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3 years ago

Given: Triangle ABC : triangle ADB. AB =24, AD =16 Find: AC

Mathematics

1 answer:

RoseWind [281]3 years ago

6 0

Answer:

Step-by-step explanation:

Given ΔABC and ΔADB, since both triangles are right angled triangles then the following are true.

From ΔADB, AB² = AD²+BD²

Given AB = 24 and AD = 16

BD² = AB² - AD²

BD² = 24²-16²

BD² = 576-256

BD² = 320

BD = $\sqrt{320}$

BD = 17.9

from ΔABC, AC² = AB²+BC²

SInce AC = AD+DC and BC² = BD² + DC² (from ΔBDC )we will have;

(AD+DC)² = AB²+ (BD² + DC²)

Given AD = 16, AB = 24 and BD = 17.9, on substituting

(16+DC)² = 24²+17.9²+ DC²

256+32DC+DC² = 24²+17.9²+ DC²

256+32DC = 24²+17.9²

32DC = 24²+17.9² - 256

32DC = 640.41

DC = $\frac{640.41}{32}$

DC = 20.01

Remember that AC = AD+DC

AC = 16+20.01

AC = 36.01

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3 years ago

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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

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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)

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

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Answer:

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