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NeX [460]
3 years ago
11

Jacob and Jordan are training for track season. Jacob did 39 sit-ups in 30 seconds. Jordan did 59 sit-ups in 50 seconds. who wil

l likely do more sit-ups in 2 1/2 minutes
Mathematics
2 answers:
scoray [572]3 years ago
7 0
Jordan bc it a bigger number
Naddik [55]3 years ago
3 0
Jordan cuz it is bigger

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Solve for x in the equation x/2 = 17 user: solve for x in the equation x/2 = 17 user: apply the rules for order of operations to
Finger [1]
Well to solve for x you simply need to multiply both sides of the equation to isolate x, the answer will be x=34. the second question requires bedmas (brackets, exponents, division and multiplication, addition and subtraction)
It will be 224 no matter if you multiply or divide first 
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3 years ago
What are the first 4 prime numbers
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2, 3, 5 and 7 the the first 4 prime numbers
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Nookie1986 [14]

Answer:

B

Step-by-step explanation:

7pi2

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3 years ago
Sam, Maya, and John playing a match, Sam has Half of what Maya has, and maya has fewer than John with three points. Their points
Ksivusya [100]

so john is x and if Maya has the same as john but -3 then

x+x-3

but sam is still missing and she has half of Maya which is 0.5x - 1.5

so the equation is

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and then you add 3 and 1.5 to 68 which is

2.5x=72.5

and you divide by 2.5 which is 29

John has 29 points


7 0
3 years ago
If a test of H subscript 0 colon space mu subscript D equals 0 space v s. space H subscript a colon space space mu subscript D g
lara [203]

Answer:

p_v =P(t_{(n-1)}>t_{calculated}) =0.0601

The p value on this case is given  by the problem.

If we compare the p value with a significance level assumed \alpha=0.05, we see that p_v > \alpha and we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before is less or equal than 0.

Step-by-step explanation:

A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations (This problem) we can use it.  

Let put some notation  

x=test value before , y = test value after

The system of hypothesis for this case are:

Null hypothesis: \mu_y- \mu_x \leq 0

Alternative hypothesis: \mu_y -\mu_x >0

The first step is calculate the difference d_i=y_i-x_i

The second step is calculate the mean difference  

\bar d= \frac{\sum_{i=1}^n d_i}{n}

The third step would be calculate the standard deviation for the differences, and we got:

s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}

The 4 step is calculate the statistic given by :

t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=t_{calculated}

The next step is calculate the degrees of freedom given by:

df=n-1

Now we can calculate the p value, since we have a right tailed test the p value is given by:

p_v =P(t_{(n-1)}>t_{calculated}) =0.0601

The p value on this case is given  by the problem.

If we compare the p value with a significance level assumed \alpha=0.05, we see that p_v > \alpha and we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before is less or equal than 0.

4 0
3 years ago
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