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3 years ago

What is 13 minus 4 5/7

Mathematics

2 answers:

marysya [2.9K]3 years ago

8 0

Answer: <em>8 2/7 </em> i think :)

kolbaska11 [484]3 years ago

7 0

8.2
hope i helped you out

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What is the ratio of 6 to 26 expressed as a fraction in the simplest form

kirill115 [55]

Answer:

3/13

Just divide and u will get the answer

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3 years ago

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State whether either extreme in the data set is an outlier.

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B neither are outliers

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3 years ago

Out of 100 people sampled, 42 had kids. Based on this, construct a 99% confidence interval for the true population proportion of

maksim [4K]

Answer:

The 99% confidence interval for the true population proportion of people with kids is (0.293, 0.547).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

Out of 100 people sampled, 42 had kids.

This means that $n = 100, \pi = \frac{42}{100} = 0.42$

99% confidence level

So $\alpha = 0.01$ , z is the value of Z that has a p-value of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.42 - 2.575\sqrt{\frac{0.42*0.58}{100}} = 0.293$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.42 + 2.575\sqrt{\frac{0.42*0.58}{100}} = 0.547$

The 99% confidence interval for the true population proportion of people with kids is (0.293, 0.547).

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2 years ago

Circle A has a circumference of 36(3.14) meters. What is the radius of circle A?

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3 years ago

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Create a matrix that is equal to F+G. The first matrix below is named F and the second matrix below is named G. Name the new mat

likoan [24]

F+G:

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Then, add the elements that occupy the same position:

$H=\begin{bmatrix}{-1.8+1.32} & {-8.6+(-1.9)} & {} \\ {2.85+2.25} & {-1.4+0.0} & {} \\ {-1.8+(-6.2)} & {5.1+1.4} & {}\end{bmatrix}$

Solve

$H=\begin{bmatrix}{-0.48} & {-10.5} & {} \\ {5.1} & {-1.4} & {} \\ {-8} & {6.5} & {}\end{bmatrix}$

So, we find the element at address h31:

$H=\begin{bmatrix}{h11} & {h12} & {} \\ {h21} & {h22} & {} \\ {h31} & {h32} & {}\end{bmatrix}$

In this case, position h31 is - 8.0

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1 year ago