Answer:
C. ± 2.326 years.
Step-by-step explanation:
We have the standard deviation for the sample. So we use the t-distribution to solve this question.
We have that to find our 
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of 
.
So it is z with a pvalue of 
, so ![z = 2.326/tex]Now, find the width of the interval[tex]W = z*\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=z%20%3D%202.326%2Ftex%5D%3C%2Fp%3E%3Cp%3E%3Cstrong%3ENow%3C%2Fstrong%3E%2C%20find%20the%20width%20of%20the%20interval%3C%2Fp%3E%3Cp%3E%5Btex%5DW%20%3D%20z%2A%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
In which 
is the standard deviation of the population and n is the size of the sample.
In this question:
So
The correct answer is:
C. ± 2.326 years.
Answer:
Hey Jeffrey i got to ur school...
lol
Step-by-step explanation:
C
Answer:
70 times
you didn't attach a picture so i can't help with the other part
Step-by-step explanation:
Answer:
For the top table:
[x] | 3.1 | 2.5 | 1.2 | 0.9 | 0.14 | 0.06 | 0.02 |
[y] | 15.5 | 12.5 | 6 | 4.5 | 0.7 | 0.3 | 0.1 |
For the bottom table:
k = 5
[x] | 3.1 | 2.5 | 1.2 | 0.9 | 0.14 | 0.06 | 0.02 |
[y] | 15.5 | 12.5 | 6 | 4.5 | 0.7 | 0.3 | 0.1 |
Answer:
The expressions which equivalent to 
are:
⇒ B
⇒ C
Step-by-step explanation:
Let us revise some rules of exponent
Now let us find the equivalent expressions of
A.
∵ 4 = 2 × 2
∴ 4 = 
∴ 
= 
- By using the second rule above multiply 2 and (n + 2)
∵ 2(n + 2) = 2n + 4
∴ = 
B.
∵ 4 = 2 × 2
∴ 4 = 2²
∴ = 2² × 
- By using the first rule rule add the exponents of 2
∵ 2 + n + 1 = n + 3
∴ =
C.
∵ 8 = 2 × 2 × 2
∴ 8 = 2³
∴ = 2³ × 
- By using the first rule rule add the exponents of 2
∵ 3 + n = n + 3
∴ =
D.
∵ 16 = 2 × 2 × 2 × 2
∴ 16 = 
∴ 
= ×
- By using the first rule rule add the exponents of 2
∵ 4 + n = n + 4
∴ = 
E.
is in its simplest form
The expressions which equivalent to are:
⇒ B
⇒ C
