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3 years ago

What is the value of k + 12, if k=6?

Mathematics

2 answers:

Makovka662 [10]3 years ago

5 0

18
literally you do 6+12

deff fn [24]3 years ago

4 0

Answer:

Step-by-step explanation:

If k = 6, the equation would be 6 + 12, which is equal to 18.

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Aliun [14]

I would say (I am abbreviating square root to sqrt)

1. 4 sqrt(8488) = 4 sqrt(2x2x2122) = 8 sqrt(2122)

2. 3 sqrt(6936) = 3 sqrt(6x34x34) = 102 sqrt (6)

3. 2 sqrt(8648) = 2 sqrt(2x2x2162) = 4 sqrt(2162)

3 0

2 years ago

A rectangular pan has a length that is the width. The total area of the pan is 432 in.2. What is the width of the cake pan? A =

spayn [35]

Answer:

Let width = w

length = 4w/3

Area = 432 = length*width = w*4w/3

4w^2/3 = 432

w^2 = 432*3/4 = 324

w = 18

so the width of the cake pan is 18 inches.

4 0

3 years ago

A programmer plans to develop a new software system. In planning for the operating system that he will use, he needs to estimat

Vedmedyk [2.9K]

Using the z-distribution, we have that:

a) A sample of 601 is needed.

b) A sample of 93 is needed.

c) A. Yes, using the additional survey information from part (b) dramatically reduces the sample size.

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which z is the z-score that has a p-value of $\frac{1+\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence level, hence $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so $z = 1.96$ .

For this problem, we consider that we want it to be within 4%.

Item a:

The sample size is n for which M = 0.04.

There is no estimate, hence $\pi = 0.5$

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.04 = 1.96\sqrt{\frac{0.5(0.5)}{n}}$

$0.04\sqrt{n} = 1.96\sqrt{0.5(0.5)}$

$\sqrt{n} = \frac{1.96\sqrt{0.5(0.5)}}{0.04}$

$(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.5(0.5)}}{0.04}\right)^2$

$n = 600.25$

Rounding up:

A sample of 601 is needed.

Item b:

The estimate is $\pi = 0.96$ , hence:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.04 = 1.96\sqrt{\frac{0.96(0.04)}{n}}$

$0.04\sqrt{n} = 1.96\sqrt{0.96(0.04)}$

$\sqrt{n} = \frac{1.96\sqrt{0.96(0.04)}}{0.04}$

$(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.96(0.04)}}{0.04}\right)^2$

$n = 92.2$

Rounding up:

A sample of 93 is needed.

Item c:

The closer the estimate is to $\pi = 0.5$ , the larger the sample size needed, hence, the correct option is A.

For more on the z-distribution, you can check brainly.com/question/25404151

8 0

2 years ago

Jess wants 180 cupcakes for her party. Jess has already made 60 berry cupcakes, and 48

Pavel [41]

Answer:

Hi! The answer to your question is 72 Cupcakes

Step-by-step explanation:

$60+48 = 108$

$180 - 108 = 72$

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Hope this helps!!

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5 0

2 years ago

Help please asap thanks

sleet_krkn [62]

It would be B

Hope this helps! ;)

4 0

3 years ago