Simplify the following:
-5×1×11×4
-5×1 = -5:
-5×11×4
-5×11 = -55:
-55×4
-55×4 = -220:
Answer: -220
Solution :
To claim to be tested is whether "the mean salary is higher than 48,734".
i.e. μ > 48,734
Therefore the null and the alternative hypothesis are

and 
Here, n = 50

s = 3600
We take , α = 0.05
The test statistics t is given by


t = 2.15
Now the ">" sign in the
sign indicates that the right tailed test
Now degree of freedom, df = n - 1
= 50 - 1
= 49
Therefore, the p value = 0.02
The observed p value is less than α = 0.05, therefore we reject
. Hence the mean salary that the accounting graduates are offered from the university is more than the average salary of 48,734 dollar.
2x+4=3x+1 they've substituted y =2x+4 into the second eq.
I got 18. i did this by using the SOH CAH TOA method. In this case you have the the angle and the hypotenuse, and you’re looking for the adjacent side, therefore you need to use CAH
Answer:
29.714
Step-by-step explanation:
12/3 (52/7) = 4 *( 52/7) = 29.714