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3 years ago

6

The heights of 200 adults were recorded and divided into two categories. Which two-way frequency table correctly shows the margi

nal frequencies?

1 answer:

Temka [501]3 years ago

7 0

Answer: c

Step-by-step explanation:

C

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Which best models a plan a. floor b. pebble c. jump rope d. stick

Setler79 [48]

Its .A.

hope this helps

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3 years ago

Rochelle works part-time at the college museum, which pays her at a rate of $10 per hour. She also makes small bracelets that sh

katrin [286]

We know for our problem that Rachel makes $10 per hour. Since $x$ represent the number of hours that she works, $10x$ will be the total amount that she will make for working $x$ hours. We also know that she sells bracelets for $5 each. Since $y$ represents the number of of bracelets that she sells, $5y$ will be her total revenue for selling $y$ bracelets. We also know that she needs to earn at least $200 a week to cover her expenses, so the sum of $10x$ and $5y$ must be equal or greater than 200:

$10x+5 \geq 200$

We can conclude that <span>the graphs that shows the inequality that represents this situation, with its solution region shaded is:</span>

4 0

3 years ago

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)

= 1 - 0.64058 = 0.35942

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

6 0

3 years ago

Hi please help,i really need it and so appreciate it:)

nadya68 [22]

Hey there!

12 ≥ 5f - 18

5f - 18 ≤ 12

ADD 18 to BOTH SIDES

5f - 18 + 18 ≤ 12 + 18

SIMPLIFY IT!

5f ≤ 30

DIVIDE 5 to BOTH SIDES

5f/5 ≤ 30/5

SIMPLIFY IT!

f ≤ 6

Therefore, your answer is: f ≤ 6

Good luck on your assignment & enjoy your day!

~Amphitrite1040:)

7 0

2 years ago

Which of the following statements is true about the three quadrilaterals?

OLEGan [10]

Answer:

M and O are similar but not congruent

4 0

2 years ago