Suppose a basketball team had a season of games with the following characteristics:
60% of all the games were at-home games. Denote this by H (the remaining were away games).
25% of all games were wins. Denote this by W (the remaining were losses).
20% of all games were at-home wins.
Of the at-home games, we are interested in finding what proportion were wins. In order to figure this out, we need to find:
A P(H)
B P(W)
C P(H and W)
D P(H | W)
E P(W | H)

Answer:

N*r^n

Step-by-step explanation:

Let the initial number of cases at the start of year 1 be represented by N.

From the given information, we know that the number of cases increases by the same factor each year. Let this factor be represented by r.

Then, at the end of year 1, the number of cases would be N*r, since it has increased by a factor of r.

Similarly, at the end of year 2, the number of cases would be Nrr, or N*r^2.

At the end of year 3, the number of cases would be Nrrr, or Nr^3.

We can use this pattern to write a general expression for the number of cases after n years:

N * r^n

where N is the initial number of cases, r is the common factor by which the number of cases increases each year, and n is the number of years elapsed.

The answer and workout is provided in the attachment.

Step-by-step explanation:

hope this will help u

Answer: what do you mean? I need more info-

Step-by-step explanation:

I can answer it with more info :)

With the given parameters, the width of the 80% CI for the normal distribution's mean is roughly 1.5.

The following formula can be used to determine the width of an 80% confidence interval (CI) for the mean of a normal distribution with unknown variance, the sample mean 9, sample variance 6, and sample size 15:

width = t * (s / √(n))

where s is the sample standard deviation (the square root of the sample variance), n is the sample size, and t is the value from the t-distribution with n-1 degrees of freedom for an 80% CI (from a t-table or calculator).

We must first determine the sample standard deviation:

s = √(6) ≈ 2.45

Then, we can find the worth of t from a t-table or mini-computer for a 80% CI with 14 levels of opportunity (15-1):

Lastly, these numbers can be used to calculate the 80% CI width:

With the given parameters, the width of the 80% CI for the normal distribution's mean is roughly 1.5.

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Right

so 7 is the hypotenuse because it is the biggest. so you have to use 6 and 4 in the formula to see if they equal 7.

(a)²+(b)²=c²

(6)²+(4)²=c²

36+16=c²

(square root) 52=c²

the square root of 52 is 7

so therefore it is a right triangle.

n = ceil(sqrt((2008/4phi)[90 - arccos(k(2 cos(phi/2008)))]))

       2 [cos(alpha) sin(alpha) + cos(4 alpha) sin(2 alpha) + cos(9 alpha) sin(3 alpha) + ... + cos(n^2 alpha) sin(n alpha)]

     = [sin(2 alpha) + sin(8 alpha) + sin(18 alpha) + ... + sin(n^2 alpha)].

        sin(2 alpha) + sin(8 alpha) + sin(18 alpha) + ... + sin(n^2 alpha)

     = (sin(2 alpha) - sin(2n^2 alpha))/(1 - sin(2 alpha))

     = (2 sin(n^2 alpha) cos(n^2 alpha))/(2 cos(alpha) - 1)

     = [sin(2n^2 phi/2008)]/(2 cos(alpha) - 1)

        [sin(2n^2 phi/2008)]/(2 cos(alpha) - 1)

        sin(2n^2 phi/2008) = k(2 cos(alpha) - 1)

        cos(90 - 2n^2 phi/2008) = k(2 cos(alpha) - 1)

       90 - 2n^2 phi/2008 = ±arccos(k(2 cos(alpha) - 1))

       n^2 = (2008/4phi)[90 ± arccos(k(2 cos(alpha) - 1))]

       n = ceil(sqrt((2008/4phi)[90 - arccos(k(2 cos(phi/2008)))]))

We can find the smallest positive integer n by incrementing the value of k starting from 1 until ceil(k^2*alpha) is greater than or equal to n.

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Answer:

  (a)  No, because for each input there is not exactly one output.

Step-by-step explanation:

You want to know if the relation shown in the graph is a function.

