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balandron [24]
3 years ago
14

What would be a good estimate of the mean for the data below? Please help I will give brainly!!!

Mathematics
1 answer:
Nastasia [14]3 years ago
6 0

Answer:

90

Step-by-step explanation:

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What is the answer.
Alenkasestr [34]

       √11 √5  =  √(11 * 5)  =  √55  =  approx  7.42 (rounded)
4 0
4 years ago
Terday. Priya successfully made 6 free throws. Today, she made 75% as
tatuchka [14]

Answer:

10

Step-by-step explanation:

Yesterday, Priya successfully made 6 free throws.

Today, she made 75% as many as yesterday.

We are asked to calculate the number of successful free throws that Priya made today.

So, the number will be 6 (1 + \frac{75}{100} ) = 10.5.

Since it is the number of successful free throws, so it will be 10 throws. ( Answer )  

4 0
3 years ago
In order to test for the significance of a regression model involving 3 independent variables and 47 observations, the numerator
matrenka [14]

Answer:

The correct option is  C

Step-by-step explanation:

From the question we are told that

     The number of independent variables is  n =  3

     The number of observation is  z =  47

     

Since n is are independent variables then their degree of freedom is  3

   The  denominator(i.e z) degrees of freedom is evaluated as

            Df(z) =  z - (n +1)

          Df(z) =  47 - (3 +1)

          Df(z) =  43

So for the numerator (n) the degree of freedom is  Df(n) = 3

So for the denominator(i.e z) the degree of freedom is  Df(z) = 43

4 0
3 years ago
100 Points!!!!!!!!!
artcher [175]

Answer:

its u=(-2,1)

Step-by-step explanation:

8 0
3 years ago
Rachel was an 85 average after four test. what does she need to earn on the next test if she wants to raise her average to exact
irakobra [83]

she took four tests with one more to take totals 5 tests

 88 x 5 = 440

 all her tests need to equal at least 440 to get an 88 average

since she has 85 average after 4, 85*4 = 340, so her 4 tests so far have equaled 340

440-340 = 100

 she will need to get a 100 on her 5th test to have an 88 average

4 0
3 years ago
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