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3 years ago

9

Sophia vegetable garden is divided into 12 equal sections.She plants carrots in 8 of section .Write two fractions that are equiv

alent to the part of Sophia’s garden that is planted with carrots.

2 answers:

Ksju [112]3 years ago

7 0

8/12 which is equal to 2/3

stepan [7]3 years ago

4 0

The unsimplified fraction is 8/12 which can be reduced to 2/3

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Molly was on a long 136 mile road trip. The first part of the trip there was lots of traffic, she only averaged 16 mph. The seco

Mazyrski [523]

Answer:

In traffic, she drove for 3 hours

and After the traffic cleared she drove for 2 hours.

Explanation:

Given that the road trip was 136 miles;

$d=136$

The first part of the trip there was lots of traffic, she only averaged 16 mph;

$v_1=16$

The second part of the trip there was no traffic so she could drive 44 mph;

$v_2=44$

She traveled for a total of 5 hours;

$t=5$

let x represent the time in traffic when she traveled at 16 mph

$t_1=x$

the time the traffic is clear would be;

$t_2=t-t_1=5-x$

Recall that distance equals speed multiply by time;

$d=v_1t_1_{}_{}^{}+v_2t_2$

substituting the values;

$136=16x+44(5-x)$

solving for x;

$\begin{gathered} 136=16x+220-44x \\ 44x-16x=220-136 \\ 28x=84 \\ x=\frac{84}{28} \\ x=3 \end{gathered}$

So;

$\begin{gathered} t_1=3\text{ hours} \\ t_2=5-x=5-3=2 \\ t_2=2\text{ hours} \end{gathered}$

Therefore, In traffic, she drove for 3 hours

and After the traffic cleared she drove for 2 hours.

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1 year ago

3х – 8y = — 16<br> =<br> M<br> So yeah I kinda need help

marishachu [46]

Answer:

3x-8y-16 = 0

Step-by-step explanation:

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3 years ago

Help Me??????????????

tekilochka [14]

Statistical questions: A, C, E, because they all talk about students, where B and D talk about adults and people in general. Frank just wants to know about students.

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3 years ago

Please answer this question

USPshnik [31]

Answer:

C - The relationship represents a function because each input gives one unique output.

Step-by-step explanation:

To see if it is a function we can do the line test where we have a vertical line and "drag" it along the line. As long as it only hits the line once in every situation it is a function! We can see this passes and therefore the answer is c.

(why is it not d? you can have a non-linear line, but still have a function)

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2 years ago

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