A relation is a function if its graph passes the vertical line test. That is, a vertical line cannot intercept the graph of the relation at more than one point.

The points (-1, -2) and (-1, 3) will both be intercepted by the vertical line x = -1. This tells us the relation is not a function, because it has two outputs for that input.

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Answer:   Based on the problem statement, we can define a recurrence relation as follows:

t(n) = t(n-1) + t(n-2) + t(n-4)

This means that the number of ways to construct a tower of height n cm can be obtained by considering the possible heights of the bottom block in the tower. If the bottom block has height 1 cm, then the remaining blocks make up a tower of height (n-1) cm, for which there are t(n-1) ways to construct it. If the bottom block has height 2 cm, then the remaining blocks make up a tower of height (n-2) cm, for which there are t(n-2) ways to construct it. If the bottom block has height 4 cm, then the remaining blocks make up a tower of height (n-4) cm, for which there are t(n-4) ways to construct it.

Since we are assuming an unlimited supply of blocks of each size, we can use these blocks repeatedly to construct towers of different heights. Also, we can use dynamic programming to compute the values of t(n) for each integer n from 1 to 25, by using the recurrence relation above and the base cases:

t(0) = 1 (there is only one way to construct a tower of height 0 cm, which is to not use any blocks)

t(n) = 0 for n < 0 (there is no way to construct a tower of negative height)

Using these, we can compute the values of t(n) for n = 1, 2, ..., 25, as follows:

t(0) = 1

t(1) = t(0) = 1

t(2) = t(1) + t(0) = 2

t(3) = t(2) + t(1) = 3

t(4) = t(3) + t(2) + t(0) = 6

t(5) = t(4) + t(3) + t(1) = 10

t(6) = t(5) + t(4) + t(2) = 19

t(7) = t(6) + t(5) + t(3) = 32

t(8) = t(7) + t(6) + t(4) = 61

t(9) = t(8) + t(7) + t(5) = 104

t(10) = t(9) + t(8) + t(6) = 195

t(11) = t(10) + t(9) + t(7) = 332

t(12) = t(11) + t(10) + t(8) = 626

t(13) = t(12) + t(11) + t(9) = 1065

t(14) = t(13) + t(12) + t(10) = 2002

t(15) = t(14) + t(13) + t(11) = 3405

t(16) = t(15) + t(14) + t(12) = 6403

t(17) = t(16) + t(15) + t(13) = 10946

t(18) = t(17) + t(16) + t(14) = 20618

t(19) = t(18) + t(17) + t(15) = 350

Step-by-step explanation:

Answer: y - 10.26

Step-by-step explanation: Add 5.4 on both sides and that leaves you with y

the t-score for a 99.8% confidence interval with a sample size of 5 is 4.604.

The t-score for a 99.8% confidence interval with a sample size of 5 can be found using a t-distribution table or calculator.

Since we have a sample size of 5, the degrees of freedom (df) will be n - 1 = 4.

From the t-distribution table or calculator, the t-score for a 99.8% confidence interval with 4 degrees of freedom is approximately 4.604.

Therefore, the t-score for a 99.8% confidence interval with a sample size of 5 is 4.604.

the complete question is :

what is the value of the t score for a 99.8% confidence interval if we take a sample of size 5?

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Answer:

4

Step-by-step explanation:

Answer:

[tex]x\leq 6[/tex]

Step-by-step explanation:

ANSWER-

Let the nonadhacent anglesof complementary angle be x and y where, x=79 and y =?

WE KNOW,

x+y=90

or,79+y=90

or, y=90-79

:. y=11,,

triangle C

Step-by-step explanation:

If you draw it out, it looks unique

The area of the table top is approximately 28.27 square feet. To find the area of a circular table top with a given circumference, we can use the formula A = πr², where r is the radius.

