If nori made 2% in interest on $5,000 and her brother Sean made 1% in interest on $10,000, who made more money in interest?

To find the area of the sector NPQ, we first need to find the measure of the central angle that intercepts the arc PQ. We know that the measure of angle NPQ is 104 degrees, and since it is an inscribed angle, its measure is half the measure of the central angle that intercepts the same arc. Therefore, the central angle measure is 208 degrees.

To find the area of the sector, we use the formula:

Area of sector = (central angle measure/360) x pi x radius^2

We know that the radius of circle P is NP = 9 units. Plugging in the values, we get:

Area of sector NPQ = (208/360) x pi x 9^2
= (0.5778) x 81pi
= 46.99 square units

Rounding to the nearest hundredth, the area of the sector NPQ is 47.00 square units.

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If the base and slant height both are divided by a factor of 3, the surface area will get multiplied by factor, option b, 3².

Here we are given that the square pyramid has a base of 3m and a slant height of 6 m.

The surface area formula for a square pyramid with square edge a and slant height h is

a² + 2a√(a²/4  + h²)

Here, a = 3 and h = 6. Hence we get

3² + 2X3√(3²/4  + 6²)

= 46.108

Now the base and slant height are multiplied by 3. Hence we will get

9a² + 6a√(9a²/4  + 9h²)

414.972

Now, dividing both obtained we will get

414.972/46.108

= 9

= 3²

Hence, it should be multiplied by 3².

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Answer:

1) 1/2 + 1/4 = 2/4 + 1/4 = 3/4

2) To add these fractions, you need to find a common denominator. The smallest common multiple of 7 and 9 is 63, so we can write:

3/7 * 9/9 + 2/9 * 7/7 = 27/63 + 14/63 = 41/63

3) To add these fractions, you need to find a common denominator. The smallest common multiple of 5 and 15 is 15, so we can write:

3/5 * 3/3 + 1/15 * 1/1 = 9/15 + 1/15 = 10/15

But we can simplify this fraction by dividing both the numerator and denominator by 5:

10/15 = 2/3

4) To add these fractions, you need to find a common denominator. The smallest common multiple of 9 and 8 is 72, so we can write:

1/9 * 8/8 + 7/8 * 9/9 = 8/72 + 63/72 = 71/72

5) To add these fractions, you need to find a common denominator. The smallest common multiple of 7 and 21 is 21, so we can write:

6/7 * 3/3 + 2/21 * 1/1 = 18/21 + 2/21 = 20/21

6) To add these fractions, we need to find a common denominator first. The smallest number that both 6 and 10 divide into is 30. So, we convert 4/6 to 20/30 by multiplying both the numerator and denominator by 5, and we convert 2/10 to 3/15 by multiplying both the numerator and denominator by 3. Now we have:

20/30 + 3/15 = (20x1 + 3x2)/(30x2) = 23/60

Therefore, 4/6 + 2/10 = 23/60.

7) To add these fractions, we need to find a common denominator first. The smallest number that both 11 and 22 divide into is 22. So, we convert 1/11 to 2/22 by multiplying both the numerator and denominator by 2, and we convert 3/22 to 3/22 (it is already in terms of 22). Now we have:

2/22 + 3/22 = (2 + 3)/22 = 5/22

Therefore, 1/11 + 3/22 = 5/22.

8) To add these fractions, we need to find a common denominator first. The smallest number that both 4 and 20 divide into is 20. So, we convert 1/4 to 5/20 by multiplying both the numerator and denominator by 5, and we convert 8/20 to 8/20 (it is already in terms of 20). Now we have:

5/20 + 8/20 = (5 + 8)/20 = 13/20

Therefore, 1/4 + 8/20 = 13/20.

9) To add these fractions, we need to find a common denominator first. The smallest number that both 7 and 9 divide into is 63. So, we convert 4/7 to 24/63 by multiplying both the numerator and denominator by 3, and we convert 2/9 to 14/63 by multiplying both the numerator and denominator by 7. Now we have:

24/63 + 14/63 = (24 + 14)/63 = 38/63

Therefore, 4/7 + 2/9 = 38/63.

