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3 years ago

Two lines intersect and two of the vertical.angles measure 37 what is the measure of the other two vertical angles

2 answers:

4 0

Lets start of with what we know.
• There are two 27 degree angles 27 + 27 = 52 is the sum of the angles.
•There are 360 degrees all around the intersection

So, we can find out the sum of the 2 angles that are unknown by subtracting.
360-52=308

So, if the sum of the unknown 2 angles are 308, we can divide by 2 to find the measure of the unknown angles.
308/2=154

Andre45 [30]3 years ago

4 0

When line intersect the two vertical angle adds up to 180 degrees
To find the other vertical angle:

180 -37=143 degrees

So for two vertical angles
we take (180*2)-(37*2)=286 degrees

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Two integers, a and b have a product of 15. what is the least possible sum of a and b?

Neporo4naja [7]

If they're integers, that means they can't be fractions (such as 15/2 and 2).

So the only couples of integers that, when multiplied, result in 15 are:

1 and 15
5 and 3.

(there are no other possibilities as both 5 and 3 are not divisible further).
The smaller of the two sums (15+1; 5+3) is 5+3=8.

The least possible sum of a and b is 8.

7 0

3 years ago

Suppose we roll a fair die and let X represent the number on the die. (a) Find the moment generating function of X. (b) Use the

Likurg_2 [28]

Answer:

(a) moment generating function for X is $\frac{1}{6}\left(e^{t}+e^{2 t}+e^{2 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$

(b) $\mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}$

Step-by step explanation:

Given X represents the number on die.

The possible outcomes of X are 1, 2, 3, 4, 5, 6.

For a fair die, $P(X)=\frac{1}{6}$

(a) Moment generating function can be written as $M_{x}(t)$ .

$M_x(t)=\sum_{x=1}^{6} P(X=x)$

$M_{x}(t)=\frac{1}{6} e^{t}+\frac{1}{6} e^{2 t}+\frac{1}{6} e^{3 t}+\frac{1}{6} e^{4 t}+\frac{1}{6} e^{5 t}+\frac{1}{6} e^{6 t}$

$M_x(t)=\frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$

(b) Now, find $E(X) \text { and } E\((X^{2})$ using moment generating function

$M^{\prime}(t)=\frac{1}{6}\left(e^{t}+2 e^{2 t}+3 e^{3 t}+4 e^{4 t}+5 e^{5 t}+6 e^{6 t}\right)$

$M^{\prime}(0)=E(X)=\frac{1}{6}(1+2+3+4+5+6)$

$\Rightarrow E(X)=\frac{21}{6}$

$M^{\prime \prime}(t)=\frac{1}{6}\left(e^{t}+4 e^{2 t}+9 e^{3 t}+16 e^{4 t}+25 e^{5 t}+36 e^{6 t}\right)$

$M^{\prime \prime}(0)=E(X)=\frac{1}{6}(1+4+9+16+25+36)$

$\Rightarrow E\left(X^{2}\right)=\frac{91}{6}$

Hence, (a) moment generating function for X is $\frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$ .

(b) $\mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}$

6 0

4 years ago

25x + 4x = x + 3x +7

Musya8 [376]

For this equation its the same as simplifying any other equation -Simplify both sides of the equation then isolate the variable- Easy then once you've do so the answer should be

7 0

3 years ago

Read 2 more answers

Pls help I dont understand. Find the inverse of this function.

lisov135 [29]

In the inverse we replace the place of x by y .

$g(x) = \sqrt[3]{x} - 3 \\ \\ x = \sqrt[3]{y} - 3 \\ \sqrt[3]{y} = x + 3 \\ y = {(x + 3)}^{3} \\ y = {(x + 3)}^{2} (x + 3) \\ y = ({x}^{2} + 6x + 9)(x + 3) \\ \\ y = {x}^{3} + 6 {x}^{2} + 9x + 3 {x}^{2} + 18x + 27 \\ \\ y = {x}^{3} + 9 {x}^{2} + 27x + 27 \\ \\ \\ g(x)^{ - 1} = {x}^{3} + 9 {x}^{2} + 27x + 27$

I hope I helped you^_^

8 0

3 years ago

"Four times the difference of -16 and 7 amounts to -92 " written as an equation is

Ira Lisetskai [31]

You put the -16 - 7 in parenthesis and then the 4 infront of that. So the equation will look like 4(-16 - 7) = -92

7 0

3 years ago