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MAXImum [283]
3 years ago
13

The table shows whether a bus pass is a child's or an adult's pass and whether it is a daily or monthly pass. One bus pass is ra

ndomly selected.
What is the probability that the pass is a daily pass?

Write the probability as a percent.

Round to the nearest tenth of a percent as needed.

Mathematics
2 answers:
OLEGan [10]3 years ago
7 0
31/61 or 50.8 percent
the daily passes added together are 31 passes for adults and children. Adding everything together gives us the denominator of the fraction 61. 31/61 = 50.8%.  

pantera1 [17]3 years ago
6 0

Answer:

The probability is \frac{31}{51} or 60.8%

Step-by-step explanation:

The given table shows the daily or monthly passes of children or adults. One bus pass is randomly selected.  

The daily passes are 8+23=31

The monthly passes are 5+15=20

Total passes in all are 31+20=51

Now, we have to find the probability that the one randomly selected pass is a daily pass.

The probability is = \frac{31}{51}

In percentage it is = \frac{31}{51} \times100 = 60.78% or rounding to nearest tenth, we get 60.8%.

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83.85% of 1-mile long roadways with potholes numbering between 22 and 58

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

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

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

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In this problem, we have that:

Mean = 49

Standard deviation = 9

Using the Empirical Rule, what is the approximate percentage of 1-mile long roadways with potholes numbering between 22 and 58?

22 = 49 - 3*9

So 22 is three standard deviations below the mean.

Since the normal distribution is symmetric, 50% of the measures are below the mean and 50% are above the mean.

Of those 50% which are below the mean, 99.7% of those are within 3 standard deviations of the mean, that is, greater than 22.

58 = 49 + 9

So 58 is one standard deviation of the mean.

Of those which are above the mean, 68% are within 1 standard deviation of the mean, that is, lesser than 58.

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5 0
3 years ago
Find the measure or angle xyz
blagie [28]

Answer:

The answer to your question is: 85°

Step-by-step explanation:

Data

m∠ xyz = 2m∠x - 9

m∠w = 38°

Process

∠x = (∠xyz + 9) / 2

∠ y = 180 - ∠xyz

∠w + ∠y + ∠x = 180

                             

Substitution              38 + (180 - ∠xyz) +   (∠xyz + 9) / 2 = 180

                                 76 + 360 - 2∠xyz + ∠xyz + 9 = 360

Simplify

                                 85 - 2∠xyz + ∠xyz = 0

                                  85 = ∠ xyz

5 0
3 years ago
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Answer:

H 0 : μ = 3.1    H 1 : μ > 3.1

And this test is defined a a right tailed test since the symbol for the alternative hypothesis H1 is >.

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The p value is:

p_v = P(t_{74} >8.66) =3.65x10^{-13}

The significance level is: 0.1 or 10%

And since the p value is very low compared to the significance level we have enough evidence to:

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Step-by-step explanation:

For this case we want to test the claim that mean GPA of night students is larger than 3.1 at the .10 significance level. The claim needs to be on the alternative hypothesis so then we have the following system of hypothesis:

H 0 : μ = 3.1    H 1 : μ > 3.1

And this test is defined a a right tailed test since the symbol for the alternative hypothesis H1 is >.

We have the following info given:

\bar X = 3.13 , s =0.03 , n =75

The statistic to check the hypothesis is given by:

t = \frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}

Replacing the info given we got:

t = \frac{3.13-3.1}{\frac{0.03}{\sqrt{75}}}= 8.66

The degrees of freedom are given by:

df = n-1=75-1 =74

The p value since is a right tailed ted is given by:

p_v = P(t_{74} >8.66) =3.65x10^{-13}

The significance level is 0.1 or 10%

And since the p value is very low compared to the significance level we have enough evidence to:

Reject the null hypothesis

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