10) To add these fractions, we need to find a common denominator first. The smallest number that both 10 and 30 divide into is 30. So, we convert 6/7 to 18/30 by multiplying both the numerator and denominator by 3, and we convert 2/30 to 1/15 by multiplying both the numerator and denominator by 15. Now we have:

18/30 + 1/15 = (18x1 + 1x2)/(30x2) = 37/30

Therefore, 6/7 + 2/21 = 37/30.

The perimeter of the semicircular region is about 28.27 meters.

We know that the semicircular region is half of a circle.

Let's assume that the radius of the circle is r meters. Then the circumference of the circle is 2πr meters, and the perimeter of the semicircular region is half of that, which is πr meters.

We are given that the radius of the circle is 9 meters, so we can plug that in to get:

The perimeter of the semicircular region = πr

= π(9)

≈ 28.27

Rounding to the nearest hundredth, the perimeter of the semicircular region is about 28.27 meters.

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Answer:

54% × 50000= 27000, the answer is 27000

The calculated value of the number of cows that can graze on 720 acres of land is 18

From the question, the statements that can be used in our computation are given as

The area of grazing needed by each cow is 40 acres

From the above statement, the equation to use is

Cows = Area of land/Unit rate of cows

By substituting the given values in the above equation, we have the following equation

Cows = 720/40

Evaluate

Cows = 18

Hence, the calculated number of cows is 18

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The area of ΔSTU is approximately 6.7 square centimeters.

To find the area of a triangle, we can use the formula A = (1/2)bh, where b is the base of the triangle and h is the height. In this case, we know that u is the base of the triangle, so we need to find the height.

To do this, we can use the sine function, which tells us that sin(6°) = h/u. Rearranging this equation, we get h = usin(6°). We can then substitute u and sin(6°) into the formula for the area to get

A = (1/2)(3.4)(3.4sin(6°)) ≈ 6.7 square centimeters.

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The histogram for the frequency table is illustrated below.

To create the frequency table, we need to count how many times each time appears in the data set. The time 9:23 appears twice, so we would put a frequency of 2 in the row corresponding to 9:23. We do this for each time in the data set.

Here is the completed frequency table:

Time Frequency

6:10 1

6:21 1

6:55 1

7:12 1

7:20 1

7:34 1

8:15 1

9:01 1

9:14 1

9:15 1

9:18 1

9:23 2

As you can see, each time appears only once or twice in the data set. This tells us that there is no dominant time that most students ran the mile in.

To create the histogram, we'll draw a bar above each time on the x-axis with a height equal to the frequency of that time. For example, there are two times of 9:23, so we'll draw a bar above 9:23 with a height of 2.

As you can see, the histogram shows a relatively even distribution of times. The most common times are around 9 minutes, but there are also several times below 8 minutes and one time below 7 minutes.

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The cost of one apple is $2.15.

To find the cost of one apple, we can set up a system of equations with the given information. Let's use A for the cost of one apple and B for the cost of one banana:

1) 4A + 9B = $12.70
2) 8A + B = $17.70

Now, we can solve this system of equations. We can multiply equation 1 by 2 to match the number of apples in equation 2:

1) 8A + 18B = $25.40

Now subtract equation 2 from the modified equation 1:

(8A + 18B) - (8A + B) = $25.40 - $17.70
17B = $7.70

Now, divide by 17 to find the cost of one banana:
B = $7.70 / 17 = $0.45

Now that we know the cost of one banana, we can substitute B in equation 2 to find the cost of one apple:

8A + ($0.45) = $17.70
8A = $17.25
A = $17.25 / 8 = $2.15

So, the cost of one apple is $2.15.

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Answer: V = 2527.2 in^3

Step-by-step explanation:

V = Bh

that is, Volume = base area x height

the base area is the hexagon, and the height is given as 12.

Think of dividing the hexagon into 6 equal triangles, with height 7.8

so the area of all 6 triangles, (effectively the area of the hexagon), will be:

6(0.5 x 9 x 7.8) = 210.6 in^2

multiply this by the height to get the volume:

210.6 x 12 = 2527.2 in^3

thats it!

V = 2527.2 in^3

Type I error in this situation would mean option a. Concluding that the bags are being under filled when they actually aren't.

