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Anni [7]
3 years ago
7

Factors of -96 that add up to -10

Mathematics
1 answer:
hjlf3 years ago
5 0
Positive six and negative sixteen
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A number that is prime, between 60-100,1s place is 2 less then 10s place.
Crank
The answer to that would be 97.

Hope this helps! :)
7 0
3 years ago
student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te
alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = \frac{1}{4}.

Thus, the probability that the student does not receive version A = 1-\frac{1}{4} = \frac{3}{4}.

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}

= (\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5

= (\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]

= (\frac{3}{4^4})[1+\frac{1}{16}]

= (\frac{3}{256})[\frac{17}{16}]

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0
3 years ago
Factories fully 6m^2+8mp
Bond [772]
Can you put a picture of the passage
8 0
3 years ago
FREE BRAINLIEST & 50 points bc im nice <3
Ostrovityanka [42]
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6 0
2 years ago
Read 2 more answers
Help Dana find the sum 346 421 152
lara [203]

Dana wants to find the sum

  3 4 6

  4 2 1

+1 5 2

________

We will not regroup this ones place because the sum of the digits in the ones place is not up to. It's just 6+1+2 = 9.

Therefore, 13a) No.

There is no regrouped tens to add, so we would not able to regroup the ones place.

Therefore, 13b) No.

We regroup the tens place because the sum of the digits in the ones place is upto to 10.

And 13c) Yes.

We add the regrouped hundred as we regrouped the ones place.

Therefore,

13 D) Yes.






7 0
3 years ago
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