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3 years ago

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C. Ms. James buys her sugar in 5-pound bags. She wants to buy enough sugar to make 12 batches of cookies for the school bake sal

e. How many bags of each type of sugar, brown and white, should she buy? Show your work. (1 pound = 16 ounces)

1 answer:

Free_Kalibri [48]3 years ago

8 0

This problem requires more information to solve.

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Z varies jointly as x and y. z=60 when x=5 and y=6. find z when x = 7 and y=3

Aneli [31]

Answer:

z = 42

Step-by-step explanation:

The question can be answered in 2 steps as follows:

Step 1: Calculation of the constant of the variation

The equation for the joint variation can be given as follows:

z = cxy ................... (1)

Where;

z = 60

c = constant = ?

x = 5

y = 6

Substituting the values into equation (1) and solve c, we have:

60 = c * 5 * 6

60 = c * 30

c = 60 / 30

c = 2

Step 2: find z when x = 7 and y = 3

Since from Step 1 c = 2, we now use equation (1) and substitute the values into it to find z as follows:

z = 2 * 7 * 3

z = 42

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3 years ago

Angles A and B are complementary. Two separate angles are shown. Angle A is (x + 22) degrees. Angle B is x degrees. What is the

HACTEHA [7]

Answer: 34

Step-by-step explanation:

(x+22) + x = 90

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3 years ago

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Help me pls I would really appreciate it

Paul [167]

The first the last and the 2nd last

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You estimated the average rental cost of a 2 bedroom apartment in Denton. A random sample of 16 apartments was taken. The sample

Zigmanuir [339]

Answer:

The new sample size required in order to have the same confidence 95% and reduce the margin of erro to $60 is:

n=28

Step-by-step explanation:

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".

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma=160)$

And the distribution for $\bar X$ is:

$\bar X \sim N(\mu, \frac{160}{\sqrt{n}})$

We know that the margin of error for a confidence interval is given by:

$Me=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The next step would be find the value of $\z_{\alpha/2}$ , $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$

Using the normal standard table, excel or a calculator we see that:

$z_{\alpha/2}=\pm 1.96$

If we solve for n from formula (1) we got:

$\sqrt{n}=\frac{z_{\alpha/2} \sigma}{Me}$

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

And we have everything to replace into the formula:

$n=(\frac{1.96(160)}{78.4})^2 =16$

And this value agrees with the sample size given.

For the case of the problem we ar einterested on Me= $60, and we need to find the new sample size required to mantain the confidence level at 95%. We know that n is given by this formula:

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

And now we can replace the new value of Me and see what we got, like this:

$n=(\frac{1.96*160}{60})^2 =27.32$

And if we round up the answer we see that the value of n to ensure the margin of error required $Me=\pm 60$ $ is n=28.

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3 years ago

What is the answer? How do I create two equations?

tatyana61 [14]

The answer may be 36 but I am not very sure

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