Answer:
By using hypothesis test at α = 0.01, we cannot conclude that the proportion of high school teachers who were single greater than the proportion of elementary teachers who were single
Step-by-step explanation:
let p1 be the proportion of elementary teachers who were single
let p2 be the proportion of high school teachers who were single
Then, the null and alternative hypotheses are:
: p2=p1
: p2>p1
We need to calculate the test statistic of the sample proportion for elementary teachers who were single.
It can be calculated as follows:
where
- p(s) is the sample proportion of high school teachers who were single ()
- p is the proportion of elementary teachers who were single ()
- N is the sample size (180)
Using the numbers, we get
≈ 1.88
Using z-table, corresponding P-Value is ≈0.03
Since 0.03>0.01 we fail to reject the null hypothesis. (The result is not significant at α = 0.01)
What is her bill?
To do this on your own, you round up the amount by 5¢ or 10¢, whichever is closer, and then take 20% of her bill
Remember to put 20% into the calculator as 0.2 and multiply 0.2 by her bill that has been rounded up
Answer:
the answer would be 36.04,
Step-by-step explanation:
im gonna just leave this blank
Look at the picture.
1)|AM| = |MB| = x
|AN| = |NC| = y
|BC| = 2y - 2x = 2(y - x)
|MN| = y - x
Therefore |BC| = 2|MN|
2)|AM| = |MB| = x
|AN| = |NC| = y
|BC| = 2y - 2x = 2(y - x)
|MN| = y - x
Therefore |BC| = 2|MN|
Answer:
roots : 4, -4, i, -i
Step-by-step explanation:
This gets a bit tricky.
We have to substitude x^2 as u in this problem.
Now to rewrite x^4 − 15x^2 − 16 = 0 with u, we get
u^2 - 15u - 16 = 0
( u - 16) (u + 1)
U = 16
U = -1
<em>This is not the end of the problem. </em>
Now we have to substitute x^2 back to u.
x^2 = 16 --> we get the roots 4 and -4
x^2 = -1 --> we get the roots i and -i
tadah!