I NEEED THIS NOW PLS HELP ME

Mathematics

2 answers:

DedPeter [7]4 years ago

6 0

Answer:

56cm²

Step-by-step explanation:

To solve this question, find the area of the triangles separately and then add all the areas to find the total area of the kite.

Attached is the image of my work

MakcuM [25]4 years ago

5 0

Answer:

56 cm²

Step-by-step explanation:

The area (A) of a kite is calculated as

A = $\frac{1}{2}$ product of the diagonals

= $\frac{1}{2}$ × d₁ × d₂

d₁ = 4 + 4 = 8 and d₂ = 10 + 4 = 14

A = $\frac{1}{2}$ × 8 × 14 = 4 × 14 = 56 cm²

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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) <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)