All categories

2 years ago

Somebody help me please

Mathematics

1 answer:

Nookie1986 [14]2 years ago

4 0

Answer:

Cannot pretty positive

You might be interested in

Name two 2 digit factors whose product is greater than 200 but less than 600

Alexxandr [17]

10 and 21, 10*21=210

3 0

3 years ago

Read 2 more answers

Pleacould you please help??

dusya [7]

Answer:

26) underroot 2=1.4

27) underroot 20=4.5

Step-by-step explanation:

8 0

2 years ago

Eight times the reciprocal of a number equals 4 times the reciprocal of 8. find the number

Burka [1]

8*(1/x) = 4*(1/8)
8/x = 4/8
64 = 4x
x = 16

4 0

2 years ago

Line t passes through points (10, 1) and (2, 6). Line u is perpendicular to t. What is the slope of line u?

diamong [38]

Let's find the slope for line t first.

We can use the given points 2,6 and 10,1 to find the slope using the slope formula.

$\frac{y.2 - y.1}{x.2 - x.1} = m$

1 - 6 / 10 - 2

-5/8

The slope is -5/8.

Because we know the slope of this line, we can find the slope of the next line instantly, as they are perpendicular.

When a slope is perpendicular to another, it is equal to the negative reciprocal.

Negative reciprocal of -5/8 = 8/5

<h3>The slope of line u is 8/5</h3>

8 0

2 years ago

The average weight of the entire batch of the boxes of cereal filled today was 20.5 ounces. A random sample of four boxes was se

zhenek [66]

Answer:

$s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And replacing we got:

$s= 0.286$

And then the estimator for the standard error is given by:

$SE= \frac{0.286}{\sqrt{4}}= 0.143$

Step-by-step explanation:

For this case we have the following dataset given:

20.05, 20.56, 20.72, and 20.43

We can assume that the distribution for the sample mean is given by:

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

And the standard error for this case would be:

$SE= \frac{\sigma}{\sqrt{n}}$

And we can estimate the deviation with the sample deviation:

$s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And replacing we got:

$s= 0.286$

And then the estimator for the standard error is given by:

$SE= \frac{0.286}{\sqrt{4}}= 0.143$

4 0

3 years ago