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stiks02 [169]
3 years ago
11

A 95% confidence interval for the proportion of students achieving a reading achievement score that is above the standard set by

the teachers for a population of third grade students is (0.43, 0.49).
The margin of error of this interval is:

Group of answer choices

0.05

0.03

0.06

None of the above
Mathematics
1 answer:
gizmo_the_mogwai [7]3 years ago
4 0
Answer:

0.03

Explanation:

First, you subtract the upper bound of the confidence interval by the lower bound of the confidence interval. Then, you divide that number by 2 to get the margin or error.

In this case: (0.49 - 0.43)/2 = 0.03
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In preparation for the Senior Prom, 8 seniors got together and decided to travel in style and rent a limousine. They investigate
Snowcat [4.5K]
<span>The limousine will cost $240.
$180 for the first three hours, plus 30 for each hour after. They need two extra hours.
So, 180+30(2)=240.
They also want to leave a $20 dollar tip.
So, 240+20=260
That is $260 dollars they need to pay altogether.
Divide that number by the number of students and you'll get how much they each must pay.
260/8=32.50

$32.50 each.

♦Brainliest please♦
</span>
4 0
4 years ago
6636=3<br> 8118=4 <br> 2212=0<br> 8688=?<br> 5971=?
pochemuha
6636 has 3 circle so 6636=3
8118 has 4 circle so 8118=4
2212 has 0 circle so 2212=0
8688 has 7 circle 8688=7
5971 has 1 circle 5971=1
6 0
2 years ago
Your class has 26 students, which represents 5% of your school's enrollment. Your friend uses the proportion 5/100 =n/26 to find
dedylja [7]
26 is a part of the whole. the proportion should be set up:

(26/x) = (5/100) It's the overall enrollment that is being asked.

If you also need to solve: Cross multiply, 26(100) = 5x, 2600 = 5x, divide both sides by 5, 520 is the enrollment
7 0
3 years ago
3. Find the balance of a savings account at the end of 15 years if the interest
mr Goodwill [35]

Answer:

The balance of the savings account in 15 years will be $1,750.24.

8 0
3 years ago
The weights (to the nearest pound) of some boxes to be shipped are found to be:. Weight 65 68 69 70 71 72 90 95 frequency 1 2 5
Lena [83]
So the mean is 72.97

We need to subtract the mean from each value and square it.
(65-72.97)^2= 63.5209
(68-72.97)^2=24.7009
(69-72.97)^2=15.7609
(70-72.97)^2=8.8209
(71-72.97)^2= 3.8809
(72-72.97)^2=0.9409
(90-72.97)^2=290.0209
(95-72.97)^2=485.3209

Now we add up the new values ( also consider their frequency) and find their mean.
Add the values
63.5209+(2 •24.7009=49.4018)+(5•15.7609=78.8045)+(8•8.8209=70.5672)+(7•3.8809=27.1663)+(3•0.9409=2.8227)+(2•290.0209=580.0418)+(2•485.3209=970.6418)= 1,842.967
Divide by total numburs to find the mean
1,842.967/ 30=61.43223333

The standar deviation is the square root of the mean so is
Square root of 61.43223333=7.837871735
Round to the nearest tenth
Standard Deviation is 7.8



7 0
3 years ago
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