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Sonja [21]
3 years ago
8

A cyclist rides at an average speed of 31 miles per hour. The cyclist’s speed, s, can differ from the average by as much as 8 mi

les per hour. The absolute value inequality mc007-1.jpg represents this situation. If the compound inequality mc007-2.jpg and mc007-3.jpg also represents this situation, what is the value of x in the compound inequality?
Mathematics
2 answers:
Artemon [7]3 years ago
8 0

Answer:

answer is C which is 8

Step-by-step explanation:

Shalnov [3]3 years ago
4 0
In this compound inequality X should be 8.
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The arc of the triangle 
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In how many ways can you park 4 different cars at a car park with 5 different parking spaces?
meriva
1. If you have 1 first then... 1,2,3,4 - 1,3,2,4 - 1,2,4,3 - 1,3,4,2 - 1,4,3,2 - 1,4,2,3
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3. And so one with 3,4
Hope this helps:)
5 0
3 years ago
The shampoo Eva likes to use costs $6 for a 24-ounce bottle. The conditioner she likes to use costs $12 for a 30-ounce bottle. H
lutik1710 [3]

Answer:

0.15

Step-by-step explanation:

6 divided by 24 equals 0.25

12 divided by 30 equals 0.4

0.4 minus 0.25 is 0.15

8 0
3 years ago
The function f(x)=5(2)^x was replaced with f(x)+k , resulting in the function graphed below. what is the value of k?
xxTIMURxx [149]
Since adding a constant to a function simply moved the graph upwards by that amount, solve f(x) for any value and see how much the graph given is different from that value of y...the simplest way in this case may be to simply find f(0) which is:

5*2^0=5

Clearly the graph at x=0 is at y=-2  so we can see that k is:

5+k=-2 so

k=-7
5 0
4 years ago
A company compiles data on a variety of issues in education. In 2004 the company reported that the national college​ freshman-to
nasty-shy [4]

Answer:

1) Randomization: We assume that we have a random sample of students

2) 10% condition, for this case we assume that the sample size is lower than 10% of the real population size

3) np = 500*0.66= 330 >10

n(1-p) = 500*(1-0.66) =170>10

So then we can use the normal approximation for the distribution of p, since the conditions are satisfied

The population proportion have the following distribution :

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})  

And we have :

\mu_p = 0.66

\sigma_{p}= \sqrt{\frac{0.66(1-0.66)}{500}}= 0.0212

Using the 68-95-99.7% rule we expect 68% of the values between 0.639 (63.9%) and 0.681 (68.1%), 95% of the values between 0.618(61.8%) and 0.702(70.2%) and 99.7% of the values between 0.596(59.6%) and 0.724(72.4%).

Step-by-step explanation:

For this case we know that we have a sample of n = 500 students and we have a percentage of expected return for their sophomore years given 66% and on fraction would be 0.66 and we are interested on the distribution for the population proportion p.

We want to know if we can apply the normal approximation, so we need to check 3 conditions:

1) Randomization: We assume that we have a random sample of students

2) 10% condition, for this case we assume that the sample size is lower than 10% of the real population size

3) np = 500*0.66= 330 >10

n(1-p) = 500*(1-0.66) =170>10

So then we can use the normal approximation for the distribution of p, since the conditions are satisfied

The population proportion have the following distribution :

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})  

And we have :

\mu_p = 0.66

\sigma_{p}= \sqrt{\frac{0.66(1-0.66)}{500}}= 0.0212

And we can use the empirical rule to describe the distribution of percentages.

The empirical rule, also known as three-sigma rule or 68-95-99.7 rule, "is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)".

On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:

• The probability of obtain values within one deviation from the mean is 0.68

• The probability of obtain values within two deviation's from the mean is 0.95

• The probability of obtain values within three deviation's from the mean is 0.997

Using the 68-95-99.7% rule we expect 68% of the values between 0.639 (63.9%) and 0.681 (68.1%), 95% of the values between 0.618(61.8%) and 0.702(70.2%) and 99.7% of the values between 0.596(59.6%) and 0.724(72.4%).

8 0
3 years ago
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