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3 years ago

What does it mean when a line of best fit has a correlation coefficient close to 0?

1 answer:

3 0

A correlation coefficient close to 0 shows that there is no correlation with the points on the graph or your "kine of best fit" does not fit the graph

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Divide number 60 into three parts in a ratio of 2:3:5

katovenus [111]

2:3:5.....added = 10

2/10(60) = 120/10 = 12 <=
3/10(60) = 180/10 = 18 <=
5/10(60) = 300/10 = 30 <=

5 0

3 years ago

Read 2 more answers

Decide if the following is a true or false statement - explain how you

NARA [144]

Answer:

true

Step-by-step explanation:

(-3) -(-4.25) = -3 + 4.25 = 1.25 (positive)

6 0

3 years ago

Suppose that LMN is isosceles with base LN suppose that M

ira [324]

Given:

In an isosceles triangle LMN, LM=MN.

$m\angle M=(3x+17)^\circ,m\angle L=(2x+36)^\circ$

To find:

The measure of the angles L, M and N.

Solution:

In triangle LMN,

$LM=MN$ (Given)

$m\angle N=m\angle L=(2x+36)^\circ$ (Base angles of an isosceles triangle are equal)

Now,

$m\angle L+m\angle M+m\angle N=180^\circ$

$(2x+36)^\circ+(3x+17)^\circ+(2x+36)^\circ=180^\circ$

$(7x+89)^\circ=180^\circ$

$(7x+89)=180$

On further simplification, we get

$7x=180-89$

$7x=91$

$x=\dfrac{91}{7}$

$x=13$

The value of x is 13. Using this value, we get

$m\angle L=(2(13)+36)^\circ$

$m\angle L=(26+36)^\circ$

$m\angle L=62^\circ$

Similarly,

$m\angle M=(3(13)+17)^\circ$

$m\angle M=(39+17)^\circ$

$m\angle M=56^\circ$

And,

$m\angle N=m\angle L$

$m\angle N=62^\circ$

Therefore, the measure of angles are $m\angle L=62^\circ,m\angle M=56^\circ,m\angle N=62^\circ$ .

5 0

3 years ago

You need to solve for X

zhenek [66]

Look at the picture.

We have the proportion:

$\dfrac{x}{2}=\dfrac{305}{122}\ \ \ \ |\cdot2\\\\x=\dfrac{305}{61}\\\\x=5\ ft$

Answer: Rayan is 5 ft tall.

5 0

3 years ago

X and y are normal random variables with e(x) = 2, v(x) = 5, e(y) = 6, v(y) = 8 and cov(x,y)=2. determine the following: e(3x 2y

andriy [413]

The result for the given normal random variables are as follows;

a. E(3X + 2Y) = 18

b. V(3X + 2Y) = 77

c. P(3X + 2Y < 18) = 0.5

d. P(3X + 2Y < 28) = 0.8729

<h3>What is normal random variables?</h3>

Any normally distributed random variable having mean = 0 and standard deviation = 1 is referred to as a standard normal random variable. The letter Z will always be used to represent it.

Now, according to the question;

The given normal random variables are;

E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8.

Part a.

Consider E(3X + 2Y)

$\begin{aligned}E(3 X+2 Y) &=3 E(X)+2 E(Y) \\&=(3) (2)+(2)(6 )\\&=18\end{aligned}$