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2 years ago

FIND THE SURFACE AREA OF PYRAMID (HELP!)

Mathematics

2 answers:

ollegr [7]2 years ago

8 0

You have to do length of the base x width of the base x height of pyramid

Vaselesa [24]2 years ago

8 0

Answer:

110 in

Step-by-step explanation:

You might be interested in

Evaluate. [(1 1/3−2/3)⋅(−3/4)^3]÷(−2) What is the value of the expression? Enter your answer as a simplified fraction in the box

AleksAgata [21]

Answer

9/64

Step-by-step explanation:

(1 1/3- 2/3) x (-3/4) divided by (-2) is the answer

bye have a nice day

7 0

2 years ago

We would like to create a confidence interval.

Vlada [557]

Answer:

c.A 90% confidence level and a sample size of 300 subjects.

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 $1-\alpha$ , we have the confidence interval with a margin of error of:

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

In which

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

In this problem

The proportions are the same for all the options, so we are going to write our margins of error as functions of $\sqrt{\pi(1-\pi)}$

a.A 99% confidence level and a sample size of 50 subjects.

$n = 50$

99% confidence interval

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The margin of error is

$M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{2.575}{\sqrt{50}}\sqrt{\pi(1-\pi)} = 0.3642\sqrt{\pi(1-\pi)}$

b.A 90% confidence level and a sample size of 50 subjects.

$n = 50$

90% confidence interval

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

The margin of error is

$M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{1.645}{\sqrt{50}}\sqrt{\pi(1-\pi)} = 0.2623\sqrt{\pi(1-\pi)}$

c.A 90% confidence level and a sample size of 300 subjects.

$n = 300$

90% confidence interval

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

The margin of error is

$M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{1.645}{\sqrt{300}}\sqrt{\pi(1-\pi)} = 0.0950\sqrt{\pi(1-\pi)}$

This produces smallest margin of error.

d.A 99% confidence level and a sample size of 300 subjects.

$n = 300$

99% confidence interval

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The margin of error is

$M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{2.575}{\sqrt{300}}\sqrt{\pi(1-\pi)} = 0.1487\sqrt{\pi(1-\pi)}$

6 0

3 years ago

Which of these is least likely to be the average salary of another of the groups?

Finger [1]

Answer:

$104,000

The probability that the average salary between two groups is the same, is actually low. It could be close, but it's quite difficult to get the same exact amount.

3 0

3 years ago

write a linear equation that intersects y=x^2 at two points. Then write a second linear equation that intersects y=x^2 at just o

Liula [17]

We know that $y = x^2$ is a parabola, concave up, with vertex in the origin $(0,0)$

So, we may use three horizontal lines for our purpose: any horizontal line above the x axis will intersect the parabola twice. The x axis itself intersects the parabola once on the vertex, while any horizontal line below the x axis won't intercept the parabola.

Here's the examples:

The horizontal line $y = 4$ intercepts the parabola twice: the system $y = x^2,\ y = 4$ is solved by $x^2=4 \implies x = \pm 2$
The horizontal line $y=0$ intercepts the parabola only once: the system is $y=x^2,\ y=0$ which yields $x^2=0\implies x=0$ which is a repeated solution
The horizontal line $y=-5$ intercepts the parabola only once: the system is $y=x^2,\ y=-5$ which yields $x^2=-5$ which is impossible, because a squared number can't be negative.