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snow_lady [41]
3 years ago
11

Someone purchase 12 books. But needs 4 books for each of the projects ar school. How many projects would they need

Mathematics
1 answer:
Fudgin [204]3 years ago
7 0
3 projects.
12/4=3. Please mark me brainliest rly wanna mark up!
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What is the application of inverse matrix?
Mila [183]

Answer:

The inverse matrix is used to solve the system of linear equations. It also tells us the consistent or inconsistent behaviour of the solution of equations.

Step-by-step explanation:

In real life,

  • Whereas in programming which is taught at the university, matrices and inverse matrices are used for coding and encrypting messages.
  • Matrices are also used in representing the real world data's like the population of people, infant mortality rate, etc.
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8 0
3 years ago
What is the value of the 30th percentile for the data set? 6283 5700 6381 6274 5700 5896 5972 6075 5993 5581 5972 6274 6075 5896
Alex17521 [72]
<h2>Answer:</h2>

5896

<h2>Step-by-step explanation:</h2>

Given data set:

6283 5700 6381 6274 5700 5896 5972 6075 5993 5581 5972 6274 6075 5896

To calculate the 30th percentile;

<em>i. Rearrange the given data set in ascending order.</em>

5581 5700 5700 5896 5896 5972 5972 5993 6075 6075 6274 6274 6283 6381

<em>ii. Use the following formula to get the index of the value at the 30th percentile.</em>

N = p(n)

<em>Where;</em>

N = the number of data points at or below the intended percentile.

p = the intended percentile.

n = the number of points in the data.

<em>In this case;</em>

p = 30 percent = 0.3

n = 14

<em>Substitute these values into the equation;</em>

N = 0.3(14)

N = 4.2

<em>iii. From the calculated index, find the value of the 30th percentile.</em>

If the index is a whole number x, the desired percentile is the average of the values at index x and x+1.

For example if the index is 4, then the desired percentile is the average of the values at index 4 and 5. In our case, that will be;

\frac{5896+5896}{2} = 5896

If the index is a decimal number, it is rounded up to the nearest integer. The value at the rounded index is the desired percentile.

For example, in our case, the index is 4.2 which when rounded up becomes 4.

Therefore, the data point or value at index 4 is the 30th percentile. From the arrangement in (ii) above, that will be 5896. i.e

5581 5700 5700 <u>5896</u> 5896 5972 5972 5993 6075 6075 6274 6274 6283 6381

8 0
3 years ago
Help with math homework pls!
Svetach [21]

the correct answer is a^6


5 0
3 years ago
The triangle's base and corresponding height are labeled.
Artyom0805 [142]
What’s something goes up but never comes down?
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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
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