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kobusy [5.1K]
2 years ago
6

What is the missing fraction on the spinner? some1 helppp

Mathematics
1 answer:
Crazy boy [7]2 years ago
3 0

Answer:

7/16

Step-by-step explanation:

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4n-(7-6n)<br><br><br> Helppp plz
belka [17]
4n - (7-6n)
4n -7 +6n (multiply minus to 7-6n)
10n -7 is the answer
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3 years ago
NEED HELP!!<br> What does x equal too?<br> 8x - 5x + 155 = 8x + 80
AlexFokin [52]

Answer:   X = 15

Step-by-step explanation:

6 0
3 years ago
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I need help with this problem, please help!!! 
OverLord2011 [107]

Answer:

Yes. (See below for explanation.)

Step-by-step explanation:

The number of servings is found by dividing the quantity available by the size of a serving. The quantity of punch is the sum of the quantities of the juices that go into the punch. The serving size of 3/4 cup is the same as 6 ounces, since a cup is 8 ounces. (3/4 × 8 oz = 6 oz)

The quantity available is (64 oz + 28 oz + 76 oz). The serving size is 6 oz. Since the units of numerator and denominator are the same, they cancel, leaving ...

... number of servings = (quantity available)/(serving size)

... = (64 +28 +76)/6 . . . . as shown in the problem statement

_____

It might not be obvious that the above ratio gives the number of servings. However, if you look at the real units, you see how it happens.

\dfrac{oz}{(\frac{oz}{serving})}= oz\dfrac{serving}{oz}=\dfrac{oz}{oz}serving=serving

3 0
3 years ago
A supervisor records the repair cost for 25 randomly selected dryers. A sample mean of $93.36 and standard deviation of $19.95 a
Feliz [49]

Answer:

The 98% confidence interval for the mean repair cost for the dryers is (83.4161, 103.3039).

Step-by-step explanation:

We have a small sample size n = 25, \bar{x} = 93.36 and s = 19.95. The confidence interval is given by  \bar{x}\pm t_{\alpha/2}(\frac{s}{\sqrt{n}}) where t_{\alpha/2} is the \alpha/2th quantile of the t distribution with n - 1 = 25 - 1 = 24 degrees of freedom. As we want the 98% confidence interval, we have that \alpha = 0.02 and the confidence interval is 93.36\pm t_{0.01}(\frac{19.95}{\sqrt{25}}) where t_{0.01} is the 1st quantile of the t distribution with 24 df, i.e., t_{0.01} = -2.4922. Then, we have 93.36\pm (-2.4922)(\frac{19.95}{\sqrt{25}}) and the 98% confidence interval is given by (83.4161, 103.3039).

3 0
3 years ago
Read 2 more answers
Compare and order the numbers below from least to greatest.<br> 4.6, 2.8, V7, V17,
wel

Answer:

2.8, 4.6, V7, V17

i hope it help n im sorry i can't explain

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2 years ago
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