A Type I error occurs when the null hypothesis (in this case, that the true mean net weight is 14 ounces) is rejected when it is actually true. In other words, it is the error of concluding that there is evidence for the alternative hypothesis (that the true mean net weight is less than 14 ounces) when there is not.

In this situation, if a Type I error is made, it would mean that the bags are being under filled (i.e. the true mean net weight is less than 14 ounces) when in reality they are not. This would lead to incorrect conclusions and potentially negative consequences for the manufacturer.

It's important to note that the probability of making a Type I error can be controlled by choosing an appropriate level of significance (usually denoted by alpha) for the hypothesis test. For example, if alpha is set at 0.05, there is a 5% chance of making a Type I error.

In summary, a Type I error in this situation would mean incorrectly concluding that the bags are being under filled when they actually are not, and the probability of making such an error can be controlled by choosing an appropriate level of significance for the hypothesis test.

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Answer:

10 adult tickets cost £75 , 20 child tickets cost £60

Step-by-step explanation:

let a be the cost of an adult ticket and c the cost of a child ticket , then

4a + c = 33 → (1)

2a + 7c = 36 → (2)

multiplying (2) by - 2 and adding to (1) will eliminate a

- 4a - 14c = - 72 → (3)

add (1) and (3) term by term to eliminate a

0 - 13c = - 39

- 13c = - 39 ( divide both sides by - 13 )

c = 3

substitute c = 3 into either of the 2 equations and solve for a

substituting into (1)

4a + 3 = 33 ( subtract 3 from both sides )

4a = 30 ( divide both sides by 4 )

a = 7.5

the cost of an adult ticket is £7.50

then 10 adult tickets cost 10 × £7.50 = £75

the cost of a child ticket is £3

the cost of 20 child tickets is 20 × £3 = £60

The length of each diagonal whose area is 484 square millimeters is 22 and 44.

Area of rhombus = 484 square millimeters

Let one diagonal of rhombus = p

other diagonal of rhombus = q

The length of one diagonal of rhombus is one-half as long as the other diagonal of rhombus

p = q/2

Area of diagonal = [tex]\frac{1}{2}d_{1}d_{2}[/tex]

Area of diagonal = [tex]\frac{1}{2}pq[/tex]

Area of diagonal = [tex]\frac{1}{2}\frac{q}{2}q[/tex]

484 = [tex]\frac{1}{4}q^{2}[/tex]

q² = 1936

q = 44 millimeter

p = q/2

p = 44/2

p = 22 millimeter

The length of each diagonal p and q is 22 mm and 44 mm respectively

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The monthly rent for the apartment would be $1238.84 in the fifth year.

The problem states that the monthly rent for an apartment is $925. To calculate the rent for the second year, we need to increase this amount by 6%.

To do this, we first need to calculate 6% of $925. We can do this by multiplying 0.06 (which is equivalent to 6%) by $925:

6% of $925 = 0.06 × $925 = $55.50

So, the rent for the second year would be:

$925 + $55.50 = $980.50

To find the rent for the third year, we need to increase the rent for the second year by 6%. We can follow the same process:

6% of $980.50 = 0.06 × $980.50 = $58.83

So, the rent for the third year would be:

$980.50 + $58.83 = $1039.33

To find the rent for any year n, we use the formula:

Rent for year n = $925 × (1 + 0.06)^n

In this formula, (1 + 0.06) represents the multiplier used to calculate the new rent each year.

For example, the multiplier for the second year is 1 + 0.06 = 1.06, and the multiplier for the third year is 1.06 × 1.06 = 1.1236 (rounded to four decimal places).

To find the rent for the fifth year, we plug in n = 5:

Rent for year 5 = $925 × (1 + 0.06)^5 = $925 × 1.3382 = $1238.84 (rounded to the nearest cent)

So, the monthly rent for the apartment would be $1238.84 in the fifth year.

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67 is the sum of handshakes given by the first and second participants.

Let n be the total number of participants in the workshop venue.

i.e, here n= 35

For the first participant, the number of handshakes is = (n-1)

= (35-1)

= 34

The number of handshakes by the second participant is also same as that of the first participant = 34

The number of handshakes given by the first and second participants together = (first participant handshake + second participant handshake - 1)

= (34 + 34 -1)

= 68-1

= 67

Hence 67 is the sum of handshakes given by the first and second participants.

