Also, explain why and how you got it right and the one that's red is wrong

PLEASE HELP BRAINLIEST WILL BE GIVEN IMMEDIATELY WITHOUT ANY WAIT

gizmo_the_mogwai [7]

y = 3/2x + b

Plug in the point (2,2) and solve for b.

2 = (3/2)(2) + b

2 = 3 + b

b = -1

Final equation is y = 3/2x - 1.

4 0

2 years ago

Last year the math club had nine sixth grade students and 4 4th grade students what is the radio of 4th grade students to the to

slega [8]

Answer:

C. 4:13

Step-by-step explanation:

there are 4 students in the 4th grade and 4+9= 13 so it is 4:13

3 0

2 years ago

Let A and B be events with =PA0.7, =PB0.3, and =PA or B0.9. (a) Compute PA and B. (b) Are A and B mutually exclusive? Explain. (

kvasek [131]

Answer:

a) $P(A \cup B) = P(A) +P(B) - P(A\cap B)$

And if we solve for $P(A \cap B)$ we got:

$P(A \cap B) = P(A) + P(B) -P(A\cup B)= 0.7+0.3-0.9 = 0.1$

b) False

The reason is because we don't satisfy the following relationship:

$P(A\cup B) = P(A) + P(B)$

We have that:

$0.9 \neq 0.3+0.7 =1$

c) False

In order to satisfy independence we need to have the following condition:

$P(A \cap B) = P(A) *P(B)$

And for this case we don't satisfy this relation since:

$0.1 \neq 0.7*0.3 = 0.21$

Step-by-step explanation:

For this case we have the following probabilities given:

$P(A) = 0.7, P(B) =0.7, P(A \cup B) =0.9$

Part a

We want to calculate the following probability: $P(A \cap B)$

And we can use the total probability rule given by:

$P(A \cup B) = P(A) +P(B) - P(A\cap B)$

And if we solve for $P(A \cap B)$ we got:

$P(A \cap B) = P(A) + P(B) -P(A\cup B)= 0.7+0.3-0.9 = 0.1$

Part b

False

The reason is because we don't satisfy the following relationship:

$P(A\cup B) = P(A) + P(B)$

We have that:

$0.9 \neq 0.3+0.7 =1$

Part c

False

In order to satisfy independence we need to have the following condition:

$P(A \cap B) = P(A) *P(B)$

And for this case we don't satisfy this relation since:

$0.1 \neq 0.7*0.3 = 0.21$

4 0

3 years ago

Which score indicates the highest relative position? I. A score of 2.6 on a test with X = 5.0 and s = 1.6 II. A score of 650 on

Zanzabum

Answer:

A score of 2.6 on a test with $\bar X$ = 5.0 and s = 1.6 and A score of 48 on a test with $\bar X$ = 57 and s = 6 indicate the highest relative position.

Step-by-step explanation:

We are given the following:

I. A score of 2.6 on a test with $\bar X$ = 5.0 and s = 1.6

II. A score of 650 on a test with $\bar X$ = 800 and s = 200

III. A score of 48 on a test with $\bar X$ = 57 and s = 6

And we have to find that which score indicates the highest relative position.

For finding in which score indicates the highest relative position, we will find the z score for each of the score on a test because the higher the z score, it indicates the highest relative position.

The z-score probability distribution is given by;

Z = $\frac{X-\bar X}{s}$ ~ N(0,1)

where, $\bar X$ = mean score

s = standard deviation

X = each score on a test

The z-score of First condition is calculated as;

Since we are given that a score of 2.6 on a test with $\bar X$ = 5.0 and s = 1.6,

So, z-score = $\frac{2.6-5}{1.6}$ = -1.5 {where $\bar X = 5.0$ and s = 1.6 }

The z-score of Second condition is calculated as;

Since we are given that a score of 650 on a test with $\bar X$ = 800 and s = 200,

So, z-score = $\frac{650-800}{200}$ = -0.75 {where $\bar X = 800$ and s = 200 }

The z-score of Third condition is calculated as;

Since we are given that a score of 48 on a test with $\bar X$ = 57 and s = 6,

So, z-score = $\frac{48-57}{6}$ = -1.5 {where $\bar X = 57$ and s = 6 }

AS we can clearly see that the z score of First and third condition are equally likely higher as compared to Second condition so it can be stated that A score of 2.6 on a test with $\bar X$ = 5.0 and s = 1.6 and A score of 48 on a test with $\bar X$ = 57 and s = 6 indicate the highest relative position.