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Usimov [2.4K]
2 years ago
9

Which equation has a solution of 8

Mathematics
1 answer:
Arada [10]2 years ago
6 0

Answer:

2

Step-by-step explanation:

because that's the answer

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a previous analysis of paper boxes showed that the standard deviation of the length is 9mm. What is the 98% confidence interval
il63 [147K]
We cannot calculate the confidence interval without the sample size.  However, for the second question, the sample size needed would be 49.

The formula we use is

n=(\frac{z_{\frac{\alpha}{2}}\times \sigma}{E})^2

To find the z-score:
Convert 98% to a decimal:  0.98
Subtract from 1:  1-0.98 = 0.02
Divide by 2:  0.02/2 = 0.01
Subtract from 1:  1-0.01 = 0.99

Using a z-table (http://www.z-table.com) we see that this value is closest to a z-score of 2.33.

Using the z-score, our standard deviation (9) and our margin of error (3), we have:
n=(\frac{2.33(9)}{3})^2=(6.99)^2=48.86\approx 49
6 0
2 years ago
[WILL MARK AS BRAINLIEST IF CORRECT]
liubo4ka [24]
A B D and E i believe
3 0
3 years ago
Read 2 more answers
I need help plzzzzzzzzzzzzjdbfhrrhebehdbzzzzzzzzzzzzzzzz
yulyashka [42]

Answer:A

Step-by-step explanation: im not too sure but i think 1-2/5=3/5 which would be red so

4 0
2 years ago
Read 2 more answers
What is the number of subsets of S= {1, 2, 3…10} that contain five element include 3 or 4 but not both?
Ivenika [448]

Answer:

140

Step-by-step explanation:

To construct a subset of S with said property, we have two choices, include 3 in the subset or include four in the subset. These events are mutually exclusive because 3 and 4 can not both be elements of the subset.

First, let's count the number of subsets that contain the element 3.

Any of such subsets has five elements, but since 3 is already an element, we only have to select four elements to complete it. The four elements must be different from 3 and 4 (3 cannot be selected twice and the condition does not allow to select 4), so there are eight elements to select from. The number of ways of doing this is {}_8C_4=70.

Now, let's count the number of subsets that contain the element 4.

4 is already an element thus we have to select other four elements . The four elements must be different from 3 and 4 (4 cannot be selected twice and the condition does not allow to select 3), so there are eight elements to select from, so this can be done in {}_8C_4=70 ways.

We conclude that there are 70+70=140 required subsets of S.

7 0
2 years ago
???????????????????????
aleksandrvk [35]
The answer is 10w.

40 divided by 4 is 10. The X and Y cancel out, leaving 4W.
6 0
3 years ago
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