which a zero can be found is -9/4 < x < -4.

Since x + 4 and 4x + 9 are linear factors of the quadratic expression, the quadratic expression can be written as:

Q(x) = k(x + 4)(4x + 9)

where k is some constant.

To find the values of x for which Q(x) = 0, we can set each factor equal to zero and solve for x:

x + 4 = 0 --> x = -4

4x + 9 = 0 --> x = -9/4

Therefore, the zeros of Q(x) are x = -4 and x = -9/4.

To find the range of x values between which a zero can be found, we need to determine the sign of Q(x) in each of the three intervals:

1. x < -9/4

2. -9/4 < x < -4

3. x > -4

For x < -9/4, both x + 4 and 4x + 9 are negative, so Q(x) = k(negative)(negative) = k(positive), which is positive.

For x > -4, both x + 4 and 4x + 9 are positive, so Q(x) = k(positive)(positive) = k(positive), which is also positive.

For -9/4 < x < -4, x + 4 is positive and 4x + 9 is negative, so Q(x) = k(positive)(negative) = k(negative), which is negative.

Therefore, the range of x values between which a zero can be found is -9/4 < x < -4.

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(a) The change in the cart's momentum is -200 kg m/s.

(b) The average force that the wall exerts on the cart is 1333.33 N.

(c) The impulse that acted on the cart is in the opposite direction to the cart's initial momentum.

(a) The change in momentum can be calculated as the final momentum minus the initial momentum. The initial momentum of the cart is zero since it was at rest, and the final momentum is calculated as (mass of cart) x (final velocity) = 100 kg x 2 m/s = 200 kg m/s. Therefore, the change in momentum is -200 kg m/s.

(b) The average force can be calculated using the impulse-momentum theorem, which states that the impulse acting on an object is equal to the change in its momentum. The impulse is calculated as (mass of cart) x (final velocity - initial velocity) = 100 kg x (2 m/s - 0 m/s) = 200 kg m/s.

The time taken for the cart to come to a stop is given as 0.15 s. Therefore, the average force exerted by the wall is 1333.33 N.

(c) The direction of the impulse is opposite to the initial momentum of the cart, which was in the direction of the hill. Since the cart was moving downhill, the impulse that acted on it was in the upward direction.

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