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Create a system of equations that can be used to determine the rental cost of each truck for the 1-day move. Be sure to specify

bogdanovich [222]

The equation are y = 0.59x + 29.95 for company A and y = 0.79x + 19.95 for company B.

<h3>Linear equation</h3>

Linear equation is in the form:

y = mx + b

where y,x are variables, m is the rate of change and b is the initial value of y.

Let y represent the total cost of rental for each truck in x hours.

From the table, for company A:

y = 0.59x + 29.95

For company B:

y = 0.79x + 19.95

The equation are y = 0.59x + 29.95 for company A and y = 0.79x + 19.95 for company B.

Find out more on linear equation at: brainly.com/question/14323743

7 0

1 year ago

solve the linear equation 29 - 3x = 5x + 5 state the property that justifies your step and why you chose it

Debora [2.8K]

Answer:

x=3

Step-by-step explanation:

29-3x=5x+5

29=5x+5+3x

29=8x+5

29-5=8x

24÷8=x

3=x

6 0

2 years ago

Simplify the expression by combining like terms<br>5x + 5y - 3x

Tcecarenko [31]

Answer:

2x+5y

Step-by-step explanation:

5x + 5y - 3x

=2x+5y

8 0

2 years ago

Mathematics and fractions 1/2 and 6/12

77julia77 [94]

Answer:

Equivalent fractions

Step-by-step explanation:

7 0

3 years ago

A student at a four-year college claims that mean enrollment at four-year colleges is higher than at two-year colleges in the Un

WINSTONCH [101]

Answer:

Part 1: The statistic

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=35+35-2=68$

Replacing we got

$t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31$

Part 2: P value

Since is a right tailed test the p value would be:

$p_v =P(t_{68}>4.31)=0.000022 \approx 0.00002$

Comparing the p value we see that is lower compared to the significance level of 0.01 so then we can reject the null hypothesis and we can conclude that the mean for the four year college is significantly higher than the mean for the two year college and then the claim makes sense

Step-by-step explanation:

Data given

$\bar X_{1}=5135$ represent the mean for four year college

$\bar X_{2}=4436$ represent the mean for two year college

$s_{1}=783$ represent the sample standard deviation for four year college

$s_{2}=553$ represent the sample standard deviation two year college

$n_{1}=35$ sample size for the group four year college

$n_{2}=35$ sample size for the group two year college

$\alpha=0.01$ Significance level provided

t would represent the statistic (variable of interest)

System of hypothesis

We need to conduct a hypothesis in order to check if the mean enrollment at four-year colleges is higher than at two-year colleges in the United States , the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2}\leq 0$

Alternative hypothesis: $\mu_{1} - \mu_{2}> 0$

We can assume that the normal distribution is assumed since we have a large sample size for each case n>30. So then the sample mean can be assumed as normally distributed.

Part 1: The statistic

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=35+35-2=68$

Replacing we got

$t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31$

Part 2: P value

Since is a right tailed test the p value would be:

$p_v =P(t_{68}>4.31)=0.000022$

Comparing the p value we see that is lower compared to the significance level of 0.01 so then we can reject the null hypothesis and we can conclude that the mean for the four year college is significantly higher than the mean for the two year college and then the claim makes sense

6 0

2 years ago