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Anarel [89]
2 years ago
7

In a parking lot there are motorcycles and cars. You count 98 wheels, and your friend counts 30 vehicles. How many cares are the

re? How many motorcycles?
Assign two variables and equations. Please solve.
Mathematics
2 answers:
Sergeeva-Olga [200]2 years ago
7 0
X ➡ cars 4 wheels
y ➡ motorcycles 2 wheels

4x +2y = 98

x + y = 30
use either substitution or elimination to solve, the answer will be 19 cars and 11 motorcycles.
nata0808 [166]2 years ago
3 0
Car normally has 4 wheeles
motorcycles normall have 2 wheels

c=cars
m=motorcycles

4c+2m=98
c+m=30

multiply second equation by -2 and add to the original equaiton

-2c-2m=-60
4c+2m=98 +
2c+0m=38


2c=38
divide 2
c=19
sub

c+m=30
19+m=30
m=11


11 motorcycles
19 cars

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11) The Cost of maintaing a
dezoksy [38]

Answer:

a). Cost of 44 pupils = $14265

b). Least number of pupils = 31

Step-by-step explanation:

The given question is incomplete; here is the complete question.

The cost of maintaining a school is partly constant and partly varies as the number of pupils. With 50 pupils, the cost is $15,705.00 and with 40 pupils, it is $13,305.00.

(a) Find the cost when there are 44 pupils.

(b) If the fee per pupil is $360.00, what is the least number of pupils for which the school can run without a loss?

Let the equation representing the total cost of maintaining a school is,

C = ax + b

Where C = Total cost of maintaining a school

a = Fee per pupil

b = Fixed running cost

x = number of pupils

a). Cost of 50 pupils = $15705

    Equation will be,

    15705 = 50a + b -------(1)

    Cost of 40 pupils = $13305

    Equation will be,

    13305 = 40a + b --------(2)

    By subtracting equation (2) from equation (1),

    15705 - 13305 = (50a + b) - (40a + b)

    2400 = 10a

    a = 240

    From equation (1),

    b = 3705

    Equation representing the total cost will be,

    C = 240x + 3705

    If x = 44

    C = 240(44) + 3705

    C = $14265

b). If the fee per pupil 'a' = $360

    Let the number of pupils = p

    Total fee of 'p' pupils = $360p

    Total cost to run the school will be = 3705 + 240p

    For the school not to be in the loss,

    360p ≥ 3705 + 240p

    360p - 240p ≥ 3705

    120p ≥ 3705

    p ≥ \frac{3705}{120}

    p ≥ 30.875

    Therefore, to run the school without loss, number pupils should be at least 31.

3 0
2 years ago
Solve for x:<br> 3(5x + 10) + 10 = 6x - (x + 10)
34kurt

Answer:

x= -5

Step-by-step explanation:

3(5x+10)+10=6x-(x+10)

15x+30+10=6x-x-10

15x+40=5x-10

15x-5x=-10-40

10x=-50

x=-5

8 0
2 years ago
Read 2 more answers
Ryan found a pattern on the addition table . He shaded two diagonal lines that show his pattern. What is his pattern?
LenaWriter [7]

it would be adding by 2's because to get from 2 to 4 you have to add two or from 6 to 8 you have to add two also


6 0
2 years ago
Problem 2.2.4 Your Starburst candy has 12 pieces, three pieces of each of four flavors: berry, lemon, orange, and cherry, arrang
kkurt [141]

Answer:

a) P=0

b) P=0.164

c) P=0.145

Step-by-step explanation:

We have 12 pieces, with 3 of each of the 4 flavors.

You draw the first 4 pieces.

a) The probability of getting all of the same flavor is 0, because there are only 3 pieces of each flavor. Once you get the 3 of the same flavor, there are only the other flavors remaining.

b) The probability of all 4 being from different flavor can be calculated as the multiplication of 4 probabilities.

The first probability is for the first draw, and has a value of 1, as any flavor will be ok.

The second probability corresponds to drawing the second candy and getting a different flavor. There are 2 pieces of the flavor from draw 1, and 9 from the other flavors, so this probability is 9/(9+2)=9/11≈0.82.

The third probability is getting in the third draw a different flavor from the previos two draws. We have left 10 candys and 4 are from the flavor we already picked. Then the third probabilty is 6/10=0.6.

The fourth probability is getting the last flavor. There are 9 candies left and only 3 are of the flavor that hasn't been picked yet. Then, the probability is 3/9=0.33.

Then, the probabilty of picking the 4 from different flavors is:

P=1\cdot\dfrac{9}{11}\cdot\dfrac{6}{10}\cdot\dfrac{3}{9}=\dfrac{162}{990}\approx0.164

c) We can repeat the method for the previous probabilty.

The first draw has a probability of 1 because any flavor is ok.

In the second draw, we may get the same flavor, with probability 2/11, or we can get a second flavor with probability 9/11. These two branches are ok.

For the third draw, if we have gotten 2 of the same flavor (P=2/11), we have to get a different flavor (we can not have 3 of the same flavor). This happen with probability 9/10.

If we have gotten two diffente flavors, there are left 4 candies of the picked flavors in the remaining 10 candies, so we have a probabilty of 4/10.

For the fourth draw, independently of the three draws, there are only 2 candies left that satisfy the condition, so we have a probability of 2/9.

For the first path, where we pick 2 candies of the same flavor first and 2 candies of the same flavor last, we have two versions, one for each flavor, so we multiply this probability by a factor of 2.

We have then the probabilty as:

P=2\cdot\left(1\cdot\dfrac{2}{11}\right)\cdot\left(\dfrac{9}{10}\cdot\dfrac{2}{9}\right)+\left(1\cdot\dfrac{9}{11}\cdot\dfrac{4}{10}\cdot\dfrac{2}{9}\right)\\\\\\P=2\cdot\dfrac{36}{990}+\dfrac{72}{990}=\dfrac{144}{990}\approx0.145

5 0
2 years ago
A survey revealed that 31% of people are entertained by reading books, 39% are entertained by watching TV, and 30% are entertain
tester [92]
1% entertained by books
9% entertained by TV
3 0
2 years ago
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