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4 years ago

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

1 answer:

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;

$3.785\ litres = 1\ gallon$

6 0

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

Aleonysh [2.5K]

Patrick because he can run 1 lap in 90 seconds

3 0

4 years ago

You do a study of hypnotherapy to determine how effective it is in increasing the number of hours of sleep subjects get each nig

brilliants [131]

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 $t_{\alpha/2}=2.20$