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Verizon [17]
3 years ago
15

A coach buys a uniform and a basketball for each of the 15 players on the team. Each basketball costs $9.00. The coach spends $4

20 for uniforms and basketballs. What is the cost of 1 uniform in dollars?
Mathematics
1 answer:
Mnenie [13.5K]3 years ago
6 0
Call u the cost per uniform.

Since each of the 15 players gets one uniform and one basketball, we can make the following equation:

15u + 15(9) = 420

Solve this equation to get u=285/15= $19
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The weekly sales in thousands of items of a product has a seasonal sales record approximated by n=60.31+19.4sin(/24) (pi x t /24
Paladinen [302]
We are given with the equation
n = 60.31 + 19.4 sin (πt/24)
where n is the weekly sales in thousands of item
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(Uniform) Suppose X follows a continuous uniform distribution from 4 to 11. (a) Write down the PDF of X. (b) Find P(X ≤ 7). Roun
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Answer:

a) f(x)= \frac{1}{11-4}= \frac{1}{7}, 4 \leq x \leq 11

b) P(X\leq 7) = F(7) = \frac{7-4}{11-4}= 0.4286

c) P(5 < X \leq 7)= F(7) -F(5) = \frac{7-4}{7} -\frac{5-4}{7}= 0.2857

d) P(X >5 | X \leq 7)

And we can find this probability with this formula from the Bayes theorem:

P(X >5 | X \leq 7)= \frac{P(X>5 \cap X \leq 7)}{P(X \leq 7)}= \frac{P(5

Step-by-step explanation:

For this case we assume that the random variable X follows this distribution:

X \sim Unif (a=4, b =11)

Part a

The probability density function is given by the following expression:

f(x) = \frac{1}{b-a} , a \leq x \leq b

f(x)= \frac{1}{11-4}= \frac{1}{7}, 4 \leq x \leq 11

Part b

We want this probability:

P(X \leq 7)

And we can use the cumulative distribution function given by:

F(x) = \frac{x-a}{b-a}= \frac{x-4}{11-4}

And replacing we got:

P(X\leq 7) = F(7) = \frac{7-4}{11-4}= 0.4286

Part c

We want this probability:

P(5 < X \leq 7)

And we can use the CDF again and we have:

P(5 < X \leq 7)= F(7) -F(5) = \frac{7-4}{7} -\frac{5-4}{7}= 0.2857

Part d

We want this conditional probabilty:

P(X >5 | X \leq 7)

And we can find this probability with this formula from the Bayes theorem:

P(X >5 | X \leq 7)= \frac{P(X>5 \cap X \leq 7)}{P(X \leq 7)}= \frac{P(5

7 0
3 years ago
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Step-by-step explanation:

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