All categories

3 years ago

Describe a process used to draw simple random sample of size 10 from a group of 87 people.

Mathematics

1 answer:

Alina [70]3 years ago

5 0

to pick randomly uou could put peoples name in a hat and pick the amount needed

You might be interested in

This high school stadium contains a track that surrounds a soccer field. The soccer field is 100 yards long and 70 yards wide

saul85 [17]

Answer:

I have the same question, I'll try to figure it out <3

Step-by-step explanation:

3 0

3 years ago

Malik invests money in an account paying simple interest. He invests $80 and no money is added or removed from the investment. A

xeze [42]

Answer:

Step-by-step explanation:

The interest earned on the investment is the difference between the end-of-year value and the initial value:

$83.20 -80 = $3.20 . . . . simple interest per year

The interest rate is the ratio of this to the initial investment:

$3.20/$80.00 = 0.04 = 4% . . . . . simple interest rate per year

4 0

2 years ago

The number e is the base of the?

Vikki [24]

Answer:

Step-by-step explanation:

5 0

3 years ago

Solve 8[7 - (6x -6)] + 6x =0.

Rus_ich [418]

Solve for x:8 (13 - 6 x) + 6 x = 0
8 (13 - 6 x) = 104 - 48 x:
6 x + 104 - 48 x = 0
6 x - 48 x = -42 x:
-42 x + 104 = 0
Subtract 104 from both sides:
(104 - 104) - 42 x = -104
104 - 104 = 0:
-42 x = -104
Divide both sides of -42 x = -104 by -42:
(-42 x)/(-42) = -104/(-42)
(-42)/(-42) = 1:
x = -104/(-42)
The gcd of 104 and -42 is 2, so -104/(-42) = -(2×52)/(2 (-21)) = 2/2×(-52)/(-21) = (-52)/(-21):
x = (-52)/-21
Multiply numerator and denominator of (-52)/(-21) by -1:
Answer: x = 52/21

8 0

4 years ago

An engineer is comparing voltages for two types of batteries (K and Q) using a sample of 37 type K batteries and a sample of 58

inna [77]

Answer:

Step-by-step explanation:

Hello!

You want to test two samples of batteries to see is the mean voltage of these battery types are different.

Sample 1 (type K)

n₁= 37

sample mean x₁[bar]= 8.54

standard deviation S₁= 0.225

Sample 2 (Type Q)

n₂= 58

sample mean x₂[bar]= 8.69

standard deviation S₂= 0.725

1. The test hypothesis are:

H₀: μ₁ = μ₂

H₁: μ₁ ≠ μ₂

2. I'll apply the Central Limit Theorem and approximate the distribution of the sample means to normal so that I can use an approximate Z statistic for this test.

Z: (x₁[bar] - x₂[bar]) - (μ₁ - μ₂) ≈ N(0;1)

√ (S₁²/n₁) + (S₂²/n₂)

$Z_{H0}$ = (8.54 - 8.69) / [√ (0.225²/37) + (0.725²/58)]