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givi [52]
3 years ago
15

What is half of 2/3 an

Mathematics
1 answer:
jek_recluse [69]3 years ago
6 0

Answer:

0.3333

Step-by-step explanation:

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What are all the factors of 618?
Reika [66]

Answer:

1, 2, 3, 6, 103, 206, 309, 618

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What's the answer???
sertanlavr [38]
The answer is Letter "D"
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How to solve for X: X-B=A
mamaluj [8]

Answer:

Step-by-step explanation:

(x−a)(x−b)=x2−(a+b)x+ab

Now, this with the third bracket.

(x2−(a+b)x+ab)(x−c)=x3−(a+b+c)x2+(ac+bc+ab)x−abc

But there’s another way to do this, which is easier. Assume the given expression is equal to 0, then, we can form a cubic equation as

x3−(sum−of−roots)x2+(product−of−roots−taken−two−at−a−time)x−(product−of−roots) , which is essentially what we got above.

5 0
3 years ago
Read 2 more answers
What is the value of m in 3(2m-9+6m)=3m+29-7m<br> i got 1/10 but the choices are -3, 2, 5, and 120
kirill [66]
Hello,

3(2m-9+6m)=3m+29-7m
==>3(8m-9)=-4m+29
==>24m+4m=29+27
==>28m=56
==>m=2

Answer B


5 0
3 years ago
Read 2 more answers
A small business owner estimates his mean daily profit as $970 with a standard deviation of $129. His shop is open 102 days a ye
Katena32 [7]

Answer:

The probability that the shopkeeper's annual profit will not exceed $100,000 is 0.2090.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we select appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sum of values of <em>X</em>, i.e ∑<em>X</em>, will be approximately normally distributed.  

Then, the mean of the distribution of the sum of values of X is given by,  

 \mu_{x}=n\mu

And the standard deviation of the distribution of the sum of values of X is given by,  

 \sigma_{x}=\sqrt{n}\sigma

The information provided is:

<em>μ</em> = $970

<em>σ</em> = $129

<em>n</em> = 102

Since the sample size is quite large, i.e. <em>n</em> = 102 > 30, the Central Limit Theorem can be used to approximate the distribution of the shopkeeper's annual profit.

Then,

\sum X\sim N(\mu_{x}=98940,\ \sigma_{x}=1302.84)

Compute the probability that the shopkeeper's annual profit will not exceed $100,000 as follows:

P (\sum X \leq  100,000) =P(\frac{\sum X-\mu_{x}}{\sigma_{x}}

                              =P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that the shopkeeper's annual profit will not exceed $100,000 is 0.2090.

6 0
3 years ago
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