Find the value of the variable. If your answer is not an integer, leave it in simplest radical form.

A rental company charges $8 per hour to rent a kayak plus a $12 rental fee. If Margerie spent $44, for how many hours did she re

DanielleElmas [232]

4 hours she spent .............:

3 0

2 years ago

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60 PTS PLS ANSWER WILL AWARD BRAINLIEST ASAP

pishuonlain [190]

Answer:

9. a) 180 - (55+90) = a

b) a=35 degrees

10. a) 2w + 3w + 40 = 180

b) w=28

c) 2w=56 3w=84

Step-by-step explanation:

3 0

3 years ago

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The circumference of a circle exceeds the radius by 74 cm. Find the radius of the circle.

saw5 [17]

Answer:

14 cm.

Step-by-step explanation:

Let the radius of the circle = r cm.

Then circumference of circle = 2 π r.

Since circumference exceeds the radius by 74 cm

Therefore, according to the question,

$\sf \:2 π r = r+ 74$

$\sf \: = > 2 \times \frac{22}{7} \times r = r \: + 74$

$\sf \: = > \frac{44r}{7} -r = 74$

$\sf \: = > \frac{44r - 7r}{7} = 74$

$\sf \: = > \frac{37r}{7} = 74$

$\sf \: = > r = \frac{7 \times 74}{37}$

$\sf \: = > r = 14$

Hence, the radius of the circle is 14 cm.

________________________________

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2 years ago

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The scores on the GMAT entrance exam at an MBA program in the Central Valley of California are normally distributed with a mean

Kaylis [27]

Answer:

58.32% probability that a randomly selected application will report a GMAT score of less than 600

93.51% probability that a sample of 50 randomly selected applications will report an average GMAT score of less than 600

98.38% probability that a sample of 100 randomly selected applications will report an average GMAT score of less than 600

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 591, \sigma = 42$

What is the probability that a randomly selected application will report a GMAT score of less than 600?

This is the pvalue of Z when X = 600. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{600 - 591}{42}$

$Z = 0.21$

$Z = 0.21$ has a pvalue of 0.5832

58.32% probability that a randomly selected application will report a GMAT score of less than 600

What is the probability that a sample of 50 randomly selected applications will report an average GMAT score of less than 600?

Now we have $n = 50, s = \frac{42}{\sqrt{50}} = 5.94$

This is the pvalue of Z when X = 600. So

$Z = \frac{X - \mu}{s}$

$Z = \frac{600 - 591}{5.94}$

$Z = 1.515$

$Z = 1.515$ has a pvalue of 0.9351

93.51% probability that a sample of 50 randomly selected applications will report an average GMAT score of less than 600

What is the probability that a sample of 100 randomly selected applications will report an average GMAT score of less than 600?

Now we have $n = 50, s = \frac{42}{\sqrt{100}} = 4.2$

$Z = \frac{X - \mu}{s}$

$Z = \frac{600 - 591}{4.2}$

$Z = 2.14$

$Z = 2.14$ has a pvalue of 0.9838

98.38% probability that a sample of 100 randomly selected applications will report an average GMAT score of less than 600

8 0

3 years ago

Americans receive an average of 20 Christmas cards each year. Suppose the number of Christmas cards is normally distributed with

soldier1979 [14.2K]

The distribution of X is X ~ N (20 , 6) and the probability that this American will receive no more than 24 Christmas cards this year is 0.7486.

<h3>Probability</h3>

a. Distribution

X ~ N (20 , 6)

b. P(x ≤24)

= P[(x - μ ) / σ (24 - 20) / 6]

= P(z ≤0.67)

= 0.74857

=0.7486

Hence:

Probability = 0.7486

c. P(21 < x < 26)

= P[(21 - 26)/ 6) < (x - μ ) / σ < (24 - 20) / 6) ]

= P(-0.83 < z < 0.67)

= P(z < 0.67) - P(z < -0.)

= 0.74857- 0.2033

= 0.54527

Hence:

Probability =0.54527

d. Using standard normal table ,

P(Z < z) = 66%

P(Z < 0.50) = 0.66

z = 0.50

Using z-score formula,

x = z× σ + μ

x = 0.50 × 6 + 20 = 23

23 Christmas cards

Therefore the distribution of X is X ~ N (20 , 6) and the probability that this American will receive no more than 24 Christmas cards this year is 0.7486.

Learn more about probability here:brainly.com/question/24756209

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8 0

2 years ago