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3 years ago

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Last month, Grayson and Justine sold candy to raise money for their debate team. Justine sold 1 1/5 times as much candy as Grays

on did. If Grayson sold 2/3 of a box of candy, how many boxes of candy did Justine sell?

1 answer:

ale4655 [162]3 years ago

6 0

Answer: 4/5 of a box of candy

Step-by-step explanation:

Justine sold 1¹/₅ times the ²/₃ of a box of candy that Grayson sold.

Justin sold;

= 1¹/₅ * ²/₃

= 6/5 * 2/3

= 12/15

= 4/5 of a box of candy

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Solve for x. Evaluate and round your answer to 1 decimal place(tenths place). 10x=1200

Mandarinka [93]

Answer:

<h2>x = 3.1</h2>

Step-by-step explanation:

<h3> ${10}^{x} = 1200$ </h3>

<u>To solve first take logarithm to both sides</u>

That's

<h3> $log_{10}(10) ^{x} = log_{10}(1200)$ </h3><h3 /><h3> $log_{10}(10)^{x} = x log_{10}(10)$ </h3>

But

<h3> $log_{10}(10) = 1$ </h3>

So we have

$x = log_{10}(1200)$

<u>Write 1200 as a number with the factor 100</u>

That's

1200 = 100 × 12

So we have

<h3> $x = log_{10}(100 \times 12)$ </h3>

<u>Using the rules of logarithms</u>

That's

<h3> $log_{a}(x \times y) = log_{a}(x) + log_{a}(y)$ </h3>

Rewrite the expression

That's

<h3> $x = log_{10}(100) + log_{10}(12)$ </h3><h3 /><h3> $x = log_{10}(10)^{2} + log_{10}(12)$ </h3><h3> $x = 2 log_{10}(10) + log_{10}(12)$ </h3><h3> $log_{10}(10) = 1$ </h3><h3 /><h3> $x = 2 + log_{10}(12)$ </h3>

x = 3.079

So we have the final answer as

<h3>x = 3.1 to one decimal place</h3>

Hope this helps you

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3 years ago

A union of restaurant and foodservice workers would like to estimate the mean hourly wage, μ, of foodservice workers in the U.S.

pav-90 [236]

Answer:

$n=(\frac{1.960(2.25)}{0.35})^2 =158.76 \approx 159$

So the answer for this case would be n=159 rounded up to the nearest integer

Step-by-step explanation:

Assuming this complete question: A union of restaurant and foodservice workers would like to estimate the mean hourly wage, , of foodservice workers in the U.S. The union will choose a random sample of wages and then estimate using the mean of the sample. What is the minimum sample size needed in order for the union to be 95% confident that its estimate is within $0.35 of ? Suppose that the standard deviation of wages of foodservice workers in the U.S. is about $2.15 .

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".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=2.25$ represent the population standard deviation

n represent the sample size

Solution to the problem

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$ (a)

And on this case we have that ME =0.35 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.960$ , replacing into formula (b) we got:

$n=(\frac{1.960(2.25)}{0.35})^2 =158.76 \approx 159$

So the answer for this case would be n=159 rounded up to the nearest integer

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What percent of 88 is 132

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A parallelogram has a 7 inch side and a 9 inch side, and the longer diagonal is 14 inches long. find the length of the other dia

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The independent is rainy days and the dependent should be hours of tv.

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4 years ago