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erik [133]
4 years ago
13

Solve for n. –9n − 1 = –10n

Mathematics
2 answers:
mario62 [17]4 years ago
7 0

Answer:

n = -1

Step-by-step explanation:

-9n - 1 = -10n

<u>+9n        +9n</u>

-1 = n

givi [52]4 years ago
6 0

Answer:

n=1

Step-by-step explanation:

Let's solve your equation step-by-step.

−9n−1=−10n

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Step-by-step explanation:

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Option D, 5^9 is the correct answer

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4 years ago
Four times the sum of a number and ter is the same as one-hundred added to eight
AfilCa [17]

Answer:

GIVE BRAINLIEST PLZ

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x = -15

Step-by-step explanation:

1) Four times sum of a number and 10

4(x + 10)

2) is equal to 100 added to 8 times the number

4(x+10) = 8x + 100

x + 10 = 2x + 25

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7 0
3 years ago
Given that f(x) = ax – 5 and f(4) = 15, determine the value of a.
agasfer [191]

Given that f(x) = ax – 5 and f(4) = 15, then the value of a is 5

The given function is:

f(x)  =  ax  -  5

If f(4)  =  15, then x = 4

Substitute x = 4 into f(x) = ax - 5

f(4)  =  a(4)  -  5

15   =  4a   -  5

Solve for a by adding 5 to both sides

15  +  5  =  4a  -  5  +  5

20    =  4a

Divide both sides by 4

4a/4   =  20/4

a    =  5

Given that f(x) = ax – 5 and f(4) = 15, then the value of a is 5

Learn more on linear functions here: brainly.com/question/15602982

8 0
2 years ago
What is an example of when you would want consistent data and, therefore, a small standard deviation?
steposvetlana [31]

Answer:

12.1, 12.3,12.4,12.5,12.3,12.1,12.2

\bar X= \frac{12.1+12.3+12.4+12.5+12.3+12.1+12.2}{7}=12.271

And for the standard deviation we can use the following formula:

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

And after replace we got:

s = 0.1496

And as we can ee we got a small value for the deviation <1 on this case.

Step-by-step explanation:

For example if we have the following data:

12.1, 12.3,12.4,12.5,12.3,12.1,12.2

We see that the data are similar for all the observations so we would expect a small standard deviation

If we calculate the sample mean we can use the following formula:

\bar X=\frac{\sum_{i=1}^n X_i}{n}

And replacing we got:

\bar X= \frac{12.1+12.3+12.4+12.5+12.3+12.1+12.2}{7}=12.271

And for the standard deviation we can use the following formula:

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

And after replace we got:

s = 0.1496

And as we can ee we got a small value for the deviation <1 on this case.

8 0
3 years ago
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