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bekas [8.4K]
3 years ago
6

Help asap time limit ​

Mathematics
1 answer:
GrogVix [38]3 years ago
6 0

Answer:

Step-by-step explanation:

Taxable income=$ 85,968-$8500=$77,468

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Fifteen is no more than a number t divided by 5.
cupoosta [38]
The proper way to set up this equation is: 15 > t / 5
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What is the name of the relationship when the function used to fit the data is
g100num [7]

Answer

LINEAR REGRESSION

Step-by-step explanation:

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4 years ago
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Solve the system by substitution. And show your work<br><br> -9y-10=x<br><br> 5x+6y=-11
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The answer is (-1, -1).
hope this helps!

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3 years ago
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Assume that a sample is used to estimate a population mean μ μ. Find the margin of error M.E. that corresponds to a sample of si
JulijaS [17]

Answer:

The margin of error is 30.22

Step-by-step explanation:

Sample size 19(lesser than 30), and we only have the sample standard deviation. So we use the t-distribution.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 19 - 1 = 18

Now, we have to find a value of T, which is found looking at the t table, with 18 degrees of freedom(y-axis) and a confidence level of 0.995(t_{995}). So we have T = 2.878

The margin of error is:

M = T*s = 10.5*2.878 = 30.22

5 0
4 years ago
Proof by mathematical induction
Ghella [55]

Answer:

See below.

Step-by-step explanation:

I will assume that 3n is the last term.

First let  n = k, then:

Sum ( k terms) =  7k^2 + 3k

Now, the sum of k+1 terms  = 7k^2 + 3k + (k+1) th term

=  7k^2 + 3k + 14(k + 1) - 4

=   7k^2 + 17k + 10

Now 7(k + 1)^2 = 7k^2 +14 k + 7 so

7k^2 + 17k + 10

=  7(k + 1)^2  + 3k  + 3

= 7(k + 1)^2 + 3(k + 1)

Which is the  formula for the Sum of k terms with the k replaced by k + 1.

Therefore we can say if the sum formula is true for k terms then it is also true for (k + 1) terms.

But the formula is true for 1 term because  7(1)^2 + 3(1) = 10 .

So it must also be true for all subsequent( 2,3 etc) terms.

This completes the proof.

5 0
3 years ago
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