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dangina [55]
3 years ago
14

Jen can bike 12 miles in 40 minutes. How much time will she need to bike 30 miles if her speed remains the same? PLEASE HELP QUI

CK
Mathematics
2 answers:
murzikaleks [220]3 years ago
5 0

Answer: 100 minutes

Step-by-step explanation:

With the given information, we can use a proportion to figure out how long it would take for Jen to bike 30 miles.

\frac{12mi}{40min} =\frac{30mi}{x} \\

12x=30*40

12x=1200

x=100

Phoenix [80]3 years ago
4 0
Answer:
100

40 divided by 12 and then multiply it by 30.
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85.95  km

Step-by-step explanation:

Given that,

Alberto ran 45 kilometers on Monday.

On Tuesday, he runs 91% of the distance he ran on Monday.

We need to find how far did he run on Tuesday. It can be given by :

T=45+45\times \dfrac{91}{100}\\\\=85.95\ km

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Find the discriminant of the following quadratic equation then state the number of rational, irrational, and imaginary solutions
lianna [129]

Answer:

  • -6x² - 6 = -7x - 9
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<u>Discriminant:</u>

  • D = (-7)² - 4*6*(-3) = 49 + 72 = 121

<u>Since D > 0, there are 2 real solutions:</u>

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8 0
3 years ago
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The average undergraduate cost for tuition, fees, and room and board for two-year institutions last year was $13,252. The follow
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Answer:

The p-value of the test is 0.0041 < 0.05, which means that there is sufficient evidence at the 0.05 significance level to conclude that the mean cost has increased.

Step-by-step explanation:

The average undergraduate cost for tuition, fees, and room and board for two-year institutions last year was $13,252. Test if the mean cost has increased.

At the null hypothesis, we test if the mean cost is still the same, that is:

H_0: \mu = 13252

At the alternative hypothesis, we test if the mean cost has increased, that is:

H_1: \mu > 13252

The test statistic is:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question

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

In which X is the sample mean, \mu is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.

13252 is tested at the null hypothesis:

This means that \mu = 13252

The following year, a random sample of 20 two-year institutions had a mean of $15,560 and a standard deviation of $3500.

This means that n = 20, X = 15560, s = 3500

Value of the test statistic:

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

t = \frac{15560 - 13252}{\frac{3500}{\sqrt{20}}}

t = 2.95

P-value of the test and decision:

The p-value of the test is found using a t-score calculator, with a right-tailed test, with 20-1 = 19 degrees of freedom and t = 2.95. Thus, the p-value of the test is 0.0041.

The p-value of the test is 0.0041 < 0.05, which means that there is sufficient evidence at the 0.05 significance level to conclude that the mean cost has increased.

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3 years ago
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