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2 years ago

8

Rita purchased a gallon of orange juice for $2.61, a loaf of wheat bread for $1.29, and a box of cereal for $3.49. How much mone

y did Rita spend in all .

2 answers:

pogonyaev2 years ago

5 0

She spent $7.39 in all.

Katen [24]2 years ago

4 0

7.39, you just add them all together

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Gabriella made 5 waffles for breakfast. She has 25 strawberries and 15 blueberries to put on top of the waffles. She will put a

aivan3 [116]

She will use 3 blueberries on each and 5 strawberries on each hope this helps <3

5 0

3 years ago

Janice works for a salary of $2,396 per month. She has federal income tax withheld at the rate of 15 percent, Social Security ta

MAVERICK [17]

Given that Janice monthly salary is $2,396. And she has deductions of federal income tax withheld at the rate 15%, social security tax at the rate of 6.2% and medicare tax at the rate of 1.45% and health insurance premium worth 95$ per month.

Let us calculate total deductions.

Federal income tax = 15% of 2396 = 0.15*2396 =$359.4

Social security tax = 6.2% of 2396=0.062*2396 =$148.552

Medicare tax = 1.45% of 2396= 0.0145*2396=$34.742

<u>health insurance premium =$95 </u>

Total deductions = $637.694

To calculate Janice net pay we have to subtract deductions from monthly salary that is 2396-637.694 = $1758.306

Hence net pay of Janice is $1758.306 per month.

4 0

2 years ago

Please help me I need to solve for x.

Sav [38]

Answer: x=7

12x-4=0.5(22x+6)
12x-4=11x+3
-11x -11x
x=7

Explanation: use the tangent-chord angle theorem.

4 0

3 years ago

ivanzaharov [21]

Answer:

The answer is 20m=8

7 0

2 years ago

Read 2 more answers

The standard deviation of math test scores at one high school is 16.1. A teacher claims that the standard deviation of the girls

Elanso [62]

Answer:

$\chi^2 =\frac{22-1}{259.21} 166.41 =13.482$

$p_v =P(\chi^2$

In order to find the p value we can use the following code in excel:

"=CHISQ.DIST(13.482,21,TRUE)"

If we compare the p value and the significance level provided we see that $p_v >\alpha$ so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population deviation is not significantly lower than 16.1 at 1% of significance.

Step-by-step explanation:

Notation and previous concepts

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

$n=22$ represent the sample size

$\alpha=0.01$ represent the confidence level

$s^2 =12.9^2=166.41$ represent the sample variance obtained

$\sigma^2_0 =16.1^2 =259.21$ represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population deviation is smaller than 16.1 (that's equivalent to check if the population variance is lower than 259.21:

Null Hypothesis: $\sigma^2 \geq 259.21$

Alternative hypothesis: $\sigma^2$

Calculate the statistic

For this test we can use the following statistic:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.

$\chi^2 =\frac{22-1}{259.21} 166.41 =13.482$

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case 21. And since is a left tailed test the p value would be given by:

$p_v =P(\chi^2$

In order to find the p value we can use the following code in excel:

"=CHISQ.DIST(13.482,21,TRUE)"

Conclusion

If we compare the p value and the significance level provided we see that $p_v >\alpha$ so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population deviation is not significantly lower than 16.1 at 1% of significance.

3 0

3 years ago