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Morgarella [4.7K]

2 years ago

7

In the past year, Deandre watched 21 movies that he thought were very good. He watched 70 movies over the whole year. Of the mov

ies he watched, what percentage did he think were very good?

1 answer:

emmasim [6.3K]2 years ago

7 0

Answer: 30%

Step-by-step explanation: 21/70 = 21 divide 70 = 0.3 = 30%

You might be interested in

Simplify x + 6.2 + 8.5.<br><br> 14.7x<br> x + 14.7<br> 6.2x + 8.5

In-s [12.5K]

Combine the like terms.
Like terms are terms that are the same disregarding the coefficients,

In this case, add the constants.
6.2+8.5=14.7

Final answer: x+14.7

3 0

3 years ago

Read 2 more answers

To estimate the mean height μ of male students on your campus,you will measure an SRS of students. You know from government data

nexus9112 [7]

Answer:

a) $\sigma = 0.167$

b) We need a sample of at least 282 young men.

Step-by-step explanation:

Problems of normally distributed samples can be 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

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

This Zscore is how many standard deviations the value of the measure X 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.

(a) What standard deviation must x have so that 99.7% of allsamples give an x within one-half inch of μ?

To solve this problem, we use the 68-95-99.7 rule. This rule states that:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviations of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we want 99.7% of all samples give X within one-half inch of $\mu$ . So $X - \mu = 0.5$ must have $Z = 3$ and $X - \mu = -0.5$ must have $Z = -3$ .

So

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

$3 = \frac{0.5}{\sigma}$

$3\sigma = 0.5$

$\sigma = \frac{0.5}{3}$

$\sigma = 0.167$

(b) How large an SRS do you need to reduce the standard deviationof x to the value you found in part (a)?

You know from government data that heights of young men are approximately Normal with standard deviation about 2.8 inches. This means that $\sigma = 2.8$

The standard deviation of a sample of n young man is given by the following formula

$s = \frac{\sigma}{\sqrt{n}}$

We want to have $s = 0.167$

$0.167 = \frac{2.8}{\sqrt{n}}$

$0.167\sqrt{n} = 2.8$

$\sqrt{n} = \frac{2.8}{0.167}$

$\sqrt{n} = 16.77$

$\sqrt{n}^{2} = 16.77^{2}$

$n = 281.23$

We need a sample of at least 282 young men.

6 0

3 years ago

Need help with this anyone please

andriy [413]

Option B: 1

Option C: 6

Solution:

Let us first define the vertical angle theorem:

Vertical angle theorem:

<em>If two lines are intersecting each other, then the vertically opposite angles are congruent.</em>

To find angle 7 is equal to which angle:

Angle 1 and angle 7 are vertically opposite angles.

Therefore by vertical angle theorem, angle 7 = angle 1

Option B is the correct answer.

To find angle 4 is equal to which angle:

Angle 4 and angle 6 are vertically opposite angles.

Therefore by vertical angle theorem, angle 4 = angle 6

Option C is the correct answer.

5 0

3 years ago

What's the answer??

Vilka [71]

The answer is neither. i hope this helps

8 0

3 years ago

Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a

Bumek [7]

Answer:

0.381 is the probability that the number of drivers will be at most 18.

Step-by-step explanation:

We are given the following information in the question:

The number of drivers who travel between a particular origin and destination during a designated time period has a Poisson distribution with parameter μ = 20.

The Poisson distribution is the discrete probability distribution of the number of events occurring in a given time period, given the average number of times the event occurs over that time period.
The variance of Poisson distribution is equal to the mean of Poisson distribution.

a) P(number of drivers will be at most 18)

Formula:

$P(X =k) = \displaystyle\frac{\mu^k e^{-\mu}}{k!}\\\\ \mu \text{ is the mean of the distribution}$

$P( x \leq 18) =P(x=0) + P(x =1) + P(x = 2) + ... + P(x = 18)\\\\= \displaystyle\frac{20^0 e^{-20}}{0!} + \displaystyle\frac{20^1 e^{-20}}{1!} +...+ \displaystyle\frac{20^{18} e^{-20}}{18!}\\\\ = 0.381$

Thus, 0.381 is the probability that the number of drivers will be at most 18.

3 0

4 years ago