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3 years ago

What will the new vertices will be reflected across the y axis so far the coordinates will be (2,1),(8,3) and

Mathematics

1 answer:

qaws [65]3 years ago

3 0

Answer:

<h3>(9, 1) QwQ correct me if there's wrong. </h3>

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Beth and joe won first place at the local talent show and shared the 50$ price equally they also gave some money to their friend

swat32

If they each kept the same amount of the $50 prize, then we divide $50 by 2 people, giving each of them $25 winnings. Your question is then a little confusing because it could be answered in two ways. Did they each keep $20.75 out of their $25 winnings, or did the two of them keep $20.75 in a combined total? Please read your question again to see which might be the case. I'm answering on the assumption that they each kept $20.75.

STEP 1:
divide total prize money by 2

= $50 ÷ 2
= $25 each won

STEP 2:
If they each kept $20.75

= $25 won - $20.75 each kept
= $4.25 each gave away

STEP 3:
multiply amount given away by 2 people
= $4.25 * 2 people
= $8.50 total given away

ANSWER: The total given to Romy is $8.50.

Hope this helps! :)

8 0

3 years ago

Let f(x)=5x+5 find f(-1) I need your help with this. please help me

Alisiya [41]

$f(x) = 5x+5\\\\f(-1) = 5(-1) +5 = -5+5=0$

5 0

3 years ago

Some one factor 3x^2+7x-20

mylen [45]

Split it up
3x^2+12x-5x-20
group
(3x^2+12x)+(-5x-20)
factor
(3x)(x+4)+(-5)(x+4)
undistribute (x+4) like ab+ac=a(b+c)
(3x-5)(x+4)

6 0

3 years ago

The graph shows the number of paintballs a machine launches, y, in x seconds:

kow [346]

The correct answer is 2 over 12 because you can simplify it

6 0

3 years ago

A college student is taking two courses. The probability she passes the first course is 0.73. The probability she passes the sec

zhenek [66]

Answer:

b) No, it's not independent.

c) 0.02

d) 0.59

e) 0.57

f) 0.5616

Step-by-step explanation:

To answer this problem, a Venn diagram should be useful. The diagram with the information of Event 1 and Event 2 is shown below (I already added the information for the intersection but we're going to see how to get that information in the b) part of the problem)

Let's call A the event that she passes the first course, then P(A)=.73

Let's call B the event that she passes the second course, then P(B)=.66

Then P(A∪B) is the probability that she passes the first or the second course (at least one of them) is the given probability. P(A∪B)=.98

b) Is the event she passes one course independent of the event that she passes the other course?

Two events are independent when P(A∩B) = P(A) * P(B)

So far, we don't know P(A∩B), but we do know that for all events, the next formula is true:

P(A∪B) = P(A) + P(B) - P(A∩B)

We are going to solve for P (A∩B)

.98 = .73 + .66 - P(A∩B)

P(A∩B) =.73 + .66 - .98

P(A∩B) = .41

Now we will see if the formula for independent events is true

P(A∩B) = P(A) x P(B)

.41 = .73 x .66

.41 ≠.4818

Therefore, these two events are not independent.

c) The probability she does not pass either course, is 1 - the probability that she passes either one of the courses (P(A∪B) = .98)

1 - P(A∪B) = 1 - .98 = .02

d) The probability she doesn't pass both courses is 1 - the probability that she passes both of the courses P(A∩B)

1 - P(A∩B) = 1 -.41 = .59

e) The probability she passes exactly one course would be the probability that she passes either course minus the probability that she passes both courses.

P(A∪B) - P(A∩B) = .98 - .41 = .57

f) Given that she passes the first course, the probability she passes the second would be a conditional probability P(B|A)

P(B|A) = P(A∩B) / P(A)

P(B|A) = .41 / .73 = .5616

4 0

4 years ago