To find the area of the table top, we need to use the formula for the area of a circle, which is:

A = πr²

We are given the circumference of the table top, which is:

C = 2πr

We can solve for r by dividing both sides by 2π:

r = C / (2π) = 18.84 / (2 * 3.14) = 3

Now we can substitute this value for r into the formula for the area of a circle:

A = π(3)² = 9π

Using 3.14 for π, we get:

A ≈ 28.26

Rounding to the nearest hundredth, the area of the table top is approximately 28.27 square feet.

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Answer:113.04 in squared

Step-by-step explanation: a=3.14Rsquared

A=3.14 x 6squared 6 x 6=36

A=3.14 x 36

3.14 x 36 = 113.04 inches squared or 113 inches squared

To cover the circular window, you will require a piece of tinted paper measuring approximately 113 in²

The area of the circular window can be found using the formula for the area of a circle:

A = πr²

where A is the area of the circle, π is the mathematical constant pi (approximately 3.14), and r is the radius of the circle.

In this case, the radius is given as 6 inches, so we can substitute that value into the formula:

A = π (6 inches)²

A = π (36 square inches)

A ≈ 113.04 square inches

Rounding to the nearest tenth gives:

A ≈ 113 square inches

Therefore, you will need approximately 113 square inches of tinted paper to cover the circular window.

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If the number of samples is doubled, the new confidence interval would be 9.61, 10.39 or narrower (smaller) while keeping the same confidence level.

. When calculating the confidence interval, the standard error is used, along with the sample mean and the critical value from the distribution. If we have a larger sample size, we can be more confident in our estimate of the population parameter because the sample mean will more closely resemble the population mean. As a result, the confidence interval can be narrower, indicating a higher degree of precision. For Example:Suppose a sample of 50 was taken, and the mean weight of an object was 10 grams with a standard deviation of 2 grams.

At a 95 percent confidence level, the confidence interval for the mean weight would be 10 ± (1.96)(2/√50) = (9.15, 10.85)Now suppose that the sample size is doubled to 100. The standard error will be cut in half, i.e., 2/√100 = 0.2. As a result, the new confidence interval would be 10 ± (1.96)(0.2) = (9.61, 10.39). Notice that the new confidence interval is narrower, indicating a higher degree of precision while keeping the same confidence level.

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A patient's probability of recovering is 0.8, so the likelihood that they won't is 0.2.

The probability that a patient will recover from a stomach disease is 0.8. Assume that 20 persons have reportedly got this illness.

Let X = # of recoveries out of 20 patients who caught the condition while PMF for X and resolve issues. Imagine that a 20-person sample was chosen at random.

Here, Bernoulli trials are applicable.

Recall that the chance of k successes in n trials using the Bernoulli approach is given by:

P(k)=( n/k )p (power k) * (q power n−k)

where q=1-p is the probability that an attempt would fail

where q=1-p is the probability that an attempt would fail

Here, let's make recovery a success. Hence, p = 0.8, q = 0.2, and n = 20

The probability that at least 10 recoveries will occur is

P(X10) = P(10) + P(11) + P(12) +... + P (20)

The complete question will be:

The probability that a patient recovers from a stomach disease is 0.8. Suppose twenty people are known to have contracted this disease. What is the probability that at least ten recover?

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Answer:

Commutative property

The Commutative property is most simply shown with: a x b = b x a. In multiplication, the values can shift or "commute" in any order

Yes, we  would use the normal approximation to predict these results.

When it comes to hypothesis testing and confidence intervals, the normal distribution plays a crucial role.

When sample sizes are large enough, the normal distribution can be used as a reasonable approximation for the binomial distribution.

This is because the binomial distribution approaches the normal distribution as sample size increases.

To calculate the normal approximation, you will need to determine the mean and standard deviation of the binomial distribution.

The mean is np and the standard deviation is the square root of np(1-p),

where n is the sample size and p is the probability of success.

The probability of success in this case is 45%, or 0.45.

Therefore, the mean is 10 * 0.45 = 4.5 and the standard deviation is the square root of 10 * 0.45 * (1 - 0.45) = 1.37.

Now that you have the mean and standard deviation, you can use the normal distribution to make predictions about the sample.

