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3 years ago

Idk how to do this ? Like any of it

Mathematics

2 answers:

baherus [9]3 years ago

8 0

I think number one is an acute angle

Rama09 [41]3 years ago

7 0

Its complementary angles and supplementary one is 2 angles that form 90 degrees like were 5 and 6 are

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A kindergarten teacher asked the students’ parents to send their child to school with two fruits (apples and/or oranges). Below

hichkok12 [17]

Answer:

The frequency of oranges is the classroom is 47.37%.

Step-by-step explanation:

The relative frequency of oranges is the number of oranges divided by the total number of fruits.

We have that:

8 students brought 2 apples each. So there are 8*2 = 16 apples

4 students brought an apple and an orange each. So there are 16 + 4*1 = 20 apples and 4*1 = 4 oranges.

7 students brough 2 oranges each. So there are 4 + 2*7 = 18 oranges.

There are 18 oranges, and 20+18 = 38 fruits in total.

So the frequency of oranges in the classrom is

$F = \frac{18}{38} = 0.4737$

5 0

3 years ago

While watching a football game, Jordan decided to list yardage gained as positive integers and yardage lost as negative integers

koban [17]

Answer:

There is a loss of 2 yards.

Step-by-step explanation:

To find the overall gain or loss, we add or subtract there values.

Initially, we have 0 yards.

Records 14, so 0 + 14 = 14, that is, for now a gain of 14

Then loses 7, so 14 - 7 = 7, for now a gain of 7.

Then, 7 - 9 = - 2, so in all, there is a loss of 2 yards.

3 0

2 years ago

4-1/x divided by 16-1/xsquared

Vladimir [108]

4 - 1/x (16)-1/x 2
4x-18/x

3 0

2 years ago

PLS HELLP PLS PLS

Andrew [12]

Answer:

it would be 100 rounded because 14x8=112 and 112 is rounded to 100 to the nearest hundreth. Please subscribe to Amiredagoat Yt

Step-by-step explanation:

6 0

2 years ago

A population has a standard deviation of 5.5. What is the standard error of the sampling distribution if the sample size is 81?

VladimirAG [237]

Answer:

$\sigma = 5.5$

And we have a sample size of n =81. We want to estimate the standard error of the sampling distribution $\bar X$ and for this case we know that the distribution is given by:

$\bar X \sim N(\mu ,\frac{\sigma}{\sqrt{n}})$

And the standard error would be:

$\sigma_{\bar x}= \frac{\sigma}{\sqrt{n}}$

And replacing we got:

$\sigma_{\bar x}=\frac{5.5}{\sqrt{81}}= 0.611$

Step-by-step explanation:

For this case we know the population deviation given by:

$\sigma = 5.5$

And we have a sample size of n =81. We want to estimate the standard error of the sampling distribution $\bar X$ and for this case we know that the distribution is given by:

$\bar X \sim N(\mu ,\frac{\sigma}{\sqrt{n}})$

And the standard error would be:

$\sigma_{\bar x}= \frac{\sigma}{\sqrt{n}}$

And replacing we got:

$\sigma_{\bar x}=\frac{5.5}{\sqrt{81}}= 0.611$

4 0

2 years ago