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The complete question is =

A workshop venue has 35 participants . Each participant shakes hands with each each and every other participant . How is the sum of. Handshakes that will bemade bythe first and second participant

Mario can afford to buy a maximum of 12 hamburgers from the local store.

Let's assume Mario can buy "h" hamburgers.

The cost of each hamburger is $2.90, and Mario wants to buy "h" hamburgers, so the total cost of hamburgers would be 2.90h.

He is also buying one packet of hamburger rolls, which costs $4.75.

Therefore, the total amount he can spend at the store must be less than or equal to his budget of $39.55.

Putting it all together, the inequality representing this situation is:

2.90h + 4.75 ≤ 39.55

To find out how many hamburgers Mario can afford to buy, we need to solve the inequality:

2.90h + 4.75 ≤ 39.55

Subtracting 4.75 from both sides of the inequality:

2.90h ≤ 34.80

Next, divide both sides of the inequality by 2.90:

h ≤ [tex]\frac{34.80 }{ 2.90}[/tex]

Simplifying the right side:

h ≤ 12

Therefore, Mario can afford to buy a maximum of 12 hamburgers from the local store.

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If the terminal side of angle λ intersects the unit circle at point (-0. 358, 0. 934), the approximate decimal value of cot(λ) is -0.383.

To find the approximate decimal value of cot(λ), we first need to determine the values of x and y on the unit circle. Since the given point (-0.358, 0.934) lies on the unit circle, we can use the Pythagorean theorem to find the missing side:

x² + y² = 1
(-0.358)² + (0.934)² = 1
0.128 + 0.872 = 1
1 = 1

Therefore, we have x = -0.358 and y = 0.934. Since cot(λ) = cos(λ)/sin(λ), we can use the values of x and y to calculate the cosine and sine of λ:

cos(λ) = x = -0.358
sin(λ) = y = 0.934

Substituting these values into the formula for cot(λ), we get:

cot(λ) = cos(λ)/sin(λ) = -0.358/0.934 ≈ -0.383

Therefore, the approximate decimal value of cot(λ) is -0.383.

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On performing an ANOVA test the p-value obtained is less than the chosen significance level, hence it is verified that the sample standard deviations use of ANOVA allow the the means to compare the population means. Since ANOVA test shows significant difference therefore, it suggests that these factors play a role in influencing the evaluation of the product.

To verify that the sample standard deviations use of ANOVA allows means to compare the population means, discuss the terms ANOVA, sample standard deviation, population means.

1. ANOVA (Analysis of Variance): ANOVA is a statistical method used to compare the means of multiple groups to determine if there's a significant difference between them.

2. Sample Standard Deviation: Sample standard deviation is a measure of how spread out the values in a sample are. It helps estimate the population standard deviation, which is necessary for calculating the F statistic in ANOVA.

a. Calculate the sample means and standard deviations for each group.
b. Perform an ANOVA test using calculated means and standard deviations.
c. Interpret results: If p-value obtained from the ANOVA test is less than the chosen significance level (e.g., 0.05), it means there is a significant difference between population means.

Regarding the effect of the subject's gender and attractiveness of the confederate on the evaluation of the product, if the ANOVA test shows a significant difference, it suggests that these factors play a role in influencing the evaluation of product. You can further analyze the data by performing post-hoc tests to identify which specific groups differ significantly.

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Answer:

So Monica is 9 years old.

To find Amita's age, we substitute x into the expression for Amita's age:

Amita's age = 9 + 3 = 12

Therefore, Amita is 12 years old.

The  evaluated probability that Sue travel to France this summer is 0.67, under the condition that she just returned from her winter vacation in Mexico.

For the required problem we have to apply  Bayes' theorem.

Let  us consider that A is the event that Sue goes to France in the summer and B be the event that Sue goes to Mexico in the winter.

