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2 years ago

During his last 5 jobs, an electrician used the following amounts of cable:

Mathematics

1 answer:

ella [17]2 years ago

7 0

The total amount of cable used is 47.4m

Explanation:

Given:

Amounts of cable used that has to be rounded off to 1 decimal place:

12.995 m ≈ 13.0 m

7.505 m ≈ 7.5m

10.08 m ≈ 10.1 m

6.95 m ≈ 7.0 m

9.75 m ≈ 9.8 m

Total amount of cable used = 13m + 7.5m + 10.1m + 7m + 9.8m

= 47.4m

Therefore, total amount of cable used is 47.4m

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Given regular pentagon ABCDE, what are the coordinates of vertex C?

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Suppose that a jewelry store tracked the amount of emeralds they sold each week to more accurately estimate how many emeralds to

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Answer:

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

And for this case the 95% confidence interval is given by (2.13; 2.37)

We have a point of estimate for the sample mean with this formula:

$\bar X = \frac{Upper+ Lower}{2}= \frac{3.37+2.13}{2}= 2.75$

And for the margin of error we have the following estimation:

$ME= \frac{Upper -Lower}{2}= \frac{3.37-2.13}{2}= 0.62$

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".

$\bar X$ represent the sample mean

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

And for this case the 95% confidence interval is given by (2.13; 2.37)

We have a point of estimate for the sample mean with this formula:

$\bar X = \frac{Upper+ Lower}{2}= \frac{3.37+2.13}{2}= 2.75$

And for the margin of error we have the following estimation:

$ME= \frac{Upper -Lower}{2}= \frac{3.37-2.13}{2}= 0.62$

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2 years ago

Animals: The label on a bottle of pet vitamins lists dosage guidelines. What dosage would you give to each of these animals? (a)

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Answer:

a) 1 tsp.

b) 1.5 tsp.

c) 2 tsp.

Step-by-step explanation:

Complete Correct Question

Animals: The label on a bottle of pet vitamins lists dosage guidelines. What dosage would you give to each of these animals? (a) a 18 kg adult dog (b) a 6 kg cat (c) 18 kg pregnant dog

- The dosage listed for adult dog is 1/2 tsp per 9 kg dog weight.

- The dosage listed for adult cat is 1/4 tsp per 1 kg weight.

- The dosage listed for pregnant dog is 1/2 tsp per 4.5 kg dog weight.

Solution

a) 18 kg adult dog

The dosage listed for adult dog is 1/2 tsp per 9 kg dog weight.

9 kg dog weight = (1/2) tsp

18 kg adult dog = (18/9) × (1/2) = 1 teaspoon.

b) 6 kg cat

The dosage listed for adult cat is 1/4 tsp per 1 kg weight.

1 kg cat = (1/4) tsp

6 kg cat = (6/1) × (1/4) = (6/4) = 1.5 teaspoons.

c) 18 kg pregnant dog

The dosage listed for pregnant dog is 1/2 tsp per 4.5 kg dog weight.

4.5 kg dog weight = (1/2) tsp

18 kg pregnant dog = (18/4.5) × (1/2) = 2 teaspoons.

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