Answer:
The ordered pair is not a solution of the equation
Step-by-step explanation:
we know that
If a ordered pair is a solution of a linear equation, then the ordered pair must satisfy the linear equation
we have
Substitute the value of x and the value of y of the given ordered pair in the linear equation and analyze the results
For x=-1, y=15
----> is not true
so
the ordered pair not satisfy the equation
therefore
The ordered pair is not a solution of the equation
Answer:
Step-by-step explanation:
Assuming this info from R
hist(gifted$count)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 21.00 28.00 31.00 30.69 34.25 39.00
## Sd
## [1] 4.314887
Data given and notation
represent the mean
represent the sample standard deviation
sample size
represent the value that we want to test
represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is different than 32, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
(1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
P-value
The first step is calculate the degrees of freedom, on this case:
Since is a two sided test the p value would be: