All categories

3 years ago

PLEASE ANSWER I WILL GIVE BRAINLIEST TO WHOEVER ANSWERS GOOD PLEASE

Mathematics

2 answers:

antoniya [11.8K]3 years ago

6 0

Answer:

First answer is correct

Step-by-step explanation:

amid [387]3 years ago

4 0

Answer:

OK.

Now

The points that Make this triangle are

A(-2, 3)

B(-3, 1)

C(-1, 1)

When we Translate 6units right and 2 units Up...

It means

Say we have point (x,y)

we translate 6 units right(Referring to the x coordinate) and 2 units up(referring to y coordinate)

So after translation

Point (x,y) would be (x+6 , y +2)

So Lets apply this to the points given above

A(-2,3) Would be (-2+6 , 3+2) = (4,5)

B(-3,1) would be ( -3+6 , 1+2) = (3,3)

C(-1,1) would be (-1+6 , 1+2 ) = (5, 3)

So We have It

Our Answer

(3,3) (5,3) ( 4,5)

THE FIRST OPTION IS LEGIT!!!

You might be interested in

Which equation can be used to calculate the surface area of the triangular prism net show below?

deff fn [24]

Answer:

the answer is a

Step-by-step explanation:

4 0

3 years ago

In order to develop a more appealing cheeseburger, a franchise uses taste tests with 12 different buns, 7 different cheeses,

Readme [11.4K]

Answer:

It will take a week to taste all the cheeseburgers possible.

Step-by-step explanation:

Given:

Number of buns of different types = 12

Number of cheeses of different types = 7

Number of lettuce of different types = 4

Number of tomatoes of different types = 3

Time taken to eat one cheeseburger = 10 minutes

To find:

Total time taken to eat all the possible cheeseburgers = ?

Solution:

First of all, we need to find out what all possible number of cheeseburgers can be made using the given number of ingredients.

Total number of possible ways to make cheeseburgers = Number of choices of buns $\times$ Number of choices of cheeses $\times$ Number of choices of Lettuces $\times$ Number of choices of tomatoes

And this number of choices are nothing but the number of types of ingredients available.

So, total number of cheeseburgers possible = 12 $\times$ 7 $\times$ 4 $\times$ 3 = 1008

Time taken to eat 1 cheeseburger = 10 minutes

Time taken to eat 1008 cheeseburger = 10 $\times$ 1008 = 10080 minutes = 168 hours = 7 days

i.e. It will take a week to taste all the cheeseburgers possible.

3 0

4 years ago

18+ (7 - 4) x 3 I’m very confused and need help:(

Roman55 [17]

Start with 7-4=3
Then 3x3=9
Then 18+9=27
Answer 27

5 0

3 years ago

Read 2 more answers

What is the midpoint of the line segment with endpoints (-2, -2) and (4,6)?

IRINA_888 [86]

Answer:

(1,2)

Step-by-step explanation:

To find the midpoint, add the x coordinates and divide by 2

(-2+4)/2 = 2/2 =1

add the y coordinates and divide by 2

(-2+6)/2 = 4/2 = 2

The midpoint is

(1,2)

3 0

3 years ago

A professor has been teaching accounting for over 20 years. From her experience she knows that 75% of her students do homework r

Nata [24]

Answer:

a) $P(B|A) =\frac{P(A and B)}{P(A)}=\frac{0.65}{0.75}=0.867$

b) In order to two events can be considered as independent we need to have this condition:

$P(A and B) = P(A)*P(B)$

If we check the condition we see this:

$P(A)*P(B)= 0.75*0.7=0.525 \neq P(A and B)$

So since we don't have the condition satisfied the events "do the homework regularly" and "pass the course" are not independent.

Other way to check independence is with the following two conditions:

$P(B|A)=P(B)$ or $P(A|B)=P(A)$ , and we see that $P(B|A)\neq P(B)$ for our case.

Step-by-step explanation:

Data given

Lets define the following events

A: Represent the students who do homework regularly

B: Represent the students who pass the course

75% of her students do homework regularly

70% of her students pass the course

65% of her students do homework and also pass the course

For this case we have some probabilities given:

P(A) =0.75, P(B) =0.70 and P(A and B)=0.65

Solution to the problem

Part a

On this case we want to find this probability:

P(B|A) the conditional probability (Pass the course given that he does the homework regularly).

By defintion of conditional probability we know that:

$P(B|A) =\frac{P(A and B)}{P(A)}$

And now we can replace in the conditional formula like this:

$P(B|A) =\frac{P(A and B)}{P(A)}=\frac{0.65}{0.75}=0.867$

Part b

In order to two events can be considered as independent we need to have this condition:

$P(A and B) = P(A)*P(B)$

If we check the condition we see this:

$P(A)*P(B)= 0.75*0.7=0.525 \neq P(A and B)$

So since we don't have the condition satisfied the events "do the homework regularly" and "pass the course" are not independent.

Other way to check independence is with the following two conditions:

$P(B|A)=P(B)$ or $P(A|B)=P(A)$ , and we see that $P(B|A)\neq P(B)$ for our case.

6 0

3 years ago