Six times the sum of a number and twelve is forty.

1 answer:

4 0

X-the number

6(x + 12) = 40 |use distributive property: a(b + c) = ab + ac

6x + 72 = 40 |subtract 72 from both sides

6x = -32 |divide both sides by 6

x = -32/6

x = -16/3

x = -5 1/3

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Consider trapezoid LMNO.

jenyasd209 [6]

The side LO is congruent to the side MN, the diagonal LN is congruent to the diagonal MO, and the angle L is congruent to the angle M in an isosceles trapezoid, denoted by the symbols LMNO.

What are the conditions for an Isosceles Trapezoid?

The conditions listed below demonstrate that any trapezoid is an isosceles trapezoid:

The length of both legs is the same.
2nd condition: The base angles are of equal proportion.
The length of the diagonals is the same.

When these conditions are met by the given trapezoid LMNO, it will be referred to as an isosceles trapezoid. Hence, the following conditions of trapezoid LMNO need to be fulfilled,

LN ≅ MO

LO ≅ MN

∠L ≅ ∠M

Learn more about a trapezoid here:

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70 miles per hour

Step-by-step explanation:

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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) Now, here sample of NBA players is 26 so n = 26.

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)