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Katyanochek1 [597]
3 years ago
14

What must be added to 3x-7 to make x^2+4x-1

Mathematics
2 answers:
marta [7]3 years ago
3 0
3x-7+c=x^2+4x-1
find c
minus 3x from both sides
-7+c=x^2+x-1
add 7 to both sides
c=x^2+x+6

x^2+x+6 must be added
Gemiola [76]3 years ago
3 0
The term that can be added is the difference between the two expressions you have:

x^2+4x-1-(3x-7)

x^2+4x-1-3x+7

x^2+x+6
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You deposit 2000 in account A, which pays 2.25% annual interest compounded monthly. You deposit another 2000 in account b, which
stellarik [79]
To model this situation, we are going to use the compound interest formula: A=P(1+ \frac{r}{n} )^{nt}
where
A is the final amount after t years 
P is the initial deposit 
r is the interest rate in decimal form 
n is the number of times the interest is compounded per year
t is the time in years 

For account A: 
We know for our problem that P=2000 and r= \frac{2.25}{100} =0.0225. Since the interest is compounded monthly, it is compounded 12 times per year; therefore, n=12. Lets replace those values in our formula:
A=2000(1+ \frac{0.0225}{12} )^{12t}

For account B:
P=2000, r= \frac{3}{100} =0.03, n=12. Lest replace those values in our formula:
A=2000(1+ \frac{0.03}{12} )^{12t}

Since we want to find the time, t, <span>when  the sum of the balance in both accounts is at least 5000, we need to add both accounts and set that sum equal to 5000:
</span>2000(1+ \frac{0.0225}{12} )^{12t}+2000(1+ \frac{0.03}{12} )^{12t}=5000

Now that we have our equation, we just need to solve for t:
2000[(1+ \frac{0.0225}{12} )^{12t}+(1+ \frac{0.03}{12} )^{12t}]=5000
(1+ \frac{0.0225}{12} )^{12t}+(1+ \frac{0.03}{12} )^{12t}= \frac{5000} {2000}
(1.001875)^{12t}+(1.0025 )^{12t}= \frac{5}&#10;{2}
ln(1.001875)^{12t}+ln(1.0025 )^{12t}=ln( \frac{5} {2})
12tln(1.001875)+12tln(1.0025 )=ln( \frac{5} {2})
t[12ln(1.001875)+12ln(1.0025 )]=ln( \frac{5} {2})
t= \frac{ln( \frac{5}{2} )}{12ln(1.001875)+12ln(1.0025 )}
17.47

We can conclude that after 17.47 years <span>the sum of the balance in both accounts will be at least 5000.</span>
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A market surveyor wishes to know how many energy drinks teenagers drink each week. They want to construct a 98% confidence inter
frosja888 [35]

Answer:

The minimum sample size required to create the specified confidence interval is 1024.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.98}{2} = 0.01

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.01 = 0.99, so z = 2.327

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

What is the minimum sample size required to create the specified confidence interval

This is n when M = 0.08, \sigma = \sqrt{1.21} = 1.1.

M = z*\frac{\sigma}{\sqrt{n}}

0.08 = 2.327*\frac{1.1}{\sqrt{n}}

0.08\sqrt{n} = 2.327*1.1

\sqrt{n} = \frac{2.327*1.1}{0.08}

(\sqrt{n})^{2} = (\frac{2.327*1.1}{0.08})^{2}

n = 1024

The minimum sample size required to create the specified confidence interval is 1024.

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Answer:

-5x^{3} + 65x + 60

<h3>Step-by-step explanation:</h3>
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