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Rina8888 [55]
3 years ago
13

The city of Jasper has a water tower that holds 1,325,000 gallons of water. Express this number in scientific notation.

Mathematics
2 answers:
VladimirAG [237]3 years ago
7 0
The answer is B) 1.325 x 10^6
VladimirAG [237]3 years ago
3 0
1.325 x 10^6. B is the correct option.
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10%.

Step-by-step explanation:

These 2 probabilities are independent so we multiply them:

Prob(It rains both days) =  0.40 * 0.25

= 0.10

= 10%.

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Solve for x.<br> (17x –23)<br> °<br> (8x –4)<br> °<br> (3x + 17)°
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What is the greatest common factor of 81, 135, and 216?
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The greatest common factor is 27. Hope this helps!
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Suppose there is a population of test scores on a large, standardized exam for which the mean and standard deviation are unknown
RoseWind [281]

Answer:

The confidence interval based on the sample with the larger standard deviation would be wider.

Step-by-step explanation:

The confidence interval expresses the range of values in which the true mean can exist in with a certain level.of confidence.

It is usually calculated thus,

Confidence interval = (Sample mean) ± (Margin of error)

So, it is evident that the width of the range is determined by the Margin of Error.

Margin of error = (critical value) × (Standard error of the mean)

The critical value depends on solely the sample size, and the confidence level. This would be he same for the two distributions being considered. Hence, they would have the same critical value.

But Standard Error of the mean, σₓ, is given as

σₓ = (σ/√n)

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n = Sample size (equal for both data set)

This shows that the distribution with the higher standard deviation has a standard error of the mean.

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Hope this Helps!!!

3 0
3 years ago
A university cafeteria line in the student center is a
nirvana33 [79]

Answer:

Lamda= 4 students/min, µ= 5 students/min  

P= Lamda/µ= 4/5= 0.8

a.) Probability that system is empty= P0= 1-P= 1-0.8= 0.2

b.) Probability of more than 2 students in the system= ∑(n=3 to inf) P^n*P0= (1-P)*(1/(1-P) – (1-P) –(1-P)*P –(1-P)*P^2)= (.2)*(5-  - .2 - (.8)*.2 – (.2)*.8^2))= 0.848

Probability of more than 3 students in the system= ∑(n=4 to inf) P^n*P0= (1-P)*(1/(1-P) – (1-P) –(1-P)*P –(1-P)*P^2 – (1-P)*P^3)= 0.768

c.) W(q)= Waiting time in Queue= lamda/µ(µ- lamda)= 4/5(1)= 0.8 minutes

d.) L(q)= lamda*W(q)= 4*.8= 3.2 students

e.) L(System)= lamda/(µ-lamda)= 4 students.

f.) If another server with same efficiency as the 1st one is added, then µ= 6 sec/student= 10 students/min.

P= 4/10= 0.4

Probability that system is empty= P0= 1-.4= 0.6

W(q)= 4/10(10-4)= 0.0667 minutes

L(q)= Lamda*W(q)= 4*.0667=0.2668

L(system)= Lamda/(µ-lamda)= 4/6= .667

Step-by-step explanation:

6 0
3 years ago
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