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Leto [7]
3 years ago
10

A football team carried out a report to see the impact of stretching on preventing injury. Of the 32 footballers in the squad 25

stretch regularly. Of those who stretch, 3 got injured last year. There was a total of 8 injured players last year. The results can be presented in a frequency tree. What fraction of players are not stretching regularly?
Mathematics
1 answer:
Oxana [17]3 years ago
8 0

Answer:

1/16

Step-by-step explanation:

Total = 32 footballers

25 stretch regularly.

3 injure last year

now, 22 stretch regularly.

out of 32, 8 are injured. Therefore, 32-8=24 should stretch regularly but only

22 stretch regularly. Therefore, 24-22 =2 are not stretching regularly.

fraction of players are not stretching regularly = 2/32 =1/16

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-2 - (8)<br> I don’t know what is the answer can someone help me
MrMuchimi

Answer:

its 16

Step-by-step explanation:

because if you take off the parenthathesis you get -2-8 then you multiply it and since they are both negative you get a positive, so that means you get positive 16

4 0
3 years ago
Mary logged into work at 8:10 am and logged out at 3:45 pm. How many hours did she work? (Her employer rounds to the nearest qua
IRINA_888 [86]
First we count how many hours is inbetween 8 and 3  which is 7

then subtract 10 for the first time

6:50 then add 45 for the second time

7:35

round down so she worked 7 1/2 hours
3 0
3 years ago
Read 2 more answers
The government of Preon (a small island nation) was voted in at the last election with 58% of the votes. That was 2 years ago, a
mel-nik [20]

Answer:

a) z=\frac{0.684 -0.58}{\sqrt{\frac{0.58(1-0.58)}{114}}}=2.250  

b) p_v =2*P(z>2.250)=0.0244  

If we compare the p value and the significance level given we see that p_v we have enough evidence to reject the null hypothesis at 5% of significance.

Step-by-step explanation:

Data given and notation

n=114 represent the random sample taken

\hat p=0.684 estimated proportion of people that their approval rating might have changed

p_o=0.58 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Hypothesis

We need to conduct a hypothesis in order to test the claim that true proportion of people that their approval rating might have changed is 0.58 or no.:  

Null hypothesis:p=0.58  

Alternative hypothesis:p \neq 0.58  

Part a

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.684 -0.58}{\sqrt{\frac{0.58(1-0.58)}{114}}}=2.250  

Part b: Statistical decision  

The significance level provided \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

p_v =2*P(z>2.250)=0.0244  

If we compare the p value and the significance level given we see that p_v we have enough evidence to reject the null hypothesis at 5% of significance.

7 0
3 years ago
If 16 counters are 4/9, then what is the ONE?
notsponge [240]
36...

<span>divide your 16 by 4, so that you have 1/9. You then multiply the resulting number(4) by your denominator, 9. 4 times 9 = 36</span>
6 0
2 years ago
Read 2 more answers
ASAP MAJOR HELP NEEDED. I really need help. I am very confused on how to do this please help.
alexandr402 [8]

Because the 2 sides of the triangle are the same, they are both 21, the two bottom angles would also be the same.


The 3 inside angles of a triangle must equal 180 degrees.


The top angle is given as 38, so the remaining two angles must equal 180 - 38 = 142 degrees.


Since the 2 bottom angles would be the same, divide 142 by 2 to get x.

X = 142 / 2 = 71 degrees.


6 0
3 years ago
Read 2 more answers
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