Answer:
Option C - Reject H0. We have convincing evidence that the mean salary offer for accounting graduates of this university is higher than the national average of $48,722.
Step-by-step explanation:
First of all let's define the hypothesis;
Null hypothesis;H0; μ = $48,722
Alternative hypothesis;Ha; μ > $48,722
Now, let's find the test statistic for the z-score. Formula is;
z = (x' - μ)/(σ/√n)
We are given;
x' = 48,722
μ = 49,870
σ = 3900
n = 50
Thus;
z = (49870- 48722)/(3900/√50)
z = 2.08
So from online p-value calculator as attached, using z = 2.08 and α = 0.05 ,we have p = 0.037526
This p-value of 0.037526 is less than the significance value of 0.05,thus, we reject the claim that that the mean salary offer for accounting graduates of this university is higher than the national average of $48,722
I agree with you. I added the mixed numbers and got 4 1/10 and that’s not an option.
Let x be the digit in the tens place and y be the digit in the ones place.
so, the digit is xy
<span>
The ten's digit of a two digit number is 1 more than 4 times the units' digit.
</span>x = 4y + 1
<span>63 is subtracted from the number, the order of the digits is reversed
</span>10x + y - 63 = 10y + x
9x - 9y = 63
x = 4y + 1 ------------ (1)
9x - 9y = 63 ---------- (2)
Sub (1) into (2)
9(4y + 1) - 9y = 63
36y + 9 - 9y = 63
27y = 63 - 9
27y = 54
y = 2 ------- sub into (1)
x = 4(2) + 1 = 9
x = 9, y = 2
The number is 92
If your asking for the answer it’d be
x= -1