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postnew [5]
3 years ago
6

Equation or expression ​

Mathematics
2 answers:
ICE Princess25 [194]3 years ago
8 0
1. Equation
2. Expression
3. Expression
4. Equation
marishachu [46]3 years ago
3 0

Answer:

The first is equation the second is expression the third is expression and the fourth is equation

Step-by-step explanation:

equations have equal signs and expressions dont

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70 bc the distance from b and c is 14 then multiply that by five cause that is the distance between c and d
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3 years ago
3(b - 15) =9 solve for b
shepuryov [24]
B = 18 because 18-5 is equal to 3 and 3x3 is equal to 9.
4 0
3 years ago
PV = nRT<br> solve for R
sukhopar [10]

Answer:

you need to have values to solve for this

Step-by-step explanation:

8 0
4 years ago
An economist wants to estimate the mean income for the first year of work for college graduates who have had the profound wisdom
suter [353]

Answer:

We need at least 601 incomes.

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{1-0.95}{2} = 0.025

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 1-0.025 = 0.975, so z = 1.96

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.

How many such incomes must be found if we want to be 95% confident that the sample mean is within $500 of the true population mean?

We have to find n, for which M = 500, \sigma = 6250. So

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

500 = 1.96*\frac{6250}{\sqrt{n}}

500\sqrt{n} = 1.96*6250

\sqrt{n} = \frac{1.96*6250}{500}

(\sqrt{n})^{2} = (\frac{1.96*6250}{500})^{2}

n = 600.25

Rounding up

We need at least 601 incomes.

3 0
4 years ago
-30 + 14.7<br><img src="https://tex.z-dn.net/?f=%20-%2030%20%2B%2014.7" id="TexFormula1" title=" - 30 + 14.7" alt=" - 30 + 14.7"
Murrr4er [49]

Answer:

-15.3


Step-by-step explanation:


8 0
3 years ago
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