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kogti [31]
3 years ago
9

Jeremy walked 6/8 of the way to school and ran the rest of the way.what fraction in simplest form shows the part of the way the

Jeremy walked
Mathematics
1 answer:
lilavasa [31]3 years ago
7 0

Answer:

3/4

Step-by-step explanation:

Because 6 and 8 are both divisible by the same number, in order to simplify, thats what you do. since the largest number that they can be divided by is 2, you do just that. 6/8 divided by 2 is 3/4.

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What is the distance between the points (−3, −5) and (−3, −7) ?
mario62 [17]

Step-by-step explanation:

d = √(x2-x1)²+(y2-y1)²

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5 0
2 years ago
In a large lecture class, the professor announced that the scores on a recent exam were normally distributed with a range from 5
iVinArrow [24]

Answer:

The standard deviation of the scores on a recent exam is 6.

The sample size required is 25.

Step-by-step explanation:

Let <em>X</em> = scores of students on a recent exam.

It is provided that the random variable <em>X</em> is normally distributed.

According to the Empirical rule, 99.7% of the normal distribution is contained in the range, <em>μ </em>± 3<em>σ</em>.

That is, P (<em>μ </em>- 3<em>σ </em>< <em>X</em> < <em>μ </em>+ 3<em>σ</em>) = 0.997.

It is provided that the scores on a recent exam were normally distributed with a range from 51 to 87.

This implies that:

P (51 < <em>X</em> < 87) = 0.997

So,

<em>μ </em>- 3<em>σ </em>= 51...(i)

<em>μ </em>+ 3<em>σ </em>= 87...(ii)

Subtract (i) and (ii) to compute the value of <em>σ</em> as follows:

<em>   μ </em>-     3<em>σ </em>=    51

(-)<em>μ </em>+ (-)3<em>σ </em>= (-)87

______________

-6<em>σ </em>= -36

<em>σ</em> = 6

Thus, the standard deviation of the scores on a recent exam is 6.

The (1 - <em>α</em>)% confidence interval for population mean is given by:

CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

The margin of error of this interval is:

MOE = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

Given:

MOE = 2

<em>σ</em> = 6

Confidence level = 90%

Compute the <em>z</em>-score for 90% confidence level as follows:

z_{\alpha/2}=z_{0.10/2}=z_{0.05}=1.645

*Use a <em>z</em>-table.

Compute the sample required as follows:

MOE = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\\2=1.645\times \frac{6}{\sqrt{n}}\\n=(\frac{1.645\times 6}{2})^{2}\\n=24.354225\\n\approx 25

Thus, the sample size required is 25.

4 0
3 years ago
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