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3 years ago

Is y=6x+2 proportional

Mathematics

2 answers:

gizmo_the_mogwai [7]3 years ago

7 0

Yes y=6x+2 is proportional

padilas [110]3 years ago

3 0

Answer: yes they are proportional

Step-by-step explanation:

The variables y and x are directly proportional to each other.

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Circle O has a circumference of 36tt cm. what is length of the radius r

allochka39001 [22]

Answer:

5.73cm

Step-by-step explanation:

Given parameters:

Circumference of the circle = 36cm

Unknown

Length of the radius = ?

Lets represent the radius by r

Solution

The circumference of a circle is the defined as the perimeter of a circle. The formula is given as:

Circumference of a circle = 2πr

Since the unknown is r, we make it the subject of the formula:

r = $\frac{circumference of the circle}{2π}$

r = $\frac{36}{2 x 3.142}$ = $\frac{36}{6.284}$ = 5.73cm

3 0

3 years ago

Read 2 more answers

Why is the slope between any two points on a straight line always the same?

mixas84 [53]

Answer:

because it never changes.

7 0

3 years ago

Can anyone help me i really need it rn<br>

Hunter-Best [27]

<u>Answer:</u>

11. g

12. f

13. e (add all the frequencies)

14. d

15. c (eg. 17.5 -8.5 = 9)

16. b

17. a

Let me know if you have any questions.

Hope this helps!

4 0

2 years ago

Which one of the relationships described is not proportional?

ivann1987 [24]

Answer:

Marianne is 15 years old and her brother is 5 years old; is the relationship between their ages proportional?

Step-by-step explanation:

i just did it

8 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

4 years ago