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3 years ago

Question

Mathematics

1 answer:

enot [183]3 years ago

7 0

Answer:

The percent increase in pay is 50%.

Step-by-step explanation:

We need to find the percent increase from $24 to $36.

To solve this, we first must ask ourselves the question

36 is what percent of 24?

Y = P% of 24

36 = X · 24

To get X alone, we divide 36 by 24. That's 1.50. So now we know 36 is 150% of 24. Knowing 24 is 100%, we can find the percent increase. Well, how do we get from 100% (24) to 150% (36)? By increasing by 50%. Therefore, the percent increase in pay is 50%.

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Please, please help with this question!! I don't understand it!

Kazeer [188]

Answer:

Step-by-step explanation:

Find the area of each shape, then multiply by 20, then round to nearest hundred.

Half circle:

=πr²/2

=50π

=157.08

Large square:

=L•w

=20(22)

=440

Small square:

=L(w)

=12(3)

=36

Triangle:

=b(h)/2

=8(3)/2

=12

Btu req:

=(157.08+440+36+12)(20)

=(645.08)(20)

=12901.6

=12900 BTU

3 0

3 years ago

Using place value strategy to divide<br>3,600÷9?

Crazy boy [7]

Answer:

The answer is 400 :)

Step-by-step explanation:

hope that helped a little

5 0

3 years ago

Enter the ratio as a fraction in lowest terms 14 minutes to 18 minutes

Mademuasel [1]

Answer:

Part A is 7/16 of the whole
Part B is 9/16 of the whole

Solution:
Assuming there are no other parts,
the Whole = A + B is the denominator:
Whole = 14 + 18 = 32

Part A = 14 and Part B = 18 are numerators for each fraction.

The fractions are then:
14/32 and 18/32

Meaning:
Part A is 14/32 of the whole
Part B is 18/32 of the whole

Reducing the fractions, it is also true that:
Part A = 7/16 of the whole
Part B = 9/16 of the whole

7 0

3 years ago

Read 2 more answers

Luda [366]

Answer:

a) H0: There is no association between level of education and TV station preference (Independence)

H1: There is association between level of education and TV station preference (No independence)

b) $\chi^2 = \frac{(15-10)^2}{10}+\frac{(15-20)^2}{20}+\frac{(10-10)^2}{10}+\frac{(5-10)^2}{10}+\frac{(25-10)^2}{10}+\frac{(10-20)^2}{20} =33.75$

c) $\chi^2_{crit}=5.991$

d) Since the p value is lower than the significance level we enough evidence to reject the null hypothesis at 5% of significance, and we can conclude that we have dependence between the two variables analyzed.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

High school Some College Bachelor or higher Total

Public Broadcasting 15 15 10 40

Commercial stations 5 25 10 40

Total 20 40 20 80

We need to conduct a chi square test in order to check the following hypothesis:

Part a

H0: There is no association between level of education and TV station preference (Independence)

H1: There is association between level of education and TV station preference (No independence)

The level os significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

Part b

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total col * total row}{grand total}$

And the calculations are given by:

$E_{1} =\frac{20*40}{80}=10$

$E_{2} =\frac{40*40}{80}=20$

$E_{3} =\frac{20*40}{80}=10$

$E_{4} =\frac{20*40}{80}=10$

$E_{5} =\frac{40*40}{80}=20$

$E_{6} =\frac{20*40}{80}=10$

And the expected values are given by:

High school Some College Bachelor or higher Total

Public Broadcasting 10 20 10 40

Commercial stations 10 10 20 40

Total 20 30 30 80

Part b

And now we can calculate the statistic:

$\chi^2 = \frac{(15-10)^2}{10}+\frac{(15-20)^2}{20}+\frac{(10-10)^2}{10}+\frac{(5-10)^2}{10}+\frac{(25-10)^2}{10}+\frac{(10-20)^2}{20} =33.75$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(rows-1)(cols-1)=(2-1)(3-1)=2$

Part c

In order to find the critical value we need to look on the right tail of the chi square distribution with 2 degrees of freedom a value that accumulates 0.05 of the area. And this value is $\chi^2_{crit}=5.991$

Part d

And we can calculate the p value given by:

$p_v = P(\chi^2_{3} >33.75)=2.23x10^{-7}$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(33.75,2,TRUE)"

Since the p value is lower than the significance level we enough evidence to reject the null hypothesis at 5% of significance, and we can conclude that we have dependence between the two variables analyzed.

7 0

4 years ago

101001 base 2-11100 base 2=

igor_vitrenko [27]

D-41
My friend . Have a good day

6 0

3 years ago