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strojnjashka [21]
3 years ago
10

I need help thanks!!!!!

Mathematics
1 answer:
irga5000 [103]3 years ago
3 0
Is this question asking for descriptions?
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A number reckoned on the basis of a whole divided into 100 parts is known as a
VladimirAG [237]

The answer for the question is: Percentage.

The explanation for this answer is shown below:

By definition, a percentage is a number that is computed on the basis of a whole divided into one hundred parts. Therefore, based on this definition, you can write a percent as percentage (with the symbol %), as a fraction (with a numerator and a denominator) and you can also express it as a decimal, as following:

20%=\frac{1}{5}=0.2

8 0
3 years ago
Helpp me please
kiruha [24]

Answer:

d. RP

Step-by-step explanation:

Point p is not on the line.

Therefore RP is not a possible name for the illustrated line

7 0
2 years ago
Pattie buys a handbag that costs 240 dollars and a scarf that costs 86 dollars. If sales tax is 7%, what is the total purchase p
Klio2033 [76]

Answer:

$348.82

Step-by-step explanation:

240 +86=326

326× 7%= 22.82

326+22.82= $348.82

7 0
2 years ago
If a,b,c are in arithmetic sequence then a = ...​
andriy [413]

Answer:

If a, b and c are in arithmetic progression, that means that a + c = 2b. So a + b, a + c and b + c are also in arithmetic progression.

This might help.

8 0
3 years ago
Read 2 more answers
In August 2003, 56% of employed adults in the United States reported that basic mathematical skills were critical or very import
Serga [27]

Answer:

Yes

Step-by-step explanation:

First, suppose that nothing has changed, and possibility p is still 0.56. It's our null hypothesis. Now, we've got Bernoulli distribution, but 30 is big enough to consider Gaussian distribution instead.

It has mean μ= np =  30×0.56=16.8

standard deviation s = √npq

sqrt(30×0.56×(1-0.56)) = 2.71

So 21 is (21-16.8)/2.71 = 1.5494 standard deviations above the mean. So the level increased with a ˜ 0.005 level of significance, and there is sufficient evidence.

7 0
3 years ago
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