Answer: Z is less than Zc ∴ 1.342 < 1.96
Therefore, Null hypothesis is not Rejected.
There is no sufficient evidence to claim that students turning in their test first score is significantly different from the mean.
Step-by-step explanation:
Given that;
U = 75
X = 78
standard deviation α = 10
sample size n = 20
population is normally distributed
PROBLEM is to test
H₀ : U = 75
H₁ : U ≠ 75
TEST STATISTIC
since we know the standard deviation
Z = (X - U) / ( α /√n)
Z = ( 78 - 75 ) / ( 10 / √20)
Z = 1.3416 ≈ 1.342
Now suppose we need to test at ∝ = 0.05 level of significance,
Then Rejection region for the two tailed test is Zc = 1.96
∴ Reject H₀ if Z > Zc
and we know that Z is less than Zc ∴ 1.342 < 1.96
Therefore, Null hypothesis is not Rejected.
There is no sufficient evidence to claim that students turning in their test first score is significantly different from the mean.
Answer:
A
Step-by-step explanation:
Answer:
x = 13
Step-by-step explanation:
Inscribed Angle = 1/2 Intercepted Arc
30 = 1/2 (5x-5)
Multiply each side by 2
2*30 = 2* 1/2 ( 5x-5)
60 = 5x-5
Add 5 to each side
60+5 = 5x-5+5
65 = 5x
Divide each side by 5
65/5 = 5x/5
13 = x
HI THERE!
i can help you :)
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let the number of raisins (in pounds) be R and the number of peanuts (in pounds be p.
since the bag is 30 pounds, so R+P =30
the total cost of the bag is 30 pounds* 4.05 per pound = $121.50 .
now we can use the per-pound costs to write an equation . . .
5R+3.8P=121.5
Now just solve that first equation. . (R+P=30) for R. R= 30- P
now use that to substitute in the other equation
5R+3.8P=121.5
5(30-p) + 3.8P=121.5
solve for p
150- 5P+ 3.8p = 121.5
150-1.2p=121.5
150-1.2p=121.5
-1.2p=-28.5
P=23.75
now use that to find r
R+p =30
r+23.75=30
R=6.25
(hope this helps sorry it took so long to type )
have a good day !
