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Sonbull [250]
3 years ago
9

effren recorded the number of students in each class level at highschool.based on his findings,what is the probability that a ra

ndomly selected student will be a freshman?

Mathematics
1 answer:
eimsori [14]3 years ago
4 0
The answer is C) 3/10
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Can somebody please help me out please my grads are bad
Schach [20]

Step-by-step explanation:

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7 0
3 years ago
I equals 3.14, T equals 5.71, P equals three years. <br><br> I divided by T divided by P equals?
Lunna [17]
3.14/5.71=0.18330414466 years hope this hepls
5 0
3 years ago
Scores are 78,81,83, and 76 what must she earn on the fifth test to earn an average test score of 81
iogann1982 [59]
The average value of a data set is equal to the sum of each entry divided by the number of entries. In this case, we have 5 scores, so we can find set up the following equation where x is the fifth score:

\dfrac{78+81+83+76+x}{5} = 81

Now solve for x:

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318 + x = 405

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The fifth score must be 87.
6 0
3 years ago
Read 2 more answers
The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airp
motikmotik

Answer:

The 95% confidence interval for the population mean rating is (5.73, 6.95).

Step-by-step explanation:

We start by calculating the mean and standard deviation of the sample:

M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{50}(6+4+6+. . .+6)\\\\\\M=\dfrac{317}{50}\\\\\\M=6.34\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{49}((6-6.34)^2+(4-6.34)^2+(6-6.34)^2+. . . +(6-6.34)^2)}\\\\\\s=\sqrt{\dfrac{229.22}{49}}\\\\\\s=\sqrt{4.68}=2.16\\\\\\

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=6.34.

The sample size is N=50.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

s_M=\dfrac{s}{\sqrt{N}}=\dfrac{2.16}{\sqrt{50}}=\dfrac{2.16}{7.071}=0.305

The degrees of freedom for this sample size are:

df=n-1=50-1=49

The t-value for a 95% confidence interval and 49 degrees of freedom is t=2.01.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_M=2.01 \cdot 0.305=0.61

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 6.34-0.61=5.73\\\\UL=M+t \cdot s_M = 6.34+0.61=6.95

The 95% confidence interval for the mean is (5.73, 6.95).

4 0
3 years ago
PLEASE HELP!! I'll make brainliest ​
Gekata [30.6K]
Sorry people send you links instead of actually helping you
5 0
3 years ago
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