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3 years ago

What is the ratio of 96:132 in simplest form?

Mathematics

2 answers:

Anika [276]3 years ago

8 0

Divide

96/12 = 8
132/12 = 11

8/11 is the final answer

Aleks04 [339]3 years ago

4 0

Treat ast fraction
96/132
divide by 2,3,5,7 or factors starting from 2
96/2=48
132/2=66

48/66
divide 2
48/2=24
66/2=33

24/33
divide 3 since cannot divide 33 by 2
24/3=8
33/3=11
8/11
8:11
answer is 8:11

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Answer Explanation:

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3 years ago

Write an equation of a line with the given slope and y-intercept. m = –5, b = –3. A. y = –5x – 3 . B. y = –5x + 3 . C. y = 5x –

True [87]

Y = mx + b
y = -5x -3

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3 years ago

Read 2 more answers

Twice the sum of a number and 5 is equal to three times the difference of the number and 7. Find the number.

Alexandra [31]

Step-by-step explanation:

Let the required number be x.

According to the given information:

$2(x + 5) = 3(x - 7) \\ \\ \therefore \: 2x + 10 = 3x - 21 \\ \\ \therefore \: 10 + 21 = 3x -2x \\ \\ \therefore \: 31 = x \\ \\ \huge \red { \boxed{\therefore \: x = 31}}$

Hence the required number is 31.

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3 years ago

In a random sample of 80 teenagers, the average number of texts handled in a day is 50. The 96% confidence interval for the mean

Nastasia [14]

Answer:

a) $\bar X =\frac{46+54}{2}=50$

And the margin of error is given by:

$ME= \frac{54-46}{2}= 4$

The confidence level is 0.96 and the significance level is $\alpha=1-0.96=0.04$ and the value of $\alpha/2 =0.02$ and the margin of error is given by:

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

We can calculate the critical value and we got:

$z_{\alpha/2} = 2.05$

And if we solve for the deviation like this:

$\sigma = ME * \frac{\sqrt{n}}{z_{\alpha/2}}$

And replacing we got:

$\sigma =4 *\frac{\sqrt{80}}{2.05} =17.45$

b) $ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=2.05 *\frac{17.45}{\sqrt{160}}=2.828$

And as we can see that the margin of error would be lower than the original value of 4, the margin of error would be reduced by a factor $\sqrt{2}$

Step-by-step explanation:

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

$\mu$ population mean (variable of interest)

$\sigma$ represent the population standard deviation

n=80 represent the sample size

Solution to the problem

Part a

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

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

For this case we can calculate the mean like this:

$\bar X =\frac{46+54}{2}=50$

And the margin of error is given by:

$ME= \frac{54-46}{2}= 4$

The confidence level is 0.96 and the significance level is $\alpha=1-0.96=0.04$ and the value of $\alpha/2 =0.02$ and the margin of error is given by:

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

We can calculate the critical value and we got:

$z_{\alpha/2} = 2.05$

And if we solve for the deviation like this:

$\sigma = ME * \frac{\sqrt{n}}{z_{\alpha/2}}$

And replacing we got:

$\sigma =4 *\frac{\sqrt{80}}{2.05} =17.45$

Part b

For this case is the sample size is doubled the margin of error would be:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=2.05 *\frac{17.45}{\sqrt{160}}=2.828$

And as we can see that the margin of error would be lower than the original value of 4, the margin of error would be reduced by a factor $\sqrt{2}$

5 0

3 years ago