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gulaghasi [49]
3 years ago
13

Mrs. Wilbur can grade a set of 30 test in 2 hours. What is Miss wilbur rate per hour?

Mathematics
2 answers:
shusha [124]3 years ago
8 0
Answer: 15 tests per hour
otez555 [7]3 years ago
5 0

Answer: 15 tests per hour.


Step-by-step explanation: Mrs. Wilbur did 30 tests in 2 hours, so you would divide to find the rate per hour. 15 should be your answer.


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There are 12 tickets needed. Just add all the 4's together.
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3 years ago
1) Lithium isotope rations are important to medicine, the 6Li/7Li ratio in a standard reference material was measured several ti
uysha [10]

Answer:

1) 0.0826052-2.776\frac{0.000013424}{\sqrt{5}}=0.082588    

0.0826052+2.776\frac{0.000013424}{\sqrt{5}}=0.0826219    

b) ME= 2.776\frac{0.000013424}{\sqrt{5}}=0.0000166653

And we want 2/3 of the margin of error so then would be: 2/3 ME = 0.00001111

The margin of error is given by this formula:

ME=z_{\alpha/2}\frac{s}{\sqrt{n}}    (1)

And on this case we have that ME =0.00001111016 and we are interested in order to find the value of n, if we solve n from equation (1) we got:

n=(\frac{z_{\alpha/2} s}{ME})^2   (2)

Replacing we got:

n=(\frac{2.776(0.000013424)}{0.00001111})^2 =11.25 \approx 12

So the answer for this case would be n=12 rounded up to the nearest integer

Step-by-step explanation:

Information given

0.082601, 0.082621, 0.082589, 0.082617, 0.082598

We can calculate the sample mean and deviation with the following formulas:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X=0.0826052 represent the sample mean

\mu population mean

s=0.000013424 represent the sample standard deviation

n=5 represent the sample size  

Part 1

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

The degrees of freedom, given by:

df=n-1=5-1=4

The Confidence level is 0.95 or 95%, and the significance would be \alpha=0.05 and \alpha/2 =0.025, the critical value would be using the t distribution with 4 degrees of freedom: t_{\alpha/2}=2.776

Now we have everything in order to replace into formula (1):

0.0826052-2.776\frac{0.000013424}{\sqrt{5}}=0.082588    

0.0826052+2.776\frac{0.000013424}{\sqrt{5}}=0.0826219    

Part 2

The original margin of error is given by:

ME= 2.776\frac{0.000013424}{\sqrt{5}}=0.0000166653

And we want 2/3 of the margin of error so then would be: 2/3 ME = 0.00001111

The margin of error is given by this formula:

ME=z_{\alpha/2}\frac{s}{\sqrt{n}}    (1)

And on this case we have that ME =0.00001111016 and we are interested in order to find the value of n, if we solve n from equation (1) we got:

n=(\frac{z_{\alpha/2} s}{ME})^2   (2)

Replacing we got:

n=(\frac{2.776(0.000013424)}{0.00001111})^2 =11.25 \approx 12

So the answer for this case would be n=12 rounded up to the nearest integer

3 0
3 years ago
20 POINS FOR ANSWER!!
inna [77]

Answer: the margin of error is 0.02

Step-by-step explanation:

Confidence interval for population proportion is written as

Sample proportion ± margin of error

If the confidence interval for this population proportion is (0.52, 0.56.), it means that the lower boundary is 0.52 and the upper boundary is 0.56

Let x represent the sample proportion and y represent the margin of error. Therefore,

x - y = 0.52

x + y = 0.56

Subtracting the second equation from the first equation, it becomes

- 2y = - 0.04

y = - 0.04/-2

y = 0.02

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Answer:

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Step-by-step explanation: I used a calculator ahaha.

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10%=30 15%=20 Hope this helps :)
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