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3 years ago

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Marie is saving money for home repairs. So far, she has saved $1,558. She needs at least $2,158 for the repairs. She plans to ad

d $60 per week to her current savings until she can afford the repairs. given the situation, which inequality models the number of additional weeks Marie needs to continue saving to afford the home repairs? Plot the answer on a number line.

1 answer:

Minchanka [31]3 years ago

5 0

Answer:

1,558 +60x = 2158

Step-by-step explanation:

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Ethan is playing in a soccer league that has 6 teams (including his team). Each team plays every other team twice during the reg

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Answer:

There are going to be 31 matches played in the soccer league.

Step-by-step explanation:

The soccer league has 6 teams, so if every team plays against the others twice, there are going to be played 30 matches:

-Team 1: v Team 2 (2), v Team 3 (2), v Team 4 (2), v Team 5 (2), v Team 6 (2)

-Team 2: v Team 3 (2), v Team 4 (2), v Team 5 (2), v Team 6 (2)

-Team 3: v Team 4 (2), v Team 5 (2), v Team 6 (2)

-Team 4: v Team 5 (2), v Team 6 (2)

-Team 5: v Team 6 (2)

-Team 6: -

If there is a final championship game after the 30 regular season matches, there are going to be 31 matches played in the league.

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3 years ago

The ratio of boys to girls in a group is 7:1. If there are 66 more boys than girls, work out how many girls there are

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Answer:11

Step-by-step explanation:

Boys : girls=7:1

Sum of ratio=7+1=8

Let them number of girls be y

Then they number of boys=y+66

Total number of pupils=y+y+66

Total number of pupils=2y+66

Number of girls=(girls ratio)/(sum of ratio) x (total number of pupils)

y=1/8 x (2y+66)

Cross multiply

y x 8=2y + 66

8y=2y + 66

Collect like terms

8y-2y=66

6y=66

Divide both sides by 6

6y/6=66/6

y=11

The number of girls is 11

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2 years ago

What is the solution for the equation? 14.8 = n minus 0.3

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Answer:

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Step-by-step explanation:

14.8=n-.3

you'd add .3 to other side so it'd be 15.1

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Step-by-step explanation:

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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