Answer:
The 80% confidence interval for the the population mean nitrate concentration is (0.144, 0.186).
Critical value t=1.318
Step-by-step explanation:
We have to calculate a 80% confidence interval for the mean.
The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.
The sample mean is M=0.165.
The sample size is N=25.
When σ is not known, s divided by the square root of N is used as an estimate of σM:

The degrees of freedom for this sample size are:

The t-value for a 80% confidence interval and 24 degrees of freedom is t=1.318.
The margin of error (MOE) can be calculated as:
Then, the lower and upper bounds of the confidence interval are:

The 80% confidence interval for the population mean nitrate concentration is (0.144, 0.186).
Answer:
There is not sufficient evidence to support the claim.
Step-by-step explanation:
The claim to be tested is:
The mean respiration rate (in breaths per minute) of students in a large statistics class is less than 32.
To test this claim the hypothesis can be defined as follows:
<em>H₀</em>: The mean respiration rate of students is 32, i.e. <em>μ</em> = 32.
<em>Hₐ</em>: The mean respiration rate of students is less than 32, i.e. <em>μ</em> < 32.
The sample mean respiration rate of students is 31.3.
According to the claim the sample mean is less than 32.
The sample mean value is not unusual if the claim is true, and the sample mean value is also not unusual if the claim is false.
Thus, there is not sufficient evidence to support the claim.
we are given
absolute value of 2 and 2 3rd and negative 9 over 4
Firstly , we need write it in terms of expression
2 and 2 3rd is

so, we get as

now, we can simplify it
Firstly , we will find common denominator


we can see that
all denominators are same
so, we can combine numerators


..............Answer
<span>−4.0 is an integer.
hope it helps</span>