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2 years ago

Use long division to write

Mathematics

1 answer:

julia-pushkina [17]2 years ago

5 0

Answer:

In this photo the steps are given.

The difference between these answers are that 4/9 is a repetitive decimal and that 5/8 is a finite decimal.

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Find the solution set of the system of linear equations represented by the augmented matrix. (if there is no solution, enter no

PIT_PIT [208]

There is no system to solve for

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3 years ago

Use special right triangle find the exact value of x. X=?

swat32

Answer: 12

Step-by-step explanation:

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3 years ago

The Fall Festival charges $0.75 per ticket for the rides. Kendall bought 18 tickets for rides and spent a total of $33.50 at the

PIT_PIT [208]

Part (a)

The variable y is the dependent variable and the variable x is the independent variable.

Part (b)

The cost of one ticket is $0.75. Therefore, the cost of 18 tickets will be:

$0.75\times 18=13.5$ dollars

Now, we know that Kendall spent her money only on ride tickets and fair admission and that she spent a total of $33.50.

Therefore, the price of the fair admission is: $33.50-$13.50=$20

If we use y to represent the total cost and x to represent the number of ride tickets, the linear equation that can be used to determine the cost for anyone who only pays for ride tickets and fair admission can be written as:

$y=0.75x+20$ ......Equation 1

Part (c)

The above equation is logical because, in general, the total cost of the rides will depend upon the number of ride tickets bought and that will be 0.75x. Now, even if one does not take any rides, that is when x=0, they still will have to pay for the fair admission, and thus their total cost, y=$20.

Likewise, any "additional" cost will depend upon the number of ride tickets bought as already suggested. Thus, the total cost will be the sum of the total ride ticket cost and the fixed fair admission cost. Thus, the above Equation 1 is the correct representative linear equation of the question given.

4 0

3 years ago

The formula for the volume of a sphere is a power function:

soldier1979 [14.2K]

Answer:

r≥0

V(r) ≥0

Step-by-step explanation:

V(r) = 4/3 * Pi* r^3

The domain is the input values for r

The radius cannot be less than 0 but can be any number bigger than 0

What is the domain of the function in this situation?

r≥0

The range is the output values for the function or the volume

The volume must be greater than or equal to zero

V(r) ≥0

7 0

4 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