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vitfil [10]
3 years ago
10

What is 32/23 as a decimal rounded to the nearest hundredth

id="TexFormula1" title=" \frac{32}{23} " alt=" \frac{32}{23} " align="absmiddle" class="latex-formula">
​

Mathematics
1 answer:
Softa [21]3 years ago
7 0

Answer:

1.39

Step-by-step explanation

32/23 = 1.39130434783

round: 1.39

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The price of a gallon of unleaded gas was $2.80 yesterday. Today, the price rose to $2.87 . Find the percentage increase. Round
marin [14]
Earlier price is 2.80
Current price is 2.87

Increment is 
2.87 - 2.80
0.07

Increment percentage is
(0.07 / 2.80) * 100
2.5 %
4 0
3 years ago
-17-5(x+3)=3x plz help
katovenus [111]
  1. Start by distributing the -5 into the x and 3
  2. -17-5x-15=3x
  3. add like terms
  4. -5x-32=3x
  5. add 32 to both sides
  6. -5x=3x+32
  7. subtract 3x from both
  8. -8x=32
  9. divide both by -8
  10. x=-4

7 0
3 years ago
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For each gym memberships sold, the gym keeps $42 and the employee who sold it gets eight dollars. What is the commission of the
LenaWriter [7]

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i think 7

Step-by-step explanation:

5 0
3 years ago
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5 m<br> 12 m<br> What is the length of the hypotenuse?<br> m
Gelneren [198K]

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the length is 40m

Step-by-step explanation:

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2 years ago
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There are 2,000 eligible voters in a precinct. A total of 500 voters are randomly selected and asked whether they plan to vote f
Ann [662]

Answer:

0.7 - 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.647

0.7 + 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.753

And the 99% confidence interval would be given (0.647;0.753).

So the correct answer would be:

a. 0.647 and 0.753

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

Solution to the problem

The estimated population proportion for this case is:

\hat p = \frac{350}{500}=0.7

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.58

And replacing into the confidence interval formula we got:

0.7 - 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.647

0.7 + 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.753

And the 99% confidence interval would be given (0.647;0.753).

So the correct answer would be:

a. 0.647 and 0.753

7 0
2 years ago
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