This is equivalent to:
(2.2533/2.59)(10^8/10^4)
(0.87)(10^4) which is:
0.87X10^4 which is equal to:
0.87X10000 which is equal to:
8.7X1000 and since 1000=10^3 we can say:
8.7X10^3
Answer:
Answer is 0.5
Step-by-step explanation:
The text implies that the number of traffic accidents determines the number of ambulances required on the road. Hence, 1 traffic accident requires 1 ambulance.
The time slot and time of day remain same in both the text and the question.
The average or mean number of accidents is 1.1
QUESTION: What then is the probability that there is need for exactly 2 ambulances on that road and in that time frame?
In other words, what is the probability that there'll be 2 accidents on that road and in that time frame?
Since the mean number of accidents is 1.1, 2 is above the mean.
Probability of 1.1 accidents is 1, since that's the mean (expected value).
Probability of 2 accidents is 1/2 which is = 0.5
ok so here we go:
1) or
2 ) so the answer is (0, -4) or -4
3) y=
4) y=[tex]\frac{3}{2}x+1[tex] so the answer is (0, 1) or 1
Answer:
The probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.
Step-by-step explanation:
According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.
Then, the mean of the distribution of sample mean is given by,
And the standard deviation of the distribution of sample mean is given by,
The information provided is:
<em>μ</em> = 144 mm
<em>σ</em> = 7 mm
<em>n</em> = 50.
Since <em>n</em> = 50 > 30, the Central limit theorem can be applied to approximate the sampling distribution of sample mean.
Compute the probability that the sample mean would differ from the population mean by more than 2.6 mm as follows:
*Use a <em>z</em>-table for the probability.
Thus, the probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.