I really don’t know gang but be strong you can do it
Given:
n = 20, sample size
xbar = 17.5, sample mean
s = 3.8, sample standard deiation
99% confidence interval
The degrees of freedom is
df = n-1 = 19
We do not know the population standard deviation, so we should determine t* that corresponds to df = 19.
From a one-tailed distribution, 99% CI means using a p-value of 0.005.
Obtain
t* = 2.8609.
The 99% confidence interval is
xbar +/- t*(s/√n)
t*(s/√n) = 2.8609*(3.8/√20) = 2.4309
The 99% confidence interval is
(17.5 - 2.4309, 17.5 + 2.4309) = (15.069, 19.931)
Answer: The 99% confidence interval is (15.07, 19.93)
It is because the 35.4% divided the 100% total is B. 0.354
Answer:
The amount of weight he has lost in 3 months is 1.215 stone or 1 stone 3 pounds
Step-by-step explanation:
From the above question, we are to calculate the amount of weight he has lost in three months
1 stone = 14 pounds
Let's convert all the weight lost to stone
At the start of the diet keirin weighted 14 stone 13 pounds
14 pounds = 1 stone
13 pounds = x
Cross Multiply
14x = 13
x = 13/14
x = 0.9285714286 stone
Approximately = 0.929 stone
Hence:
14 stone 13 pounds = 14 + 0.929 = 14.929 stone
Three months later he weighted 13 stone 10 pounds
14 pounds = 1 stone
10pounds = x
Cross Multiply
14x = 10
x = 10/14
x = 0.7142857143 stone
Approximately = 0.714 stone
Hence:
13 stone 10 pounds = 13 + 0.714 = 13.714 stone
The amount of weight he has lost is calculated as:
14.929 stone - 13.714 stone = 1.215 stone or 1 stone 3 pounds
If I’m not wrong each ticket was 8.75
Because 43.75 divided by 5= 8.75:)
