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3 years ago

15

What postulate or theorem can be used to prove that these two triangles are congruent?

1 answer:

Andrei [34K]3 years ago

3 0

Step-by-step explanation:

The attached figure shows two triangles AOB and MON.

In triangles AOB and MON,

AB = MN (Given )

∠AOB = ∠MON (vertically opposite angles)

∠O = ∠O (common)

∴ ΔAOB ≅ Δ MON (ASA)

Angle-side-angle is the postulate that can be used to prove that two triangles are congruent.

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Jonah wants to construct a confidence interval using 90% confidence to estimate what proportion of silicon wafers at his factory

Nimfa-mama [501]

Answer:

170

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 of:

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

Assume:

$\pi = 0.06$

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 $Z = 1.645$ .

What is the smallest sample size required to obtain the desired margin of error?

This is n for which M = 0.03. So

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

$0.03 = 1.645\sqrt{\frac{0.06*0.94}{n}}$

$0.03\sqrt{n} = 1.645\sqrt{0.06*0.94}$

$\sqrt{n} = \frac{1.645\sqrt{0.06*0.94}}{0.03}$

$(\sqrt{n})^2 = (\frac{1.645\sqrt{0.06*0.94}}{0.03})^2$

$n = 169.6$

Rounding up, 170.

4 0

3 years ago

A coed little league team had 18 players . Even though it was raining half the girls and one third of the boys showed up for the

Genrish500 [490]

Answer:

The number of Girls players in the Team of 18 players is 6 .

Step-by-step explanation:

Given as :

The total number of player including boys and girls = 18

And $\frac{1}{2}$ times girls player + $\frac{1}{3}$ times boys player = 7

Let The number of boys player = B

The number of girls player = G

So, According to question

G + B = 18 ...1

$\frac{1}{2}$ × G + $\frac{1}{3}$ × B= 7

or, $\frac{G}{2}$ + $\frac{B}{3}$ = 7 ...2

or, Put the value of B from eq 1

or, $\frac{G}{2}$ + $\frac{18 - G}{3}$ = 7

Or, $\frac{G}{2}$ - $\frac{G}{3}$ + 6 = 7

or, $\frac{3G - 2G}{6}$ = 7 - 6

Or, $\frac{G}{6}$ = 1

∴ G = 1 × 6 = 6

Hence The number of Girls players in the Team of 18 players is 6 . Answer

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4 years ago

Can someone help me with these 3 problems. I’m stuck and it’s for a grade..

marin [14]

Axis solutions has to be coordinated

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3 years ago

Ms.mitchells coin purse has 20 coins there are 6 pennies are, 4 quarters, 3 dimes, and the remander are nickels

enot [183]

6+4+3= 13 coins

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3 years ago

What is the slope of the line

Keith_Richards [23]

You can see that for every one grid square to the right, the graph moves two grid squares up. Therefore the slope is 2.

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