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dem82 [27]
2 years ago
10

The student council raised $2,790 during its annual fundraiser. The money was shared equally among the Events Committee, the Spo

rts Committee, and the Sunshine Committee. a. The Sports Committee wants to use its share of the money to purchase t-shirts. If each t-shirt costs $12, what is the greatest number of t-shirts the Sports Committee can purchase? ANSWER NOW PLEASE!!!!
Mathematics
1 answer:
irina [24]2 years ago
6 0

Answer:

77

Step-by-step explanation:

$2790 is shares equally between 3 committees. each committee would get - $2790 / 3 = $930

If a shirt costs $12 and the Sports Committee gets $930, the committee would get $930 / $12 = 77.5

The committee cannot get half a shirt, so the greatest number of shirt that can be purchased is 77

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The sum of two positive integers, x and y, is not more than 40. The difference of the two integers is at least 20. Chaneece choo
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<span>Chaneece chooses x as the larger number and uses the inequalities y ≤ 40 – x and y ≤ x – 20 to determine the possible solutions.

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"Majesty Video Production Inc. wants the mean length of its advertisements to be 30 seconds. Assume the distribution of ad lengt
Westkost [7]

Answer:

a) \bar X \sim N(\mu=30, \frac{2}{\sqrt{16}})

b) Se=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{16}}=0.5

c) P(\bar X >31.25)=0.006=0.6\%

d) P(\bar X >28.25)=0.9997=99.97\%

e) P(28.25

Step-by-step explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variabl length of advertisements produced by Majesty Video Production Inc. We know from the problem that the distribution for the random variable X is given by:

X\sim N(\mu =30,\sigma =2)

We take a sample of n=16 . That represent the sample size.

a. What can we say about the shape of the distribution of the sample mean time?

From the central limit theorem we know that the distribution for the sample mean \bar X is also normal and is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

\bar X \sim N(\mu=30, \frac{2}{\sqrt{16}})

b. What is the standard error of the mean time?

The standard error is given by this formula:

Se=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{16}}=0.5

c. What percent of the sample means will be greater than 31.25 seconds?

In order to answer this question we can use the z score in order to find the probabilities, the formula given by:

z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}

And we want to find this probability:

P(\bar X >31.25)=1-P(\bar X

d. What percent of the sample means will be greater than 28.25 seconds?

In order to answer this question we can use the z score in order to find the probabilities, the formula is given by:

z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}

And we want to find this probability:

P(\bar X >28.25)=1-P(\bar X

e. What percent of the sample means will be greater than 28.25 but less than 31.25 seconds?"

We want this probability:

P(28.25

3 0
2 years ago
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