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Vladimir [108]
4 years ago
15

Tell whether the two ratios form a proportion. Explain. 25/80 and 5/16

Mathematics
1 answer:
IgorC [24]4 years ago
3 0
Can you give me brainliest answer? 2 more for a level up!

Okay, 25/80 and 5/16.

Divide 25/80 by 5.

25/5 = 5
80/5 = 16

5/16

The other fraction is 5/16, so the ratios are proportional!
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Is your estimate larger or smaller than the actual number of liters in a gallon? Explain how you know.
SCORPION-xisa [38]

Question:

There are 33.8 fluid ounces in a litre. There are 128 fluid ounces in a gallon. About how many litres are in a gallon?

Is your estimate larger or smaller than the actual number of litres in a gallon? Explain how you know

Answer:

It is larger

Step-by-step explanation:

Given

33.8\ fl\ oz = 1\ litre

128\ fl\ oz = 1\ gallon

Required

Estimate number of litres in a gallon

33.8\ fl\ oz = 1\ litre

128\ fl\ oz = 1\ gallon

Cross Multiply

128\ fl\ oz * 1\ litre = 33.8\ fl\ oz * 1\ gallon

128 * 1\ litre = 33.8* 1\ gallon

Divide both sides by 33.8

\frac{128 * 1\ litre}{33.8} = \frac{33.8* 1\ gallon}{33.8}

\frac{128 * 1\ litre}{33.8} = 1\ gallon

3.78698224852\ litre = 1\ gallon

Approximate

3.787\ litres = 1\ gallon

The estimate is larger than the actual litres to gallon conversion because, in standard units of conversion;

3.785\ litres = 1\ gallon

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3 years ago
Pam can run four laps around the track in 330 sec. Patrick can run two laps around the track in 180 sec. which person. Runs the
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You do a study of hypnotherapy to determine how effective it is in increasing the number of hours of sleep subjects get each nig
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Answer:

Confidence interval

8.98-2.2\frac{1.29}{\sqrt{12}}=8.16    

8.98+2.2\frac{1.29}{\sqrt{12}}=9.80  

And the margin of error would be:

ME=2.2\frac{1.29}{\sqrt{12}}=0.819

Step-by-step explanation:

For this case we have the followig dataset:

DATA: 8.2; 9.1; 7.7; 8.6; 6.9; 11.2; 10.1; 9.9; 8.9; 9.2; 7.5; 10.5

We can calculate the mean and the 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}}

And replacing we got:

\bar X= 8.98[/tex[tex]s = 1.29

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 are given by:

df=n-1=12-1=11

The Confidence level is 0.95 or 95%, the value of significance is \alpha=0.05 and \alpha/2 =0.025, and the critical value would be t_{\alpha/2}=2.20

Repplacing the info we got:

8.98-2.2\frac{1.29}{\sqrt{12}}=8.16    

8.98+2.2\frac{1.29}{\sqrt{12}}=9.80    And the margin of error would be:

ME=2.2\frac{1.29}{\sqrt{12}}=0.819

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