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3 years ago

Solve the equation. b/6 = 3 b=

Mathematics

2 answers:

andrey2020 [161]3 years ago

4 0

B = 18. Yoh would multiply 6 to 3 which is 18

Stells [14]3 years ago

4 0

Answer:

b = 18

Step-by-step explanation:

We do the inverse operation thing:

3 x 6 = 18

We can check it by filling it in:

18/6 = 3

So b = 18

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Two friends, Sasha and Maria, are growing out their hair. They plan to cut it off at a certain

34kurt

Answer:

13 months, 49 cms

Explanation:

For Sasha, multiply 3 and 13 together for the growth per month. You should get 39, then add the 10 cm Sasha already has. The sum is 49.

For Maria, multiply 13 and 1 for the growth per month. You should get 13, then add the 36 Maria already has. The sum is also 49.

Hope this helps!

8 0

3 years ago

3.) Simplify the expression by combining like

Evgesh-ka [11]

$45s^2-19-s^2+16\\=45s^2-s^2-19+16\\=s^2(45-1)-19+16\\=44s^2+(-19+16)\\=44s^2-3$

Hope that helps and hope you get better!

3 0

2 years ago

Solve the inequality for x. Show each step of the solution.10x > 2(8x – 3) – 18

Luden [163]

Answer: x<4

Given the inequality:

$10x>2(8x-3)-18$

We want to solve the inequality for x.

First, distribute the bracket on the right side of the inequality.

$\begin{gathered} 10x>2(8x)-2(3)-18 \\ 10x>16x-6-18 \\ 10x>16x-24 \end{gathered}$

Next, subtract 10x from both sides of the inequality.

$\begin{gathered} 10x-10x>16x-10x-24 \\ 0>6x-24 \end{gathered}$

Add 24 to both sides of the inequality.

$\begin{gathered} 0+24>6x-24+24 \\ 24>6x \end{gathered}$

Divide both sides of the inequality by 6.

$\begin{gathered} \frac{24}{6}>\frac{6x}{6} \\ 4>x \\ \implies x$

The solution to the inequality is x<4.

6 0

1 year 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.

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