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mina [271]
3 years ago
12

Please help I have no clue on how to do this.

Mathematics
2 answers:
GREYUIT [131]3 years ago
4 0
The second one

36/25

hope this helps
good luck!!! :))))
hammer [34]3 years ago
3 0
The answer is the second answer choice on the screen
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Lisa had a set of 12 toy cars. Then she gave her sister 3 cars. What fraction of Lisa's set did Lisa give to her sister.
Trava [24]
Lisa gave her sister 1/4, or one-fourth of her toy cars. Brainliest would really help!
5 0
3 years ago
SOLVE : 3|x-2|-10=11
Andru [333]

Answer:

The answer final

x1= -5; x2=9

Step-by-step explanation:

If you want me to expand on the equation leave me a comment :)

7 0
3 years ago
Solve x4 – 17x2 + 16 = 0.<br> Let u =<br> Select all of the zeroes of the function. i i
Soloha48 [4]

Solve \ x^4 - 17x^2 + 16 = 0

To solve this equation , we need to write it in quadratic form

ax^2+bx+c=0

To get the equation in quadratic form we replace x^2 with u

x^2 = u

x^4 - 17x^2 + 16 = 0 can be written as

(x^2)^2 - 17x^2 + 16 = 0, Replace u for x^2

So equation becomes

u^2 - 17u + 16 = 0

Now we factor the left hand side

-16  and -1  are the two factors whose product is +16  and sum is -17

(u-16) (u-1) = 0

u -16 = 0  so u=16

u-1 =0  so u=1

WE assume u = x^2, Now we replace u with x^2

u = 16 so \ it \ becomes \ x^2 =16

Now take square root on both sides , x= +4  and x=-4

u = 1 so \ it \ becomes \ x^2 =1

Now take square root on both sides , x= +1  and x=-1

So zeros of the function are -4, -1, 1, 4


8 0
3 years ago
Read 2 more answers
You have a large jar that initially contains 30 red marbles and 20 blue marbles. We also have a large supply of extra marbles of
Dima020 [189]

Answer:

There is a 57.68% probability that this last marble is red.

There is a 20.78% probability that we actually drew the same marble all four times.

Step-by-step explanation:

Initially, there are 50 marbles, of which:

30 are red

20 are blue

Any time a red marble is drawn:

The marble is placed back, and another three red marbles are added

Any time a blue marble is drawn

The marble is placed back, and another five blue marbles are added.

The first three marbles can have the following combinations:

R - R - R

R - R - B

R - B - R

R - B - B

B - R - R

B - R - B

B - B - R

B - B - B

Now, for each case, we have to find the probability that the last marble is red. So

P = P_{1} + P_{2} + P_{3} + P_{4} + P_{5} + P_{6} + P_{7} + P_{8}

P_{1} is the probability that we go R - R - R - R

There are 50 marbles, of which 30 are red. So, the probability of the first marble sorted being red is \frac{30}{50} = \frac{3}{5}.

Now the red marble is returned to the bag, and another 3 red marbles are added.

Now there are 53 marbles, of which 33 are red. So, when the first marble sorted is red, the probability that the second is also red is \frac{33}{53}

Again, the red marble is returned to the bag, and another 3 red marbles are added

Now there are 56 marbles, of which 36 are red. So, in this sequence, the probability of the third marble sorted being red is \frac{36}{56}

Again, the red marble sorted is returned, and another 3 are added.

Now there are 59 marbles, of which 39 are red. So, in this sequence, the probability of the fourth marble sorted being red is \frac{39}{59}. So

P_{1} = \frac{3}{5}*\frac{33}{53}*\frac{36}{56}*\frac{39}{59} = \frac{138996}{875560} = 0.1588

P_{2} is the probability that we go R - R - B - R

P_{2} = \frac{3}{5}*\frac{33}{53}*\frac{20}{56}*\frac{36}{61} = \frac{71280}{905240} = 0.0788

P_{3} is the probability that we go R - B - R - R

P_{3} = \frac{3}{5}*\frac{20}{53}*\frac{33}{58}*\frac{36}{61} = \frac{71280}{937570} = 0.076

P_{4} is the probability that we go R - B - B - R

P_{4} = \frac{3}{5}*\frac{20}{53}*\frac{25}{58}*\frac{33}{63} = \frac{49500}{968310} = 0.0511

P_{5} is the probability that we go B - R - R - R

P_{5} = \frac{2}{5}*\frac{30}{55}*\frac{33}{58}*\frac{36}{61} = \frac{71280}{972950} = 0.0733

P_{6} is the probability that we go B - R - B - R

P_{6} = \frac{2}{5}*\frac{30}{55}*\frac{25}{58}*\frac{33}{63} = \frac{49500}{1004850} = 0.0493

P_{7} is the probability that we go B - B - R - R

P_{7} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{33}{63} = \frac{825}{17325} = 0.0476

P_{8} is the probability that we go B - B - B - R

P_{8} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{30}{65} = \frac{750}{17875} = 0.0419

So, the probability that this last marble is red is:

P = P_{1} + P_{2} + P_{3} + P_{4} + P_{5} + P_{6} + P_{7} + P_{8} = 0.1588 + 0.0788 + 0.076 + 0.0511 + 0.0733 + 0.0493 + 0.0476 + 0.0419 = 0.5768

There is a 57.68% probability that this last marble is red.

What's the probability that we actually drew the same marble all four times?

P = P_{1} + P_{2}

P_{1} is the probability that we go R-R-R-R. It is the same P_{1} from the previous item(the last marble being red). So P_{1} = 0.1588

P_{2} is the probability that we go B-B-B-B. It is almost the same as P_{8} in the previous exercise. The lone difference is that for the last marble we want it to be blue. There are 65 marbles, 35 of which are blue.

P_{2} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{35}{65} = \frac{875}{17875} = 0.0490

P = P_{1} + P_{2} = 0.1588 + 0.0490 = 0.2078

There is a 20.78% probability that we actually drew the same marble all four times

3 0
3 years ago
If the data in the stem-and-leaf graph below were shown in a line plot, which statement’s would be true of the line plot? Please
lukranit [14]

Answer:

1) THE TOTAL NUMBER OF MARKS TELLS YOU THE SIZE OF THE DATA SET.

2)hISTOGRAM  

:)

Step-by-step explanation:

3 0
3 years ago
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