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3 years ago

12

PLEASE HELP!! NO LINKS!! GIVING 20 POINTS AWAY!!

1 answer:

shutvik [7]3 years ago

5 0

Answer:

$m\angle N = 32\degree$

$m\widehat {NQ}=106\degree$

Step-by-step explanation:

By inscribed angle theorem:

$m\angle N = \frac{1}{2} \times m\widehat {MP}$

$m\angle N = \frac{1}{2} \times 64\degree$

$m\angle N = 32\degree$

Again by inscribed angle theorem:

$m\angle P= \frac{1}{2} \times m\widehat {NQ}$

$53\degree = \frac{1}{2} \times m\widehat {NQ}$

$2\times 53\degree = m\widehat {NQ}$

$m\widehat {NQ}=106\degree$

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Firlakuza [10]

Have the kids answer questions, if correct they get to shoot a small ball into a garbage can for extra points! FUN!
~Hope This Helps! <3

4 0

3 years ago

3 1/9+2 1/4=<br> what is the answer

34kurt

<em><u>Answer:</u></em>
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Step-by-step explanation:

3 0

3 years ago

Conditions of a triangle = AAA

34kurt

Answer:

One triangle

Step-by-step explanation:

That's the answer....

3 0

3 years ago

4. A random variable X has a mean of 10 and a standard deviation of 3. If 2 is added to each value of X, what will the new mean

Ede4ka [16]

Adding 2 to each value of the random variable $X$ makes a new random variable $X+2$ . Its mean would be

$E[X+2]=E[X]+E[2]=E[X]+2$

since expectation is linear, and the expected value of a constant is that constant. $E[X]$ is the mean of $X$ , so the new mean would be

$E[X+2]=10+2=12$

The variance of a random variable $X$ is

$V[X]=E[X^2]-E[X]^2$

so the variance of $X+2$ would be

$V[X+2]=E[(X+2)^2]-E[X+2]^2$

We already know $E[X+2]=12$ , so simplifying above, we get

$V[X+2]=E[X^2+4X+4]-12^2$

$V[X+2]=E[X^2]+4E[X]+4-12^2$

$V[X+2]=(V[X]+E[X]^2)+4E[X]-140$

Standard deviation is the square root of variance, so $V[X]=3^2=9$ .

$\implies V[X+2]=(9+10^2)+4(10)-140=9$

so the standard deviation remains unchanged at 3.

NB: More generally, the variance of $aX+b$ for $a,b\in\mathbb R$ is

$V[aX+b]=a^2V[X]+b^2V[1]$

but the variance of a constant is 0. In this case, $a=1$ , so we're left with $V[X+2]=V[X]$ , as expected.

5 0

3 years ago

John has a map that shows a rectangular park with a circular pond. The scale on the map is 114 cm : 8 yd. On the map, the width

Ivenika [448]

Answer:

The area of the park is 119,966.14 cm if needed to round it would be 119,966 cm.

Step-by-step explanation:

first we would use A=length x width

so a = 614 cm x 212 cm = 130,168 cm

now we find the area of the pond using a= pi x r^2

the diameter is 114 cm so we cut that in half to find the radius which is 57 cm

a = pi x 57^2

a= 3.14 x 3249

a= 10201.86 cm

now we subtract the area of the pond from the whole area

130,168 - 10201.86 = 119,966.14 cm

The area of the park is 119,966.14 cm if needed to round it would be 119,966 cm

4 0

3 years ago