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3 years ago

Bouncing Baby Company sells their cribs for the following amounts to stores

Mathematics

1 answer:

xz_007 [3.2K]3 years ago

4 0

Answer:

Option C

Step-by-step explanation:

Since, Price of cribs (P) are proportional to the proportional to the number of cribs (N) sold,

P ∝ N

P = kN

Here 'k' is the proportionality constant

To get the value of proportionality constant,

Since, value of 10 cribs is $1320

1320 = k(10)

k = 132

Therefore, equation will be

P = 132N

Option A

If number of cribs = 6

P = 132(6) = $792

False.

Option B

For N = 22

P = 132(22) = $2904

False.

Option C

For N = 40

P = 132(40) = $5280

True

Option D

For N = 55

P = 132(55) = $7260

False

Option E

For N = 80

P = 132(80)

= $10560

False

Option F

For N = 250

P = 132(250)

= $33000

False

If the given options have been written correctly only one option, Option (C) is correct.

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What is the value of 2 3/4 ➗ 3/4

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Answer:

3 2/3 is the answer

Step-by-step explanation:

Conversion a mixed number 2 3/

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Dividing two fractions is the same as multiplying the first fraction by the reciprocal value of the second fraction. The first sub-step is to find the reciprocal (reverse the numerator and denominator, reciprocal of 3/

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Which expression is equivalent to *picture attached*

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Answer:

The correct option is;

$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$

Step-by-step explanation:

The given expression is presented as follows;

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right )$

Which can be expanded into the following form;

$\sum\limits _{n = 1}^{50} \left (4\cdot n^2 + 3 \cdot n\right ) = 4 \times \sum\limits _{n = 1}^{50} \left n^2 + 3 \times\sum\limits _{n = 1}^{50} n$

From which we have;

$\sum\limits _{k = 1}^{n} \left k^2 = \dfrac{n \times (n+1) \times(2n+1)}{6}$

$\sum\limits _{k = 1}^{n} \left k = \dfrac{n \times (n+1) }{2}$

Therefore, substituting the value of n = 50 we have;

$\sum\limits _{n = 1}^{50} \left k^2 = \dfrac{50 \times (50+1) \times(2\cdot 50+1)}{6}$

$\sum\limits _{k = 1}^{50} \left k = \dfrac{50 \times (50+1) }{2}$

Which gives;

$4 \times \sum\limits _{n = 1}^{50} \left n^2 = 4 \times \dfrac{n \times (n+1) \times(2n+1)}{6} = 4 \times \dfrac{50 \times (50+1) \times(2 \times 50+1)}{6}$

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