Zach did 1/6 of his math homework in 1/2 an hour. How much can he do in one hour?

1 answer:

8 0

Answer: 1/3
Explanation: Zach does 1/6 in 30 minutes (1/2 an hour), in an hour Zach can complete 2/6 of his homework. Simplify to 1/3.

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Sophia works in a clothing store and earns $40 per day. She earns an extra $5 for each outfit she sells. If Sophia wants to earn

andrew-mc [135]

Answer:

In order to find this answer, you will need to be able to make an inequality.

Step 1: Sophia makes $40 so start with that. 40

Step 2: Sophia also makes $5 per every outfit that she sells, so add that plus your variable (n). 40+5n

Step 3: Determine if Sophia wants to make more, equal to or more, less, or less than or equal to than what she wants to make per day. 40+5n *draw a greater than or equal to sign.

Step 4: Sophia wants to make more than or equal to what she makes per day. You will need to find what she wants to make more than. In this she wants to make more than or equal to $70.

Equation: 40+5n *greater than or equal to sign* $70

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Suppose Brianna decided that the error margin of her 95% confidence interval was too large and wanted an error margin of .05 whi

svetoff [14.1K]

Answer:

$n=\frac{0.26 (1-0.26)}{(\frac{0.05}{1.96})^2} =295.65 \approx =296$ (b)

Rounded up would be n =296

Step-by-step explanation:

Assuming this previous info: She randomly interested in the p selects 100 engineering majors, and 26 of them said they earned an A in Calc 1.

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".

Solution to the problem

The population proportion have the following distribution

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

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

If solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by: