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alukav5142 [94]
3 years ago
13

What is the product of -1 3/8 and 2 4/5

Mathematics
1 answer:
DanielleElmas [232]3 years ago
7 0
All you have to do is change the fractions to improper fractions and multiply it acrossed, then simplify.

-3 17/2
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Sally is having a problem with her puppy leaving the yard so she decides to build a new fence. The length of the yard is 10 feet
Cerrena [4.2K]

The length of the yard is 25ft.

In order to find this we first need to set up variables for the length and the width. Since we don't know anything about the width, we'll set it as x. Then, we know the length is equal to ten more than three times the width. Since the width is x, we can write the length as 3x + 10. Now we can use the formula for the perimeter of a rectangle to solve for x.

2l + 2w = P

2(3x + 10) + 2(x) = 60

6x + 20 + 2x = 60

8x + 20 = 60

8x = 40

x = 5

Now that we have a value for x, we can plug into the length equation and find the length.

3x + 10

3(5) + 10

15 + 10

25

8 0
3 years ago
A personnel manager is concerned about absenteeism. She decides to sample employee records to determine if absenteeism is distri
nlexa [21]

Answer:

df=categor-1=6-1=5

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got \chi^2_{critc}= 11.0705

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

Step-by-step explanation:

Previous concepts

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".

Solution to the problem

For this case we want to test:

H0: Absenteeism is distributed evenly throughout the week

H1: Absenteeism is NOT distributed evenly throughout the week

We have the following data:

Monday  Tuesday  Wednesday Thursday Friday Saturday    Total

 12             9                 11                 10           9            9              60

The level of 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}

The table given represent the observed values, we just need to calculate the expected values with the following formula E_i = \frac{60}{6}= 10 and the expected value is the same for all the days since that's what we want to test.

now we can calculate the statistic:

\chi^2 = \frac{(12-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(11-10)^2}{10}+\frac{(10-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(9-10)^2}{10}=0.8

Now we can calculate the degrees of freedom (We know that we have 6 categories since we have information for 6 different days) for the statistic given by:

df=categor-1=6-1=5

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got \chi^2_{critc}= 11.0705

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

4 0
3 years ago
If Mr. John plans to replace the carpet in his living room AND bedroom with carpet that cost $2.00 per square foot, how much wil
adoni [48]

Answer:

1496

Step-by-step explanation:

A = lw

22 x 16 = 352 ( bedroom)

22 x 18 = 396 (living)

352 + 396 = 748

748 x 2 = 1496

hope this helps

pls mark brainliest

5 0
2 years ago
The sum of w and fourteen is five. What is the value of w?
dmitriy555 [2]

Answer:

-9

Step-by-step explanation:

5 0
3 years ago
Can some one pls help me
DedPeter [7]

Answer:

look at the picture i have sent

6 0
3 years ago
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