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Rainbow [258]
11 months ago
11

What number is 66 2/3% of 495? Show work

Mathematics
2 answers:
son4ous [18]11 months ago
6 0

Answer:

330

Step-by-step explanation:

So to find x% of a, it can generally be expressed as: \frac{x}{100}*a where x=percentage, and a=value. All you're doing in this equation is converting the percentage into its decimal form and then multiplying it by a to find what x% is of a. So I'm assuming 66 2/3 represents 66.6666... and not 66 * 2/3. So to make this a bit easier, we can just represent this value as one fraction, by multiplying 66 by 3 and adding that to the 2, to get: \frac{200}{3}

Now let's plug the values into the equation:

\frac{\frac{200}{3}}{100} * 495

To divide the fractions, simply keep, change, flip.

\frac{200}{3}*\frac{1}{100}*459

Multiply

\frac{200}{300}*495

Simplify the fraction:

\frac{2}{3}*495

Multiply

\frac{990}{3}

Divide:

330

Ivanshal [37]11 months ago
5 0

Answer: 330.

Step-by-step explanation:

\displaystyle\\\frac{66\frac{2}{3}\%*495 }{y100\%}=\\\\ \frac{\frac{200}{3}\%*495 }{100\%} =\\\\\frac{200\%*495}{3*100\%}=\\\\ \frac{2*495}{3} =\\\\\frac{990}{3} =\\\\330.

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Draw a diagram to illustrate the problem as shown below.

Let v = speed of the westbound car, mph

Because the eastbound car travels 4 mph faster than the westbound car, its speed is (v+4) mph.

After 2 hours,
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Because they become separated by 208 miles, therefore
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3 years ago
As a store manger you would like to find out the average time it takes to unload the truck which delivers the merchandise for yo
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Answer:

The upper boundary of the 95% confidence interval for the average unload time is 264.97 minutes

Step-by-step explanation:

We have the standard deviation for the sample, but not for the population, so we use the students t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

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95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 34 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.9}{2} = 0.975([tex]t_{975}). So we have T = 2.0322

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The upper end of the interval is the sample mean added to M. So it is 204 + 60.97 = 264.97

The upper boundary of the 95% confidence interval for the average unload time is 264.97 minutes

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