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3 years ago

On her way to visit her parents,Jennifer drives 265 miles in 5 hours.What is her average rate of speed in miles per hour ?

Mathematics

1 answer:

aleksandrvk [35]3 years ago

7 0

Answer: The average rate of speed in miles per hour that Jennifer was going is 53 miles per hour. 265/5=53

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Write an equation of the line that passes through the given points. (-1,5) and (2, - 7)

Vanyuwa [196]

Answer:

The equation of the straight line is 4x +y = 1

Step-by-step explanation:

Step(i):-

Given points are (-1,5) and ( 2,-7)

Slope of the line

$m =\frac{y_{2}-y_{1} }{x_{2} -x_{1} } =\frac{-7-5}{2-(-1)} = \frac{-12}{3} = -4$

slope of the line m = -4

Step(ii):-

The equation of the straight line passing through the point (-1,5) and having slope 'm' = -4

$y - y_{1} = m( x-x_{1} )$

y - 5 = -4 ( x-(-1))

y -5 = -4 x -4

4 x + y -5 +4=0

4x +y -1 =0

Final answer:-

The equation of the straight line is 4x +y = 1

6 0

2 years ago

10=12-x what would match this equation

Eddi Din [679]

Answer:

x=2

Step-by-step explanation:

12-10=2

6 0

3 years ago

Read 2 more answers

Exhibit 9-2 The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the cu

Diano4ka-milaya [45]

Answer:

At a .05 level of significance, it can be concluded that the mean of the population is significantly more than 3 minutes.

Step-by-step explanation:

We want to test to determine whether or not the mean waiting time of all customers is significantly more than 3 minutes.

At the null hypothesis, we test if the mean is of at most 3 minutes, that is:

$H_0: \mu \leq 3$

At the alternative hypothesis, we test if the mean is of more than 3 minutes, that is:

$H_1: \mu > 3$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

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

3 is tested at the null hypothesis:

This means that $\mu = 3$

The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in the sample to check out was 3.1 minutes. The population standard deviation is known to be 0.5 minute.

This means that $n = 100, X = 3.1, \sigma = 0.5$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{3.1 - 3}{\frac{0.5}{\sqrt{100}}}$

$z = 2$

P-value of the test and decision:

The p-value of the test is the probability of finding a sample mean above 3.1, which is 1 subtracted by the p-value of z = 2.

Looking at the z-table, z = 2 has a p-value of 0.9772.

1 - 0.9772 = 0.0228

The p-value of the test is of 0.0228 < 0.05, meaning that the is significant evidence to conclude that the mean of the population is significantly more than 3 minutes.

6 0

3 years ago

Geri spent Friday, Saturday and Sunday selling a total of 24 magazines orders for her school.The amounts on those 3 days were co

ryzh [129]

The sale on Friday = x = 6

The sale on Saturday = x+2 = 6 + 2 =8

The sale on Sunday = x+4 = 6+4 =10

Step-by-step explanation:

Let x be the sale of Friday

Then

x+2 will be the sale of Saturday

x+4 will be the sale of Sunday

According to the given statement, that the total sales were 24

$x+(x+2)+(x+4) = 24\\x+x+2+x+4 = 24\\3x+6 = 24\\Subtracting\ 6\ from\ both\ sides\\3x+6-6=24-6\\3x=18\\Dividing\ both\ sides\ by\ 3\\\frac{3x}{3} = \frac{18}{6}\\x=6$