Answer:
A:
Step-by-step explanation:
***I'm pretty sure A should read "NO, because the test value 1.257 is less than the critical value 1.782. Please check the wording of the problem***
H0 : μ ≤ 400
Ha : μ > 400 (claim)
Sample mean: 6,155/13
Sample standard deviation: √44422.4359
Critical test value:
t > 1.782
t = (6,155/13 - 400)/[(√44422.4359)/√13] = 1.257
1.257 < 1.782 ; we fail to reject the null hypothesis
There is not enough evidence at the 5% level of significance to support the claim that the mean battery life is at least 400 hours.
Answer:
1.932 days (or approximatelly 1 day, 22 hours and 22 minutes)
Step-by-step explanation:
The inicial concentration is 60,000, and this concentration triples every 4 days, so we can write the equation:
P = Po * r^t
where P is the final concentration after t periods of 4 days, Po is the inicial concentration and r is the ratio that the concentration increases (r = 3)
Then, we have that:
102000 = 60000 * 3^t
3^t = 102/60 = 1.7
log(3^t) = log(1.7)
t*log(3) = log(1.7)
t = log(1.7)/log(3) = 0.483
so the number of days that will take is 4*0.483 = 1.932 days (or approximatelly 1 day, 22 hours and 22 minutes)
8 7/10 as a improper fraction is 87/10
(4,-2) is point H
answer is B. second choice
