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Xelga [282]
2 years ago
6

Write five and ninety-five hundredths as a decimal number.

Mathematics
2 answers:
kiruha [24]2 years ago
4 0

Answer: 5.95

Step-by-step explanation: I know that 5 and ninety five hundredths is written like this: 5 95/100. I turned it into a decimal by making it into a improper fraction, 595/100 then divididing the numerator by the denominator. The answer would be 5.95.

poizon [28]2 years ago
3 0

Answer:

5.95

Step-by-step explanation:

Five and ninety-five hundredths as a decimal number is 5.95.

Hoped this helped.

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3x + 6y= 180<br>What is x and y​
Diano4ka-milaya [45]

Answer and explanation:

<h2>X</h2>

3x + 6y = 180  <em>Use the original expression to find x</em>

3x = 180 - 6y  <em>Subtract 6y by both sides</em>

\frac{3x}{3} = \frac{180}{3} - \frac{6y}{3}  <em>Divide by 3 to get x</em>

x = 60 - 3y

<h2>Y</h2>

x = 60 - 3y  <em>Use the new expression to find y</em>

x - 60 = - 3y  <em>Subtract 60 by both sides</em>

\frac{x - 60}{-3} = \frac{-3y}{-3}  <em>Divide by -3 to get y</em>

-\frac{x}{3} + 20 = y

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3 years ago
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A land surveyor measured the elevations of four villages. His measurements are shown.
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2 years ago
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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