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hram777 [196]
1 year ago
15

Without drawing the graph of the function y=1.2x−7, see if the graph passes through points (-10, 5)

Mathematics
1 answer:
ad-work [718]1 year ago
7 0

Step-by-step explanation:

Substitute x with -10 and y with 5.

(5) = 1.2(10) - 7

5 = 12 - 7

This is obviously true, so the graph indeed passes through the point (-10, 5).

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2 years ago
The pie chart to the right shows how adults rate their financial shape. Suppose 4 people are chosen at random from a group of 11
vekshin1

Answer:

Step-by-step explanation:

Given that a pie chart shows data as

Excellent 7%

Good 44%

Fair 35%

Poor 13%

Other 1%

If 4 people are chosen at random we find that each person is independent of other and there are two outcomes either excellent or not excellent.

p = Prob for success = p = 7%=0.07

q = Prob for non success = 1-0.07 = 0.93

If x is a random variable representing the excellent persons then X is binomial with n =4, p = 0.07

Required probability = P(X=4)

=p^4\\=(0.07)^4\\=0.00002401

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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
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