Answer:
a: z = -1.936
b: 0.0265
d: z < -1.645
Reject H0 if z < -1.645
Step-by-step explanation:
We are given:
H0: µ = 20
HA: µ < 20
n = 60, sample mean: 19.6, σ = 1.6
Since the alternate hypothesis has a < sign in it, it is a left tailed test. The < or > sign in the alternate hypothesis points towards the rejection region.
For a: We need to calculate the test statistic for our situation. This is done with a z-score formula for samples.
For b: we need to use the z-score table to look up the p-value for the score we calculate in part a. The p-value is 0.0265. This means that there is only about a 2.65% chance that the sample values were a result of random chance.
For d: Since the significance level is 0.05, and this is a one tailed test, we have a critical value of z < - 1.645. This means that if the z-score we calculate in part a is less than -1.645, we will reject the null hypothesis
See attached photo for all the calculations!
Answer:
6k^3(2 - 5k)
Step-by-step explanation:
12k^3 - 30k^4
Factor out 6k^3
6k^3(2 - 5k)
4. Cross multiply (10•6) (15•n) 15n=60 you do the opposite operation so instead of multiply 15•n you divided 15 from n. And what u do on one side u do on both. So divide 15 from 60 and u get 4.
Answer:
idk for sure but id say 10x
Step-by-step explanation: