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Vikentia [17]
3 years ago
6

Need more help - Attached image

Mathematics
2 answers:
Naily [24]3 years ago
7 0

Answer:

D. 10x^7 + 7x^5 -2x + 9

Step-by-step explanation:

Like terms: 7x^7 and 3x^7 = 10x^7

                  5x^5 and 2x^5 = 7x^5

                  -3x and x\\ = -2x

                  7 and 2 = 9

The answer would be: 10x^7 + 7x^5 - 2x + 9

Therefore, it is <u>option D</u>

Your Welcome! Vote me brainliest!!!

mylen [45]3 years ago
4 0

Answer:B

Step-by-step explanation:

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If you can buy 1⁄3 of a box of chocolates for 6 dollars, how much can you purchase for 4 dollars?
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Step-by-step explanation:

6 / 1 / 3 = 4 / b

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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
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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

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