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3 years ago

At the beginning of a snowstorm, Sydney had 11 inches of snow on her lawn. The

Mathematics

2 answers:

Phantasy [73]3 years ago

5 0

Answer:

Snow after 3 hours: 12.5

Snow after t hours: 11+0.5t

kondor19780726 [428]3 years ago

3 0

Answer:

12.5in.

x=.5t+11

Step-by-step explanation:

First of all, create the equation so you can solve both questions at once. We know that Sydney had eleven inches of snow on her lawn, and it doesn't dissapear, so it starts with x=11. Then every hour she gets .5 inches per hour, and the amount of hours is t, so that will translate to .5t. that makes the equation x=.5t+11. To find out how much she would have in three hours, substitute 3 for t, which which goes like this: x=.5(3)+11

x=1.5+11

x=12.5

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almond37 [142]

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(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

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(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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