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3 years ago

Lydia wrote 4 1/2 pages of her science report in one hour. what was her writing rate in pages per minute?

1 answer:

7 0

To solve this question, I first converted one hour into minutes. (60 minutes)

Then I divided 4 1/2 into 60. As a decimal, the answer if 0.075. If you convert that into a lowest terms fraction, you get 3/40.

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F(x)=3x+5, g(x)=6x^2 find (fg)(x)

alexgriva [62]

Answer:

18x^3+30x^2

Step-by-step explanation:

im smart.

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3 years ago

The price of an office chair was $86. The tax rate was 6%. What was the tax on the chair?

Tamiku [17]

Answer:

X.06

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516

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05.16

Step-by-step explanation:

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3 years ago

Ron is tiling a countertop he needs to place 54 square titles in each of 8 rows to cover the counter

DiKsa [7]

He would only be able to make 6 rows with 3/4 of another

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3 years ago

One measure of an athlete’s ability is the height of his or her vertical leap. Many professional basketball players are known fo

almond37 [142]

Answer:

(1) P( $\bar X$ < 26 inches) = 0.0436

(2) P(27.5 inches < $\bar X$ < 28.5 inches) = 0.2812

Step-by-step explanation:

We are given that the mean vertical leap of all NBA players is 28 inches. Suppose the standard deviation is 7 inches and 36 NBA players are selected at random.

Firstly, Let $\bar X$ = mean vertical leap for the 36 players

Assuming the data follows normal distribution; so the z score probability distribution for sample mean is given by;

Z = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean vertical leap = 28 inches

$\sigma$ = standard deviation = 7 inches

n = sample of NBA player = 36

(1) Probability that the mean vertical leap for the 36 players will be less than 26 inches is given by = P( $\bar X$ < 26 inches)

P( $\bar X$ < 26) = P( $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{26-28}{\frac{7}{\sqrt{36} } }$ ) = P(Z < -1.71) = 1 - P(Z $\leq$ 1.71)

= 1 - 0.95637 = 0.0436

(2) <em>Now, here sample of NBA players is 26 so n = 26.</em>

Probability that the mean vertical leap for the 26 players will be between 27.5 and 28.5 inches is given by = P(27.5 inches < $\bar X$ < 28.5 inches) = P( $\bar X$ < 28.5 inches) - P( $\bar X$ $\leq$ 27.5 inches)

P( $\bar X$ < 28.5) = P( $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{28.5-28}{\frac{7}{\sqrt{26} } }$ ) = P(Z < 0.36) = 0.64058 {using z table}

P( $\bar X$ $\leq$ 27.5) = P( $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{27.5-28}{\frac{7}{\sqrt{26} } }$ ) = P(Z $\leq$ -0.36) = 1 - P(Z < 0.36)

= 1 - 0.64058 = 0.35942

Therefore, P(27.5 inches < $\bar X$ < 28.5 inches) = 0.64058 - 0.35942 = 0.2812

6 0

4 years ago

Find the missing probability. P(A)=720,P(B)=35,P(A∩B)=21100 ,P(A∪B)=?

Dimas [21]

P (A U B) = 37/50

7/20+3/5+21/100

5 0

3 years ago