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2 years ago

Prove that the angle sum of any triangle is 180°

1 answer:

3 0

I will be philosophical here...

A square is 4 -90degree angles (definition of perpendicular lines)

If we create a diagonal it splits it into two triangles which in result gives us

90degree + 90/2 degree+ 90/2 degree (because the diagonal split it in half and one full 90 by definition of perpendicular lines)

That gives us the beautiful 180 degrees

which is also what humans use to say that they’ve completely changed or did something completely different...

Mysteries of the triangle...

Six six six six six six

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A random sample of 400 Michigan State University (MSU) students were surveyed recently to determine an estimate for the proporti

Gre4nikov [31]

Answer:

a. .92

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 of the interval is:

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

The lower bound is the point estimate $\pi$ subtracted by the margin of error.

The upper bound is the point estimate $\pi$ added to the margin of error.

Point estimate:

The confidence interval is symmetric, so it is the mean between the two bounds.

In this problem:

$\pi = \frac{0.372 + 0.458}{2} = 0.415$

Sample of 400, which means that $n = 400$

Margin of error is the estimate subtracted by the lower bound. So $M = 0.415 - 0.372 = 0.043$

We have to find z.

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

$0.043 = z\sqrt{\frac{0.415*0.585}{400}}$

$z = \frac{0.043\sqrt{400}}{\sqrt{0.415*0.585}}$

$z = 1.745$

$z = 1.745$ has a pvalue of 0.96.

This means that:

$1 - \frac{\alpha}{2} = 0.96$

$\frac{\alpha}{2} = 1 - 0.96$

$\frac{\alpha}{2} = 0.04$

$\alpha = 0.08$

Confidence level:

$1 - \alpha = 1 - 0.08 = 0.92$

So the correct answer is:

a. .92

7 0

2 years ago

A storage container is in the shape of a rectangular prism. The container has a length of 10 feet, a width of 8 feet, and a heig

SVETLANKA909090 [29]

Answer:

592

Step-by-step explanation:

A=2(wl+hl+hw)

A=2(8*10+12*10+12*8)=592

8 0

2 years ago

Write an expression for the area of the triangle below.

vfiekz [6]

It should be A= 1/2bh but im not sure why you wrote those extra things. Are we supposed to use them?

7 0

2 years ago

a daily newspaper has 10225 subscribers when it began publication. six years old later it has 8200 subscribers what is the avera

seraphim [82]

Answer:

The rate of change in number of subscriber for the six years is 3.62%

Step-by-step explanation:

Given as :

The initial subscriber of newspaper = p = 10225

The subscriber of newspaper after 6 years of publish = P = 8200

The time period = t = 6 years

Let The average yearly rate of change = r%

<u>Now, According to question</u>

The subscriber of newspaper after n years = initial subscriber × $(1-\dfrac{\textrm rate}{100})^{\textrm time}$

Or, P = p × $(1-\dfrac{\textrm r}{100})^{\textrm t}$

Or, 8200 = 10225 × $(1-\dfrac{\textrm r}{100})^{\textrm 6}$

Or, $\dfrac{8200}{10225}$ = $(1-\dfrac{\textrm r}{100})^{\textrm 6}$

Or, 0.80195 = $(1-\dfrac{\textrm r}{100})^{\textrm 6}$

<u>Taking power $\dfrac{1}{6}$ both side</u>

So, $(0.80195 )^{\frac{1}{6}}$ = $((1 - \frac{r}{100} )^{6})^{\frac{1}{6}}$

Or, 0.9638 = 1 - $\dfrac{r}{100}$

Or, $\dfrac{r}{100}$ = 1 - 0.9638

Or, $\dfrac{r}{100}$ = 0.0362

Or, r = 0.0362 × 100

i.e r = 3.62

So, The rate of change in subscriber = 3.62%

Hence, The rate of change in number of subscriber for the six years is 3.62% . Answer

4 0

2 years ago

How does factoring help solve quadratic equations?

n200080 [17]

Answer:

o solve an quadratic equation using factoring : 1 . Transform the equation using standard form in which one side is zero. ... Set each factor to zero (Remember: a product of factors is zero if and only if one or more of the factors is zero).

Step-by-step explanation:

5 0

2 years ago