Answer:
a. z = 2.00
Step-by-step explanation:
Hello!
The study variable is "Points per game of a high school team"
The hypothesis is that the average score per game is greater than before, so the parameter to test is the population mean (μ)
The hypothesis is:
H₀: μ ≤ 99
H₁: μ > 99
α: 0.01
There is no information about the variable distribution, I'll apply the Central Limit Theorem and approximate the sample mean (X[bar]) to normal since whether you use a Z or t-test, you need your variable to be at least approximately normal. Considering the sample size (n=36) I'd rather use a Z-test than a t-test.
The statistic value under the null hypothesis is:
Z= X[bar] - μ = 101 - 99 = 2
σ/√n 6/√36
I don't have σ, but since this is an approximation I can use the value of S instead.
I hope it helps!
Step-by-step explanation:
here's the answer to your question
Answer:
k ≥2
Step-by-step explanation:
k+ 1/3≥7/3
Subtract 1/3 from each side
k+ 1/3-1/3≥7/3-1/3
k ≥6/3
k ≥2
Answer:
m=88
Step-by-step explanation:
First we have to isolate the variable. In this case, it would be by multiplying the reciprocal of the fraction the both sides. M will be alone and the other side should come out to be 88. Assuming you know how to multiply with fractions, of course.
Answer:
Area of rectangle = 225/2 or 112.5
Step-by-step explanation:
Given,
Consider a rectangle ABCD.
Let AC be a diagonal of rectangle of length = 15
In triangle ABC.
Sin 45° =height/hypotenuse {SinФ = height / hypotenuse}
Here, hypotenuse = diagonal of rectangle ( i.e AC = 15)
And height is AB
Therefore, sin 45° = AB/AC
or sin 45° = AB / 15
or 1/√2 = AB /15
AB = 15/√2
Similarly we can find Base (i.e BC) using cosine.
Cos 45° = Base/Hypotenuse
Cos 45° = BC / AC
or 1/√2 = BC/15
BC = 15/√2
Hence we got length of rectangle , AB= 15/√2
And width of rectangle , BC = 15/√2
Therefore, area of rectangle = Length × Width
Area of rectangle = 15/√2 × 15/√2 = 225/2
Hence, area of rectangle = 225/2 = 112.5
