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kipiarov [429]
3 years ago
9

Estimate the sum or the product.

Mathematics
1 answer:
Rashid [163]3 years ago
6 0

Answer:

See below in bold.

Step-by-step explanation:

1.  1.52(98)

Estimate is 1.5 * 100 = 150.

2. 2.04(3.97):

2 * 4 = 8.

3.  4.42 + 5.91

=  4.4 + 6

= 10.4.

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If 16+4x is 10 more than 14, what is the value of 8x?
sattari [20]
16 + 4x = 10 + 14

16 + 4x = 24.

4x = 24 - 16

4x = 8

x = 8 ÷ 4

x = 2

8x = 2 × 8

8x = 16

Final answer = 16.
3 0
3 years ago
Need help. Also what graphing utility could I use?
Vladimir79 [104]

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sadly i haven’t learned this so i can’t help you with the answers themselves but a graphing utility you could use is a graphing calculator or a graphing calculator online i know one called desmos.com just not sure if it has the intersect feature so i recommend graphing calculator

3 0
3 years ago
Math help What are the solutions​
Westkost [7]

Answer:

x² - 4x + 12

Step-by-step explanation:

x=2+2.82843i

x=2−2.82843i

6 0
2 years ago
What's 76×32÷5+(19×92)-15×2? Need help ASAP​
arsen [322]

Answer:

2204.4

Step-by-step explanation:

Okay so we want to always use order of operations: PEMDAS

Sop first we do what is in the parenthesis. 19*92 is 1748.

Now, going from left to right, we do multiplication and division. 76*32 is 2432. Then we divide that by 5 to get 486.4. And the last bit is 15*2 which is 30.

Your equation now is 486.4+1748-30.

The final answer is 2204.4

6 0
3 years ago
Read 2 more answers
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
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