The length of the other side of the yard is 16 feet
work: 160 divided by 10
hope it helps
Answer:
The 99% confidence interval for the population mean reduction in anxiety was (1.2, 8.6).
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 27 - 1 = 26
99% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 26 degrees of freedom(y-axis) and a confidence level of
. So we have T = 2.7787.
The margin of error is:

In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 4.9 - 3.7 = 1.2.
The upper end of the interval is the sample mean added to M. So it is 4.9 + 3.7 = 8.6.
The 99% confidence interval for the population mean reduction in anxiety was (1.2, 8.6).
Answer:
10 candy bars
Step-by-step explanation:
Since each of his friends needs a candy bar, you need to multiply 6 and 1 ⅔ together.
First, convert 1 ⅔ into an improper fraction: this gives us ⁵⁄₃.
Next, multiply ⁶⁄₁ and ⁵⁄₃ together. To do this, you can visualize 6 as ⁶⁄₁ (which is the same thing). Now you have ⁶⁄₁ x ⁵⁄₃.
<u>Simplify:</u>
The 6 in the numerator and the 3 in the denominator cancel out. This gives us ²⁄₁ x ⁵⁄₁ , which is 2 x 5.
2 x 5 = 10
Answer:
the pennies does not conform to the US mints specification
Step-by-step explanation:
z = (variate -mean)/ standard deviation
z= 2.5 - 2.4991 / 0.01648 = 0.0546
we are going to check the value of z in the normal distribution table, which is the table bounded by z.
checking for z= 0.0 under 55 gives 0.0219 (value gotten from the table of normal distribution)
we subtract the value of z from 0.5 (1- (0.5+0.0219)) = 0.4781 > 0.05claim
since 0.4781 > 0.05claim, therefore, the pennies does not conform to the US mints specification
the claim state a 5% significance level whereas the calculated significance level is 47.81%. therefore, the claim should be rejected