Answer:
Step-by-step explanation:
we know that
by the sum of interior angles of a triangle
so we have
we can simplify
and now we find the other angles
Artificial turf on athletic fields was first introduced in the 1960’s. It’s safety has been controversial since then. One issue
that has been investigated is wether injuries of football players tend to be more serious on artificial turf instead of grass. A study followed 24 NCAA division 1A college teams over 3 seasons.
In total, there were 1,050 injuries that occurred on field turf. Of the field turf injuries, 83.3333% were minor, and 10.857% were substantial. 77.972% of grass injuries were minor. 4.26% of injuries occurred on grass and were severe. How many injuries were severe and occurred on field turf? How many injuries were severe?
In total, there were 1,050 injuries that occurred on field turf. Of the field turf injuries, 83.3333% were minor, and 10.857% were substantial. 77.972% of grass injuries were minor. 4.26% of injuries occurred on grass and were severe. How many injuries were severe and occurred on field turf? How many injuries were severe?
2 answers:
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Let X be the weekly incomes of a large group of executives. The weekly incomes of a large group of executives follows Normal distribution with mean $2000 and standard deviation $100.
μ =2000, σ =100
We have to find z score for income $2100 i.e x=2100
Z =
=
Z = 100/100
Z = 1
The z score for income $2100 is 1
Answer:
At 5% significance level, larger proportion of military personnel students protect their profiles on social-networking sites than young adults with profiles on social-networking sites.
Step-by-step explanation:
let p be the proportion of military personnel students who restrict access to their profiles. Then null and alternative hypotheses are:
: p=0.67 (67%)
: p>0.67
We need to calculate z-statistic of sample proportion:
z= where
- p(s) is the sample proportion of military students who restrict access to their profiles (
=0.78)
- p is the proportion assumed under null hypothesis. (0.67)
- N is the sample size (100)
Then z= ≈ 2.34
The corresponding p-value is 0.0096. Since 0.0096<0.05 (significance level) we can reject the null hypothesis and conclude at 5% significance level that larger proportion of military personnel students protect their profiles on social-networking sites than young adults with profiles on social-networking sites.