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3 years ago

Solve for x 3 x − 8 ≤ 23 AND − 4 x + 26 ≥ 6 Please help thank you!!

Mathematics

1 answer:

MAXImum [283]3 years ago

8 0

To solve for x, you need to isolate/get the variable "x" by itself in the inequality:

3x - 8 ≤ 23 Add 8 on both sides

3x - 8 + 8 ≤ 23 + 8

3x ≤ 31 Divide 3 on both sides to get "x" by itself

$\frac{3x}{3} \leq \frac{31}{3}$

$x\leq \frac{31}{3}$ (x is any number less than or equal to 31/3)

-4x + 26 ≥ 6 Subtract 26 on both sides

-4x ≥ -20 Divide -4 on both sides [When you multiply/divide by a negative number to both sides of an inequality, you flip the sign </>]

$\frac{-4x}{-4} \geq \frac{-20}{-4}$ (two negative signs cancel each other out and become positive)

x ≤ 5 (x is any number less than or equal to 5)

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Answer: x= 1 , y = 7

Step-by-step explanation:

y = -6x + 13 ................. equation 1

y = 3x + 4 ................ equation 2

substitute y = 3x + 4 into equation 1 , then we have

3x + 4 = - 6x + 13

collect the like terms , we have

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substitute x = 1 into equation 2 , we have

y = 3(1) + 4

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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

$\sigma$ = 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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