Now,

P(A and B) = P(B) × P(A|B)

= 0.40

P(B) = 0.60

Therefore now we have to find P(A|B),  which means the probability that Sue traveled to France after coming from Mexico

Applying Bayes' theorem,

P(A|B) = P(B|A) × P(A) / P(B)

It is given that P(B|A) = P(A and B) / P(A), then

P(A|B) = (P(A and B) / P(A)) × P(A) / P(B)

P(A|B) = P(A and B) / P(B)

Staging the values

P(A|B) = 0.40 / 0.60

P(A|B) = 0.67

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The complete question is

The probability that Sue will go to Mexico in the winter and to France in the summer is 0. 40. the probability that she will go to mexico in the winter is 0. 60. find the probability that she will go to France this summer, given that she just returned from her winter vacation in Mexico.

You need to add 2125 milliliters of water to fill the square aquarium completely.

We need to find the volume of the square aquarium and then determine the additional water needed to fill it completely. Here are the steps:

1. Convert the given length to meters: 15 cm = 0.15 m
2. Calculate the volume of the square aquarium: Volume = length × width × height. Since it's a square aquarium, all sides are equal, so Volume = 0.15 m × 0.15 m × 0.15 m = 0.003375 cubic meters.
3. Convert the volume to milliliters: 0.003375 cubic meters × 1,000,000 mL/cubic meter = 3375 mL.
4. Calculate the additional water needed: Total volume - Current volume = 3375 mL - 1250 mL = 2125 mL.

You need to add 2125 milliliters of water to fill the square aquarium completely.

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From the given data, the statement "A higher percentage of 8th graders than 7th graders prefer Math/Science" is correct. So, correct option is C.

To calculate the relative frequencies, we need to divide the frequency of each category by the total number of responses.

For 7th grade:

Relative frequency of English: 38/116 = 0.3276

Relative frequency of History: 36/116 = 0.3103

Relative frequency of Math/Science: 28/116 = 0.2414

Relative frequency of Other: 14/116 = 0.1207

For 8th grade:

Relative frequency of English: 47/182 = 0.2582

Relative frequency of History: 45/182 = 0.2473

Relative frequency of Math/Science: 72/182 = 0.3956

Relative frequency of Other: 18/182 = 0.0989

From the data, we can see that a higher percentage of 8th graders prefer Math/Science than 7th graders. Therefore, the statement "A higher percentage of 8th graders than 7th graders prefer Math/Science" correctly describes the association between favorite subject and grade.

The statements "A higher percentage of 8th graders than 7th graders prefer History" and "A higher percentage of 7th graders than 8th graders prefer English" are incorrect as the relative frequencies for these subjects are similar in both grades.

Overall, we can conclude that the choice of favorite subject is not strongly associated with the grade level of the students.

So, correct option is C.

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Complete question is:

Calculating the relative frequencies from the data given in the table. Choose all that correctly describe an association between favorite subject and grade.

A) A higher percentage of 8th graders than 7th graders prefer History.

B) A higher percentage of 7th graders than 8th graders prefer English.

C) A higher percentage of 8th graders than 7th graders prefer Math/Science.

To determine what students want served in the cafeteria, the cook asks students in Ms. Andrew’s first period class. The sample used by the cook is known as Convenience.

The sample used is known as convenience sample. The cook only asks students in Ms. Andrews’ first period class which is a convenient and accessible group to ask but this method of sampling may not be representative of the entire student population as it only includes students in one class.

So, the results may not accurately reflect what all students want to be served in the cafeteria, hence, more representative sample could be obtained by using a simple random sample or systematic sample where  student in the population has an equal chance of being selected.

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The limit of A. x^(2^x) as x approaches 0 is 1, and the limit of B. 2^x - 1 as x approaches 0 is ln 2.

A. To find the limit of A. x^(2^x) as x approaches 0, we can take the natural logarithm of both sides and use the fact that ln(1 + a) is approximately equal to a for small values of a. This gives us:

ln(A. x^(2^x)) = 2^x ln x

ln(A. x^(2^x)) / ln x = 2^x

Taking the limit as x approaches 0, the right-hand side goes to 1, and using the continuity of the natural logarithm, we have:

ln(A) = 0

A = 1

Therefore, the limit of A. x^(2^x) as x approaches 0 is 1.

