Answer:
negative s will equal -16
The equation must equal 84, so you can eliminate B and D.
Chin charges a rate for 2 hours, then charges a reduced rate for 4 hours. There are no discounts present in his rate, so you can eliminate A.
The equation for Chin's charges can be found by the equation C. 2x + 4y = 84.
Answer: 60
Step-by-step explanation:
V = w h l = 3 x 5 x 4 = 60
Hope this helped, if not i can do the problem again :)!
Answer:
The requirements for the hypothesis test does satisfied the method for testing the claim that from two population proportions the rate of polio is less for children given the salk vaccine.
Step-by-step explanation:
The percentage of children in the treatment group was:
(201229/401974)*100 = 49.9%
The percentage of children given placebo was:
(200745/401974)*100 = 50.1%
The percentage of children that developed polio in the treatment group:
(33/200745)*100 = 0.0164%
The percentage of children that developed polio in the placebo group:
(115/201229)*100 = 0.0571%
The percentage difference between the two group:
((0.0571-0.0164)/0.0571) = 61.62%
Therefore:
The amount of children used for each group was almost divided into half of the total amount of children. The test revealed although very small percentages of the both group developed polio, 68.62% more children given placebo than the children that was given the salk vaccine. Therefore, the study shows that the rate of polio is less for children given the salk vaccine and the the hypthesis test is satisfied.
Ok, so remember that the derivitive of the position function is the velocty function and the derivitive of the velocity function is the accceleration function
x(t) is the positon function
so just take the derivitive of 3t/π +cos(t) twice
first derivitive is 3/π-sin(t)
2nd derivitive is -cos(t)
a(t)=-cos(t)
on the interval [π/2,5π/2) where does -cos(t)=1? or where does cos(t)=-1?
at t=π
so now plug that in for t in the position function to find the position at time t=π
x(π)=3(π)/π+cos(π)
x(π)=3-1
x(π)=2
so the position is 2
ok, that graph is the first derivitive of f(x)
the function f(x) is increaseing when the slope is positive
it is concave up when the 2nd derivitive of f(x) is positive
we are given f'(x), the derivitive of f(x)
we want to find where it is increasing AND where it is concave down
it is increasing when the derivitive is positive, so just find where the graph is positive (that's about from -2 to 4)
it is concave down when the second derivitive (aka derivitive of the first derivitive aka slope of the first derivitive) is negative
where is the slope negative?
from about x=0 to x=2
and that's in our range of being increasing
so the interval is (0,2)
