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2 years ago

What is the surface area of a cube with a side length of 6 m?

Mathematics

1 answer:

Ahat [919]2 years ago

4 0

Answer:

216 m²

Step-by-step explanation:

Surface area of a cube = 6a², when a = length of one side

so,

6a²

= 6×6²

= 6×36

= 216 m²

Answered by GAUTHMATH

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A student has gotten the following grades on his tests. What is the minimum grade he must get on the last test in order to have

klemol [59]

Answer:

Step-by-step explanation:

I used a calculator and if you add 76, 93, 89, 71 to 91 it equals 420 and if you divide that by 5 it is 84.

Hope this helps!

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2 years ago

A line has a slope of 5 and passes through the point

podryga [215]

Answer:

Step-by-step explanation:

The equation is mx+b=y.

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x=x coordinates

b=y intercept

y=y coordinates

For the x and y coordinates, I'll use (2,0).

5(2)+_=0

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3 years ago

Elaborar 10 ejemplos de números decimales con tres términos por ejemplo : 3.99+0.000345+322.99

maksim [4K]

Answer:

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1 year ago

Exercise 6.13 presents the results of a poll evaluating support for the health care public option in 2009, reporting that 52% of

sleet_krkn [62]

Answer:

A sample size of 6755 or higher would be appropriate.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

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

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error M is given by:

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

90% confidence level

So $\alpha = 0.1$ , z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

52% of Independents in the sample opposed the public option.

This means that $p = 0.52$

If we wanted to estimate this number to within 1% with 90% confidence, what would be an appropriate sample size?

Sample size of size n or higher when $M = 0.01$ . So

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

$0.01 = 1.645\sqrt{\frac{0.52*0.48}{n}}$

$0.01\sqrt{n} = 0.8218$

$\sqrt{n} = \frac{0.8218}{0.01}$

$\sqrt{n} = 82.18$

$\sqrt{n}^{2} = (82.18)^{2}$

$n = 6754.2$

A sample size of 6755 or higher would be appropriate.

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2 years ago

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Dafna1 [17]

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2 years ago