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Sedaia [141]
3 years ago
14

Possible my last brainliest if its right question idk may be more but this one is harsh for me

Mathematics
1 answer:
adelina 88 [10]3 years ago
4 0
Correct Solution: He read 1.5 pages per minute.


Error Analysis: You should’ve divided 3 by 2 not 300. There are 300 pages in the book total. This information is irrelevant to the question. Cole reads 3 pages in two minutes. So you divide 3 by to to find out how many pages he reads per minute. 3/2=1.5
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Find the zeros of f(x)=-3x^3+6x+5.
Tasya [4]

Answer:

x = 1.723

Step-by-step explanation:

The  zeros of a function f(x) are the points where the function crosses the x-axis. At these points, the function will have a value of zero, that is;

f(x) = 0

We simply graph the function and determine the points where it crosses the x-axis. From the attachment, f(x) crosses the x-axis at;

x = 1.723

8 0
3 years ago
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What is the slope line ?
sweet-ann [11.9K]

Answer:

-3.....

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3 0
2 years ago
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What is the gcf of 24a^3c and 3a?
Archy [21]
24a^3c = 3a(8a^2c)

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\text {GCF = }3a
6 0
3 years ago
Four teams A,B,C, and D compete in a tournament, and exactly one of them will win the tournament. Teams A and B have the same ch
beks73 [17]

Answer:

A= 0,2

B= 0,2

C= 0,4

D=0,2

Step-by-step explanation:

We know that only one team can win, so the sum of each probability of wining  is one

P(A)+P(B)+P(C)+P(D)=1

then we Know that the probability of Team A and B are the same, so

P(A)=P(B)

And that the  the probability that either team A or team C wins the tournament is 0.6, so P(A)+Pc)= 0,6, then P(C)= 0.6-P(A)

Also, we know that team C is twice as likely to win the tournament as team D, so P(C)= 2 P(D) so P(D) = P(C)/2= (0.6-P(A))/2

Now if we use the first formula:

P(A)+P(B)+P(C)+P(D)=1

P(A)+P(A)+0.6-P(A)+(0.6-P(A))/2=1

0,5 P(A)+0.9=1

0,5 P(A)= 0,1

P(A)= 0,2

P(B)= 0,2

P(C)=0,4

P(D)=0,2

4 0
3 years ago
Please help me ^^ This is due tomorrow
saveliy_v [14]
The answer is 5.

You just plug whatever number is in the f(?) part in for x!
7 0
3 years ago
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