Last week’s and this week’s low temperatures are shown in the table below. Low Temperatures for 5 Days This Week and Last Week L
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Answer:
The margin of error for the 95% confidence interval for the mean score of all such subjects is of 8.45.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 27 - 1 = 26
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 26 degrees of freedom(y-axis) and a confidence level of . So we have T = 2.0518
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
In this question:
. So
The margin of error for the 95% confidence interval for the mean score of all such subjects is of 8.45.
Vertex form is
y=a(x-h)^2+k
vertex is (h,k)
axis of symmetry is x=4, therfor h=4
y=a(x-4)^2+k
we have some points
(3,-2) and (6,-26)
input and solve for a and k
(3,-2)
-2=a(3-4)^2+k
-2=a(-1)^2+k
-2=a(1)+k
-2=a+k
(6,-26)
-26=a(6-4)^2+k
-26=a(2)^2+k
-26=a(4)+k
-26=4a+k
we have
-2=a+k
-26=4a+k
multiply first equation by -1 and add to second
2=-a-k
<u>-26=4a+k +</u>
-24=3a+0k
-24=3a
divide both sides by 3
-8=a
-2=a+k
-2=-8+k
add 8 to both sides
6=k
the equation is
y=a(x-h)^2+k
vertex is (h,k)
axis of symmetry is x=4, therfor h=4
y=a(x-4)^2+k
we have some points
(3,-2) and (6,-26)
input and solve for a and k
(3,-2)
-2=a(3-4)^2+k
-2=a(-1)^2+k
-2=a(1)+k
-2=a+k
(6,-26)
-26=a(6-4)^2+k
-26=a(2)^2+k
-26=a(4)+k
-26=4a+k
we have
-2=a+k
-26=4a+k
multiply first equation by -1 and add to second
2=-a-k
<u>-26=4a+k +</u>
-24=3a+0k
-24=3a
divide both sides by 3
-8=a
-2=a+k
-2=-8+k
add 8 to both sides
6=k
the equation is