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Tanzania [10]
4 years ago
14

What is 1/10 of 200

Mathematics
1 answer:
gladu [14]4 years ago
8 0
1/10x200
= 1/10x200/1
= 200/10
=20

the answer is 20

if u dont understand, 1/10 means a part of 200 that is divided by 10 which means the number x10 =200. So if u dont understand, you can check answer this way: n*10=200

200/10=n
n= 20
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Help please thank you!
larisa [96]

Step-by-step explanation:

In the picture it shows which quadrant is which. Now we can just mark the dots to figure it out using this representation.

W: quadrant 3

M: x axis

C: quadrant 1

K: quadrant 4

Hope that helps!

4 0
3 years ago
What is the additive inverse of the complex number 13-2
Svet_ta [14]

Answer:

-13 + j*2

Step-by-step explanation:

The additive inverse of a complex number x = a +j*b

is a number y, such that

x + y = 0

This means that

y = -x = - a - j*b

Therefore

The additive inverse of 13 - j*2 is equal to

-(13 - j*2) = -13 +j*2

5 0
3 years ago
Plz help me!
Kipish [7]

45! remember multiple is what you multiply the number by, a factor is what you have to multiply to get the number

6 0
3 years ago
Read 2 more answers
3n-5=19
alina1380 [7]

Answer:

n=8

Step-by-step explanation:

3n-5=19

3n=19+5

3n=24

n=24/3

n=8

6 0
3 years ago
In a test of the effectiveness of garlic for lowering​ cholesterol, 8181 subjects were treated with raw garlic. Cholesterol leve
xz_007 [3.2K]

Answer:

With garlic​ treatment, the mean change in LDL cholesterol is not greater than 0.

Step-by-step explanation:

The dependent <em>t</em>-test (also known as the paired <em>t</em>-test or paired samples <em>t</em>-test) compares the two means associated groups to conclude if there is a statistically significant difference amid these two means.

In this case a paired <em>t</em>-test is used to determine the effectiveness of garlic for lowering​ cholesterol.

A random sample of 81 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment.

The hypothesis for the test can be defined as follows:

<em>H₀</em>: With garlic​ treatment, the mean change in LDL cholesterol is not greater than 0, i.e. <em>d</em> ≤ 0.

<em>Hₐ</em>: With garlic​ treatment, the mean change in LDL cholesterol is greater than 0, i.e. <em>d</em> > 0.

The information provided is:

\bar d=0.40\\SD_{d}=16.2\\\alpha =0.01

Compute the test statistic value as follows:

t=\frac{\bar d}{SD_{d}/\sqrt{n}}\\\\=\frac{0.40}{16.2/\sqrt{81}}\\\\=0.22

The test statistic value is 0.22.

Decision rule:

If the <em>p</em>-value of the test is less than the significance level then the null hypothesis will be rejected and vice-versa.

Compute the <em>p</em>-value of the test as follows:

p-value=P(t_{n-1}>0.22)\\=P(t_{80}>0.22)\\=0.4132

*Use a <em>t</em>-table.

The <em>p</em>-value of the test is 0.4132.

<em>p-</em>value= 0.4132 > <em>α</em> = 0.01

The null hypothesis was failed to be rejected.

Thus, it can be concluded that with garlic​ treatment, the mean change in LDL cholesterol is not greater than 0.

3 0
3 years ago
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