B. To find the limit of B. 2^x - 1 as x approaches 0, we can use L'Hopital's Rule:

lim x→0 (2^x - 1)

= lim x→0 (ln 2 * 2^x / ln 2)

= ln 2 * lim x→0 (2^x / ln 2)

= ln 2 * (lim x→0 e^(x ln 2) / ln 2)

= ln 2 * (lim x→0 e^(x ln 2 - ln 2) / (ln 2 - ln 2))

= ln 2 * (lim x→0 e^(ln 2 * (x - 1)) / 1)

= ln 2 * e^0

= ln 2

Therefore, the limit of B. 2^x - 1 as x approaches 0 is ln 2.

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Step-by-step explanation:

remember the original trigonometric triangle inside the norm circle (radius = 1).

sine is the up/down distance from the triangle baseline or corresponding circle diameter.

cosine is the left/right distance from the center of the circle (and the point of the angle).

for larger triangles and circles all these function results need to be multiplied by the actual radius (which we skipped for the norm circle, as a multiplication by 1 is not changing anything).

when you look at the triangle with K representing the angle, we have 10 a the cosine value, 24 as the sine value and 26 as the radius.

so,

10 = cos(K) × 26

cos(K) = 10/26 = 5/13

Answer:

2.41

Step-by-step explanation:

postive = add negative = subtract

The remaining part of the given shape was drawn about to its symmetry. The complete shape obtained is similar to the Hexagon shape.

Given half shape has the following points,

The points which are missing to complete the other symmetry are:

By joining the above missing points, we can obtain the full symmetry of the shape.

To rotate the obtained shape of the Hexagon to 180° about the origin, we have to inverse the above complete symmetry points. It simply means if the above points are having positive values, we can inverse it to negative and vice-versa.

By rotating the obtained shape to 180° about the origin, we can obtain the below following points,

The images are attached below for the complete symmetry shape.

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Given question is not having enough required information, so I am attaching the image of the shape which we have to work on,

There are 4.097 x [tex]10^43[/tex] paths from (0,0) to (90,160) that pass through (80,70).

To calculate the number of paths from (0,0) to (90,160) while passing through (80,70), we need to break down the problem into smaller steps.

First, we can calculate the number of paths from (0,0) to (80,70) and then

multiply that by the number of paths from (80,70) to (90,160).

To go from (0,0) to (80,70), we need to take 80 steps to the right and 70 steps up, which gives us a total of 150 steps. The order in which we take these steps doesn't matter, so we can think of it as choosing 70 steps out of 150 to be up. This can be calculated using the binomial coefficient, which gives us (150 choose 70) = 2.364 x  [tex]10^43[/tex]

To go from (80,70) to (90,160), we need to take 10 steps to the right and 90 steps up, which gives us a total of 100 steps. Using the same method as above, the number of paths from (80,70) to (90,160) is (100 choose 10) = 17,310,309.

Multiplying these two values together, we get the total number of paths from (0,0) to (90,160) that pass through (80,70):

(2.364 x 10^34) x (17,310,309) = 4.097 x  [tex]10^43[/tex]

Therefore, there are 4.097 x [tex]10^43[/tex] paths from (0,0) to (90,160) that pass through (80,70).

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The extremum of the function is a minimum at x = -2.5

The given function is k(x)(300+10x)(5-0.2x).

To check for symmetry about the y-axis, we replace x with -x in the given function and simplify as follows:

k(-x)(300-10x)(5+0.2x)

To check for symmetry about the x-axis, we replace y with -y in the given function and simplify as follows:

k(x)(300+10x)(5-0.2x) = -k(x)(-300-10x)(5+0.2x)

To find these points, we set the function equal to zero and solve for x:

k(x)(300+10x)(5-0.2x) = 0

This equation has three solutions:

x = 0

x = -30

x = 25.

The midpoint of the line segment connecting these points is

(x1+x2) ÷ 2 = (-30+25) ÷ 2 = -2.5.

To determine the type of extremum at this point, we need to check the sign of the second derivative. The second derivative of the function is:

k(x)(-1200+x)(0.2x+15)

Since the function is symmetric about the x-axis, the second derivative will be negative at the extremum if it is maximum and positive if it is a minimum.

When x = -2.5, the second derivative is positive, which means that the function has a minimum at x = -2.5.

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Answer:

£400

Step-by-step explanation:

first find 40% of £1000

40\100*1000

=400

Therefore answer is £400