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hichkok12 [17]
2 years ago
7

The temperature was -20.5°F at 5 A.M. and rose 5 degrees per hour for the next 5 hours. Melissa says the temperature at 10 A.M.

was -5.5°F. Which statement identifies Melissa’s error and the correct answer?
Mathematics
1 answer:
Natali5045456 [20]2 years ago
6 0
It would be 1.5 degrees in 5 hours
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Allies plant has a height of 6meters. Radon’s plant grows 3/10 meters higher. How high does radon’s plant grow
kow [346]

The height of Radon plant is 6.3 meters

<em><u>Solution:</u></em>

Given that, Allies plant has a height of 6 meters

Radon’s plant grows \frac{3}{10} meters higher

To find: Height of Radon plant

From given information,

Height of Allies plant = 6 meters

Height of radon plant = \frac{3}{10} + Height of Allies plant

Substituting the known value,

\text{ Height of radon plant} = \frac{3}{10} + 6\\\\\text{ Height of radon plant} = \frac{3+60}{10}\\\\\text{ Height of radon plant} = \frac{63}{10}\\\\\text{ Height of radon plant} = 6.3

Thus Radon plant grows to height of 6.3 meters

7 0
2 years ago
A packet contains 1.5kg of oats.
Vlad1618 [11]
1.5kg is 1500 grams.
Maria uses 50g a day, so to know how many days takes eating the whole packet just divide 1500 by 50.
4 0
2 years ago
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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
Evaluate e − 1 2 f e− 2 1 ​ fe, minus, start fraction, 1, divided by, 2, end fraction, f when e = 15 e=15e, equals, 15 and f = 2
ivolga24 [154]
The answer 456 fe
Because it the benefit is doing today
7 1
2 years ago
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Myrtle Gould's total sales last month were $180,700, and her commission was $5,040. She earns a
Sophie [7]

Answer:

Step-by-step explanation:

8 0
1 year ago
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