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3 years ago

What is the area of the triangle in centimeters squared?

Mathematics

1 answer:

Goryan [66]3 years ago

6 0

Answer:

${ \tt{base = \sqrt{ {23}^{2} - {16}^{2} } = 16.5 \: cm}} \\ area = \frac{1}{2} \times base \times height \\ area = \frac{1}{2} \times 16.5 \times 16 \\ { \tt{area = 132 \: {cm}^{2} }}$

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Jace invested $380 in an account paying an interest rate of 6.2% compounded continuously. Assuming no deposits or withdrawals ar

SOVA2 [1]

Answer: 960

Step-by-step explanation:

5 0

3 years ago

The diameter of a sphere is 21.6 cm. What is the sphere's volume? Round to the nearest tenth, if necessary.

NeX [460]

By definition, the volume of a sphere is:
$V = \frac{4}{3} (\pi) (r ^ 3)
$
Where,
r: sphere radio
Substituting values we have:
$V = \frac{4}{3}(3.14)(( \frac{21.6}{2} ) ^ 3)
$
$V = 5273.99424
$ Rounding the result to the nearest tenth:
$V = 5274.0 cm ^ 3
$ Answer:
the sphere's volume is:
$V = 5274.0 cm ^ 3$

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3 years ago

What is the slope of the line through (-1,4) and (1,-2)

yawa3891 [41]

Answer:

-3

Step-by-step explanation:

Slope = change of x over change of y

6 over 2

line goes down so slope is negative

-3

5 0

4 years ago

Read 2 more answers

The teacher said we could estimate but give a good reason for that estimate

lyudmila [28]

Answer:

$11, even if you do work a job thats not that hard you should still be making a good amount of money

Step-by-step explanation:

I just used my opinion

5 0

3 years ago

Suppose that you had the following data set. 500 200 250 275 300 Suppose that the value 500 was a typo, and it was suppose to be

hodyreva [135]

Answer:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

Step-by-step explanation:

The subindex B is for the before case and the subindex A is for the after case

Before case (with 500)

For this case we have the following dataset:

500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

And the sample deviation with the following formula:

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

After case (With -500 instead of 500)

For this case we have the following dataset:

-500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

And the sample deviation with the following formula:

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

And as we can see we have a significant change between the two values for the two cases.

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

8 0

4 years ago