Since it is observed that I z I = 3.464 > 1.96 it is concluded that the null hypothesis is rejected.
Therefore we reject John's claim.
Here:
we can use 1 sample proportion test to check the claim of John
H₀ : p₀ = 0.4
H₁ : p₀ ≠ 0.4
Test statistic is
z = p - p₀/√p₀(1-p₀)n
Let X be the random variable denoting the number of heads.
Now p = X/n
= 8/50
= 0.16
p₀ = 0.4
1-p₀ = 1-0.4
= 0.6
z = -3.464
since it is observed that I z I = 3.464 > 1.96 it is concluded that the null hypothesis is rejected.
Therefore we reject John's claim.
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Remark
There is no short way to do this problem and no obvious way to get the answer other that to solve each part.
Solve
A
Multiply by 2
x + 1.6 = 2(x + 0.1) Remove the brackets
x + 1.6 = 2x + 0.1*2
x + 1.6 = 2x + 0.2 Subtract x from both sides
1.6 = x + 0.2 Subtract 0.2 from both sides
1.6 - 0.2 = x
1.4 = x
Circle A
B
Subtract 2x from both sides.
3x - 2x = 1.4
Circle B
C
Remove the brackets.
4x + 6 = 2x - 6 Add 6 to both sides
4x + 12 = 2x Subtract 4x from both sides.
12 = -2x Divide by - 2
12/-2 = x
x = - 6 Don't circle C
D
I'm going to be very scant in my solution of this. You can fill in the steps.
3x = 4.2
x = 4.2/3
x = 1.4
Circle D
A function that would represent profit based on the number of cups of lemonade is Profit = 1.5n - 14
<u>Solution:</u>
Given, Some kids are selling lemonade for $1.50 per cup at a high school baseball game.
They spent $14 on all of the items needed for the lemonade stand (cups, lemonade, table oth, sign, etc)
We have to create a function that would represent profit based on the number of cups of lemonade
Now, let the number of cups sold be "n"
Then , we know that,<em> profit = selling price – cost price </em>
Profit = number of cups sold x price per cup – cost price
Profit = n x $ 1.5 – $ 14
Profit = 1.5n – 14
Hence, the function is Profit = 1.5n - 14
Answer:
A. Men had less distress from nausea on average than women but we can not determine if this is a significant difference.
Step-by-step explanation:
Working based on the information given, the mean values of each group with with men having an average score of 1.02 and women have an average of 1.79 this reveals that distressing nausea on average is higher in women than in men . However, to test if there is a significant difference would be challenging as the information given isn't enough to make proceed with the test as the standard deviations of the two groups aren't given and no accompanying sample data is given.