Find the LCM of 18 and 45 using prime factorizations.
2 answers:
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Answer:
Truck D.
Step-by-step explanation:
Below are the rates of travel of 4 shipping trucks.
Truck A: 300 miles in 4 hours Truck B: 240 miles in 3 hours Truck C: 325 miles in 5 hours Truck D: 170 miles in 2 hours.
Speed = distance/time
Speed of Truck A,
Speed of Truck B,
Speed of Truck C,
Speed of Truck D,
It can be seen that the speed of Truck D is more out of Trucks A,B and C. Hence, Truck D is moving at a faster rate.
Answer:
the probability of a Type II error if in fact the mean waiting time u, is 9.8 minutes is 0.1251
Option A) is the correct answer.
Step-by-step explanation:
Given the data in the question;
we know that a type 11 error occur when a null hypothesis is false and we fail to reject it.
as in it in the question;
obtained mean is 9.8 which is obviously not equal to 8.3
But still we fail to reject the null hypothesis says mean is 8.3
Hence we have to find the probability of type 11 error
given that; it is right tailed and o.5, it corresponds to 1.645
so
z is equal to 1.645
z = (x-μ)/
where our standard deviation s = 3.8
sample size n = 50
mean μ = 8.3
we substitute
1.645 = (x - 8.3)/
1.645 = (x - 8.3) / 0.5374
0.884023 = x - 8.3
x = 0.884023 + 8.3
x = 9.18402
so, by general rule we will fail to reject the null hypothesis when we will get the z value less than 1.645
As we reject the null hypothesis for right tailed test when the obtained test statistics is greater than the critical value
so, we will fail to reject the null hypothesis as long as we get the sample mean less than 9.18402
Now, for mean 9.8 and standard deviation 3.8 and sample size 50
Z = (9.18402 - 9.8)/
Z = -0.61598 / 0.5374
Z = - 1.1462 ≈ - 1.15
from the z-score table;
P(z<-1.15) = 0.1251
Therefore, the probability of a Type II error if in fact the mean waiting time u, is 9.8 minutes is 0.1251
Option A) is the correct answer.