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nydimaria [60]
3 years ago
14

Shelly biked 21 miles in 4 hours. What is Shelly's average speed in miles per hour? At that same rate how many hours will it tak

e Shelly to bike 42 miles?
Mathematics
2 answers:
Liono4ka [1.6K]3 years ago
7 0

Answer:

5.25 mph , for 42 mile she would ride around  8 hours

Step-by-step explanation:

you need to divide the 21 miles by the 4 hours to find how many miles Shelly rode for one hour and that will be your answer ( 5.25 mph)

Vedmedyk [2.9K]3 years ago
5 0

So, we need to do the distance she travelled (21 miles) divided by the rate at which she travelled (4 hours). Therefore, because 21 divided by 4 is 5.25, she biked at a speed of 5.25 mph (Miles per hour). So, if she bikes 5.25 miles every hour we can divide 42 by 5.25 to get the time it would take her to travel 42 miles. Now, 42 divided by 5.25 equals 8, so that means for Shelly to bike 42 miles at her average speed of 5.25 mph, it would take her 8 hours.

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P varies directly as q. When q = 31.2, p = 20.8. Find p when q = 15.3.
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15.3 * 20.8 / 31.2

= 318.24 / 31.2

= 10.2

Answer: when q = 15.3 , p = 10.2

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2 years ago
Difference of two squares <br> Calculate this <br> 81-(3x-2)^2<br> ^2= to the power of 2
Marat540 [252]
<span>81-(3x-2)^2
= </span><span>81-(-6)^2
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= 45

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2 years ago
Taking the absolute value of a number is the same thing as taking the opposite of the number. - true or false
Dovator [93]

Answer:True

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
At a large Midwestern university, a simple random sample of 100 entering freshmen in 1993 found that 20 of the sampled freshmen
guajiro [1.7K]

Answer:

The 90% confidence interval for the difference of proportions is (0.01775,0.18225).

Step-by-step explanation:

Before building the confidence interval, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

p1 -> 1993

20 out of 100, so:

p_1 = \frac{20}{100} = 0.2

s_1 = \sqrt{\frac{0.2*0.8}{100}} = 0.04

p2 -> 1997

10 out of 100, so:

p_2 = \frac{10}{100} = 0.1

s_2 = \sqrt{\frac{0.1*0.9}{100}} = 0.03

Distribution of p1 – p2:

p = p_1 - p_2 = 0.2 - 0.1 = 0.1

s = \sqrt{s_1^2+s_2^2} = \sqrt{0.04^2 + 0.03^2} = 0.05

Confidence interval:

p \pm zs

In which

z is the z-score that has a p-value of 1 - \frac{\alpha}{2}.

90% confidence level

So \alpha = 0.1, z is the value of Z that has a p-value of 1 - \frac{0.1}{2} = 0.95, so Z = 1.645.  

The lower bound of the interval is:

p - zs = 0.1 - 1.645*0.05 = 0.01775&#10;

The upper bound of the interval is:

p + zs = 0.1 + 1.645*0.05 = 0.18225&#10;

The 90% confidence interval for the difference of proportions is (0.01775,0.18225).

6 0
3 years ago
Not sure how to go at this at all, please help.
ss7ja [257]
Here ya go answer down below

8 0
2 years ago
Read 2 more answers
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