Answer:
-2, - 1, 0, 1, 2
Step-by-step explanation:
The greater than or equal to sign (≤) demonstrates that the unknown is equal to -2 and greater, but is less than 3, so the largest integer solution is 2, not 3
D because 5 times 2 is 10 and - cuts off - so -3 times -3 would be 9 so 10+9 is 19 :)
Answer:
a) P(t>3)=0.30
b) P(t>10|t>9)=0.67
Step-by-step explanation:
We have a repair time modeled as an exponentially random variable, with mean 1/0.4=2.5 hours.
The parameter λ of the exponential distribution is the inverse of the mean, so its λ=0.4 h^-1.
The probabity that a repair time exceeds k hours can be written as:

(a) the probability that a repair time exceeds 3 hours?

(b) the conditional probability that a repair takes at least 10 hours, given that it takes more than 9 hours?
The exponential distribution has a memoryless property, in which the probabilities of future events are not dependant of past events.
In this case, the conditional probability that a repair takes at least 10 hours, given that it takes more than 9 hours is equal to the probability that a repair takes at least (10-9)=1 hour.


Answer: (-2y+3)(y+11)
Step-by-step explanation:
33 is 3×11 so expect something like
(y+3)(y+11) = y^2+14y+33
leading term has coefficient -2 so
one factor is like (-2y+__)
Try (-2y+3)(y+11) = -2y^2-19y+33
which happens to be the answer
Answer:
a. We reject the null hypothesis at the significance level of 0.05
b. The p-value is zero for practical applications
c. (-0.0225, -0.0375)
Step-by-step explanation:
Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.
Then we have
,
,
and
,
,
. The pooled estimate is given by
a. We want to test
vs
(two-tailed alternative).
The test statistic is
and the observed value is
. T has a Student's t distribution with 20 + 25 - 2 = 43 df.
The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value
falls inside RR, we reject the null hypothesis at the significance level of 0.05
b. The p-value for this test is given by
0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.
c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)
, i.e.,
where
is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So
, i.e.,
(-0.0225, -0.0375)