If you want to find the probability that exactly 5 students will go to prom, for example, you would use the formula for the normal distribution with a mean of 4.5 and a standard deviation of 1.37.

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The length of the hypotenuse is 2 times the square root of 5. The Pythagorean theorem can be used to determine the length of the hypotenuse.

To graph the right triangle with the given points as the hypotenuse, we first plot the points on a coordinate plane . The two points form the endpoints of the hypotenuse, which is the line segment connecting them. We can find the length of this line segment using the distance formula:

distance = [[tex]\sqrt{(x2 - x1)^2 + (y2 - y1)^2]}[/tex]

In this case, we have:

distance = [[tex]\sqrt{(-5 - (-3))^2 + (-6 - (-4))^2}[/tex]]

distance = [tex]\sqrt{(-5 - (-3))^2 + (-6 - (-4))^2}[/tex]]

distance = [[tex]\sqrt{4 + 4}[/tex]]

distance = [[tex]\sqrt{8}[/tex]]

We can simplify [tex]\sqrt{8}[/tex] by factoring out the perfect square factor of 4:

distance = [[tex]\sqrt{4 * 2}[/tex]]

distance = [tex]\sqrt{4} *\sqrt{2}[/tex]

distance = 2 * [tex]\sqrt{2}[/tex]

Thus, the distance between the two points is 2 * [tex]\sqrt{2}[/tex] ] units.

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Answer:

multiply BD and AC and you got it

Step-by-step explanation:

if yuo put ABD triangle over CEB you got a sqare, if you do it again you got a aqare 20x10

The value 1.106 represents the slope of the regression line or the rate of change of y with respect to x.

In the given regression equation y = 3.915(1.106)x:

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Answer:

Part A is 8/13 of the whole

Part B is 5/13 of the whole

Step-by-step explanation:

Assuming there are no other parts,

the Whole = A + B is the denominator:

Whole = 40 + 25 = 65

Part A = 40 and Part B = 25 are numerators for each fraction.

The fractions are then:

40/65 and 25/65

Meaning:

Part A is 40/65 of the whole

Part B is 25/65 of the whole

Reducing the fractions, it is also true that:

Part A = 8/13 of the whole

Part B = 5/13 of the whole

Arranged from smallest to largest, the given values are 0.2, 3/20, 1/8, 11%, and 13%.

To compare these values, we need to convert the percentages to decimals. We can do this by dividing them by 100. So, 11% becomes 0.11 and 13% becomes 0.13.

Next, we can convert 3/20 and 1/8 to decimals by dividing them using a calculator. We get:

3/20 = 0.15

1/8 = 0.125

Now, we can arrange these values in ascending order:

0.2 < 0.125 < 0.15 < 0.11 < 0.13

Therefore, the values arranged in order from smallest to largest are 0.2, 3/20, 1/8, 11%, and 13%.

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Answer:

a. A' = {3, 5, 7, 9} (complement of A)

b. A∩B = {2} (intersection of A and B, which contains only the even prime number 2)

c. A∪B = {2, 4, 6, 8, 3, 5, 7} (union of A and B, which contains all even numbers and all prime numbers between 2 and 9)

Nonparametric tests can be useful in situations where the data may not follow a specific distribution or where the assumptions of a parametric test are not met.

The nonparametric tests mentioned in your question do not assume any specific probability distribution for the data. Hence, they are called nonparametric tests. These tests are used when the assumptions required for parametric tests (e.g., normality) are not met, or when the data is measured on ordinal or nominal scales rather than continuous ones.

The Wilcoxon rank-sum test, sign test, and Wilcoxon signed-rank test are used to compare two independent or dependent samples. The Kruskal-Wallis test and Friedman test are used to compare three or more independent or dependent samples.

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Answer:

biquadratic polynomial

Step-by-step explanation:

it has degree 4

Answer:Tham I can’t even se the letters you should have posted each question one by one

Step-by-step explanation:

good luck

Step-by-step explanation:

Find the area of the yellow circle using pie times radius squared. Then find the area of the entire circle using the same formula then take the answer for the area of the entire circle - area of the yellow circle