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Daniel [21]
3 years ago
15

Sam gave jen 1/2 of his jujubes. Jen ate 1/2 of the jujubes and gave the rest to klye. Kyle kept 8 of the jujubes and gave the l

ast 10 to Kim. How many jujubes did Jen eat?
Mathematics
1 answer:
USPshnik [31]3 years ago
7 0

Answer:

Jen ate 18 jujubes

Step-by-step explanation:

Start with Kim. She was given 10. There is no remainder.

Kyle kept 8 from the total he was given, so Kyle had 18.

Let the number of Jujubes given to Jen = x

(1/2) x = 18    Multiply both sides by 2

2(1/2)x = 18*2

x = 36

Jen ate 1/2 of 36 which was 18.

Though you are not asked, Sam must have begun this with 2*36 = 72 Jujubes.


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What is 1 5/6 + 1/2 ?<br><br> a.<br> 14/6<br> b.<br> 17/12<br> c.<br> 17/6<br> d.<br> 25/12
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1 and 5/6=1+5/6=6/6+5/6=11/6

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Compared to the tone of Passage 2, the tone of
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The correct answer is option e) tempered.

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Passage 1 discusses recycling as a socio-cultural phenomena, whereas passage 2 is a recycling critique.

Therefore, the tone of passage 1 is a little more upbeat than that of passage 2.

As a result, we may determine that tempered is the most appropriate tone for this passage 1.

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2 years ago
A doctor wants to estimate the mean HDL cholesterol of all​ 20- to​ 29-year-old females. The number of subjects needed to estima
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Answer:

n=(\frac{1.64(14.7)}{3})^2 =64.57 \approx 65

So the answer for this case would be n=65 rounded up to the nearest integer

And the sample size would decrease by 160-65=95 subjects

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

ME represent the margin of error

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

2) Solution to the problem

Since the new Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that z_{\alpha/2}=1.64

Assuming that the deviation is known we can express the margin of error is given by this formula:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

And on this case we have that ME =\pm 3 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

Replacing into formula (b) we got:

n=(\frac{1.64(14.7)}{3})^2 =64.57 \approx 65

So the answer for this case would be n=65 rounded up to the nearest integer

And the sample size would decrease by 160-65=95 subjects

8 0
3 years ago
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