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murzikaleks [220]

2 years ago

5

A speed limit on the highway is 65 mph a vehicle is traveling 14 mph below the limit what is the speed of the vehicle is traveli

ng

2 answers:

Bess [88]2 years ago

8 0

65-14=51 mph

Answer: 51 mph

xxTIMURxx [149]2 years ago

3 0

Answer:

51 mph.

Step-by-step explanation:

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Dominik [7]

Please go into more detail about the question then u will be able to help you

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2 years ago

What is the answer to this question

astra-53 [7]

Answer: I think the answer is B

Step-by-step explanation:

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2 years ago

Howard Slater age 44 an actor expected his income to the coin future years he decides to take out a 20-year payment life policy

SpyIntel [72]

Answer:

Individuals end to continue paying the premiums of the automobile insurance as a habit. However, serious thoughts and putting in element of strategizing helps to reduce the premium in most cases. At times, there is a sudden like on the part of the insurer even for a flawless driver.

A good look up and research of the insurance websites can be of real help in comparing whether a better deal is offered by the other insurance companies, or whether a certain change in the policy or small adjustments of the term would give benefit to the customer.

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5 0

3 years ago

The distribution of SAT II Math scores is approximately normal with mean 660 and standard deviation 90. The probability that 100

gayaneshka [121]

Using the <em>normal distribution and the central limit theorem</em>, it is found that there is a 0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.

After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

In this problem:

The mean is of 660, hence $\mu = 660$ .

The standard deviation is of 90, hence $\sigma = 90$ .

A sample of 100 is taken, hence $n = 100, s = \frac{90}{\sqrt{100}} = 9$ .

The probability that 100 randomly selected students will have a mean SAT II Math score greater than 670 is <u>1 subtracted by the p-value of Z when X = 670</u>, hence:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{670 - 660}{9}$

$Z = 1.11$

$Z = 1.11$ has a p-value of 0.8665.

1 - 0.8665 = 0.1335.

0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.

To learn more about the <em>normal distribution and the central limit theorem</em>, you can take a look at brainly.com/question/24663213

7 0

1 year ago

Maddie bought 3 2/3 yards of satin ribbon at the craft store. She wanted to make hair bows for the cheer squad. Each bow require

kotegsom [21]

Answer:

4

Step-by-step explanation:

Given that:

Total yards of satin bought by Maddie = $3\frac{2}{3}$ yards

Number of yards required make one ribbon = $\frac{2}{3}$ yards

To find:

Number of complete bows that can be made ?

Solution:

First of all, let us learn about converting the mixed fraction to fractional form.

Formula:

$a\dfrac{b}{c} = \dfrac{a\times c+b}{c}$

Therefore, total yards of satin bought by Maddie can be written in fractional form as:

$3\dfrac{2}{3} = \dfrac{3\times 3+2}{3} = \dfrac{11}{3}$

One hair bow requires $\frac{2}{3}$ yards of ribbon.

Number of hair bows that can be made, can be found by dividing the total ribbon length by ribbon required for making one hair bow.

Answer is:

$\dfrac{\frac{11}{3}}{\frac{2}{3}}\\\Rightarrow \dfrac{11\times 3}{2\times 3}\\\approx 4$

4 hair bows can be made.

4 0

2 years ago