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Oksana_A [137]
2 years ago
8

a hatchling turtle has a shell with a 1-inch diameter. the shell keeps a similar shape as the turtle grows. compare the volume a

nd surface area of the hatchling's shell to the shell when it's diameter is 4 inches
Mathematics
1 answer:
earnstyle [38]2 years ago
5 0
Answer would be 4 multiple
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In basketball, Shaquille makes 40% of his free throw attempts. He goes to the free throw line to shoot one and one ( if he makes
Anna11 [10]
Situations:
1) he makes 1st shot + he makes 2nd shot
P(A) = 0.4 · 0.4 = 0.16  ( 16 % )
2) he makes 1 st shot + he misses 2nd shot
P(B) = 0.4 · 0.6 = 0.24   ( 24 % )
3) he misses 1st shot and he has no more attempts
P (C) = 0.6 ( 60 %)
0.16 · 2 + 0.24 · 1 + 0.6 · 0 = 0.32 + 0.34 = 0.56
The expected value is 0.56 points.
6 0
2 years ago
Read 2 more answers
Sam is 3 years younger than Susan. Susan is 40 years old. If Sam is s years old, the equation that relates his age with Susan's
dedylja [7]

Answer:

Sam is 37 years old.

Step-by-step explanation:

Susan is 40. Sam is just 3 years younger, making him/her 37.

40 = S + 3

(S = 37)

I hope this helped at all ^-^ Please tll me if I made an error or left something out.

5 0
2 years ago
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
What is the surface area of the regular pyramid below
Alenkinab [10]

S=14\times14+4(\frac{14}{2}\times18) \\S=196+4(7\times18) \\S=196+4\times126 \\S=196+504 \\S=\boxed{700}

5 0
2 years ago
Read 2 more answers
Look at Andrea’s running journal.
german

First align the decimal points and the numbers,  then add the extra 0's if needed. Lastly, add and the total answer is 14.225.

7 0
2 years ago
Read 2 more answers
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