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Oksanka [162]
3 years ago
6

What does a probability distribution​ indicate? choose the correct answer below.

Mathematics
1 answer:
Sedaia [141]3 years ago
6 0
The correct answer is C) both a and b.

A probability distribution shows all of the possible values of the experiment and the probabilities that go with them.
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How do you model 604 divided 4 and 796 divided 6 and 449 divided 5
cupoosta [38]
604 divided by 4 is 151. 796 divided by 6 is 132.6666666... 449 divided by 5 is 89.8
5 0
3 years ago
W.8
lesya692 [45]

Answer:

.

Step-by-step explanation:

7 0
2 years ago
24+[5x12x5)-2] 21<br> Someone help me
artcher [175]
6,282? Because 5x12x5= 300 -2

298x21 = 6,258 + 24
=6,282
8 0
3 years ago
Find the length of the unknown side. Round your answer to the nearest tenth.
Paraphin [41]

Answer:

<em>The correct option is:  11.5</em>

Step-by-step explanation:

Suppose, the length of the unknown side is  x

So, here the length of hypotenuse is 14 and the lengths of other two sides are 8 and x.

As the given triangle is right angle triangle, so <u>using Pythagorean theorem</u>, we will get.........

8^2+x^2= 14^2\\ \\ 64+x^2=196\\ \\ x^2=196-64=132\\ \\ x=\sqrt{132}=11.489...\approx 11.5

<em>(Rounded to the nearest tenth)</em>

So, the length of the unknown side is 11.5

6 0
3 years ago
Read 2 more answers
A researcher wants to determine whether the number of minutes adults spend online per day is related to gender. A random sample
suter [353]

Answer:

The expected frequency for the cell E2,2 is = 33

Step-by-step explanation:

The given data is

Gender               | 0-30| 30-60| 60-90| 90+       Totals

Male                    | 23    |    35   |   76    |  46        180

<u> Female               |   31   |    42   |   46    |  16         135    </u>

<u>Totals                    54         77        122    62          315  </u>

<u />

The expected frequency for the cell E2,2 is :

Expected for the (30-60 box for females)   =  Total of (30-6)/ (total )* females

                                                                      = ( 77/315)135= 33

Here p= 77/315 and n= 135

therefor X= pn = 33

χ²=  (33-42)²/33= 2.455  ( for the single value of E2,2=33)

Expected for the (30-60 box for males)   =  Total of (30-6)/ (total )* males

                                                                      = ( 77/315)180= 44

χ²=  (44-42)²/44= 0.09

The critical region is  χ² (0.05) 3 ≥  χ²= 11.34

Let the null and alternate hypothesis be

H0:the number of minutes spent online per day is not related to gender

against

Ha: the number of minutes spent online per day is related to gender

The single value of χ² for E2,2 = 2.45 is less than the critical value of 11.34.  

6 0
2 years ago
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