Can anyone please help me it’s only one question please help me

Ms. Grant’s car gets between 20 and 22 miles per gallon, inclusive. The gasoline she uses costs between $4.20 and $4.50 per gall

maria [59]

Answer:

You need to be more clearer with the money amount not like 4.20 - 4.50

Step-by-step explanation:

7 0

3 years ago

-0.5,1.25,-1/3,0.5,-5/3 ordered from least to greatest

vampirchik [111]

Answer:

Step-by-step explanation:

-5/3,-0.5,-1/3,0.5,1.25

8 0

3 years ago

Multiply using FOIL or Table method (x + 4) (x + 5)

mylen [45]

Answer:

${x}^{2} + 9x + 20$

Step-by-step explanation:

1) Use FOIL method: (a + b) (c + d) = ac + ad + bc + bd.

${x}^{2} + 5x + 4x + 20$

2) Collect like terms.

${x}^{2} + (5x + 4x) + 20$

3) Simplify.

${x}^{2} + 9x + 20$

Therefor, the answer is x² + 9x + 20.

5 0

3 years ago

PLEASE ANSWER ASAP

Vlada [557]

Answer:

c

Step-by-step explanation:

For an inverse function you take f(x) and turn that to y then you switch x and y. So you should start off with y= 5x -8. so then you switch x and y so x= 5y - 8. So then you solve. You subtract 8 from both sides and divide by 5 and you get F(x) inverse = x + 8/5

8 0

3 years ago

The degenerative disease osteoarthritis most frequently affects weight-bearing joints such as the knee. Two random samples from

Vinvika [58]

Answer:

$t=\frac{(810-770)-0}{\sqrt{\frac{67^2}{13}+\frac{56^2}{19}}}}=1.771$

$df=n_1 +n_2 -2=13+19-2=30$

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

$p_v =P(t_{30}>1.771)=0.0434$

Comparing the p value with the significance level $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for group 1 is significantly higher than the mean for the group 2

Step-by-step explanation:

Data given

$\bar X_{1}=810$ represent the mean for sample 1

$\bar X_{2}=770$ represent the mean for sample 2

$s_{1}=67$ represent the sample standard deviation for 1

$s_{2}=56$ represent the sample standard deviation for 2

$n_{1}=13$ sample size for the group 2

$n_{2}=19$ sample size for the group 2

$\alpha=0.05$ Significance level provided

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for category 1 is higher than the mean for category 2, the system of hypothesis would be:

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

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

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:

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

And the degrees of freedom are given by $df=n_1 +n_2 -2=13+19-2=30$

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

With the info given we can replace in formula (1) like this:

$t=\frac{(810-770)-0}{\sqrt{\frac{67^2}{13}+\frac{56^2}{19}}}}=1.771$

P value

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

$p_v =P(t_{30}>1.771)=0.0434$

Comparing the p value with the significance level $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for group 1 is significantly higher than the mean for the group 2

4 0

3 years ago