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2 years ago

HELP ME PLEASEEEEE QUESTION IS IN IMAGE

Mathematics

2 answers:

Daniel [21]2 years ago

8 0

Answer:

I think it is the first answer but I am not 100% sure..

Step-by-step explanation:

SVETLANKA909090 [29]2 years ago

7 0

Answer:

ACB-ADC-CDB

Step-by-step explanation:

So the most obvious one is ACB

B is the farther point, C is the right angled point, and A is the one left (lol)

The other triangles would be named ADC and CDB (following the far-right-left pattern)

Then we get

ACB-ADC-CDB

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PQRS is an isosceles trapezoid and m angle Q=127. Find M angle P.

Allisa [31]

PQRS is an isosceles trapezoid and m < Q=127

so m<Q = m<P =127

answer
127

7 0

3 years ago

Read 2 more answers

From 7x^2+5xy-3y^2 subtract the sum of 2x^2+3xy-5y^2 and x^2-7xy-4y^2

Crazy boy [7]

Answer:

4x^2 +9xy +6y^2

Step-by-step explanation:

Combine like terms. I find it useful to factor out the variable part of the expression so I can see what coefficients are being combined.

7x^2+5xy-3y^2 - (2x^2+3xy-5y^2 + x^2-7xy-4y^2)

= (7 -2 -1)x^2 +(5 -3+7)xy +(-3 +5 +4)y^2

= 4x^2 +9xy +6y^2

5 0

3 years ago

A candidate for a US Representative seat from Indiana hires a polling firm to gauge her percentage of support among voters in he

UkoKoshka [18]

Answer:

a) The minimum sample size is 601.

b) The minimum sample size is 2401.

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 is:

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

For this problem, we have that:

We dont know the true proportion, so we use $\pi = 0.5$ , which is when we are are going to need the largest sample size.

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

a. If a 95% confidence interval with a margin of error of no more than 0.04 is desired, give a close estimate of the minimum sample size that will guarantee that the desired margin of error is achieved. (Remember to round up any result, if necessary.)

This is n for which M = 0.04. So

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

$0.04 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.04\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.04}$

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

$n = 600.25$

Rounding up

The minimum sample size is 601.

b. If a 95% confidence interval with a margin of error of no more than 0.02 is desired, give a close estimate of the minimum sample size necessary to achieve the desired margin of error.

Now we want n for which M = 0.02. So

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

$0.02 = 1.96\sqrt{\frac{0.5*0.5}{n}}$