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lilavasa [31]
3 years ago
7

Ricardo earns a base salary of $500 a week plus a weekly commission he has earned a commission of $325, $460, $280, $400, $380,

and $250 each week for the last six weeks.
Find the mean, median, mode, and range for the commision Ricardo earned during these six weeks. Round answers to the nearest tenth.

What commission must Ricardo earn during the seventh week to raise his mean weekly commision to $370.70?

What commission must Ricardo earn during the seventh week to raise his mean weekly earnings to $850?

Mathematics
1 answer:
Vinvika [58]3 years ago
4 0

Answers:

1)

Mean = $349.2

Median = $352.5

Range = $210

2)

$499.90

3)

$355

I have attached a photo of the work written for this question:

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Which of the following is the expansion of (3c + d2)6?
vovangra [49]

Answer: Option A) 729c^6 + 1,458c^5d^2 + 1,215c^4d^4 + 540c^3d^6 + 135c^2d^8 + 18cd^{10} + d^{12} is the correct expansion.

Explanation:

on applying binomial theorem,  (a+b)^n=\sum_{r=0}^{n} \frac{n!}{r!(n-r)!} a^{n-r} b^r

Here a=3c, b=d^2 and n=6,

Thus, (3c+d^2)^6=\sum_{r=0}^{6} \frac{6!}{r!(6-r)!} (3c)^{n-r} (d^2)^r

⇒ (3c+d^2)^6= \frac{6!}{(6-0)!0!} (3c)^{6-0}.(d^2)^0+\frac{6!}{(6-1)!1!} (3c)^{6-1}.(d^2)^1+\frac{6!}{(6-2)!2!} (3c)^{6-2}.(d^2)^2+\frac{6!}{(6-3)!3!} (3c)^{6-3}.(d^2)^3+\frac{6!}{(6-4)!4!} (3c)^{6-4}.(d^2)^4+\frac{6!}{(6-5)!5!} (3c)^{6-5}.(d^2)^5+\frac{6!}{(6-6)!6!} (3c)^{6-6}.(d^2)^6

⇒(3c+d^2)^6= \frac{6!}{(6-)!0!} (3c)^6.d^0+\frac{6!}{(5)!1!} (3c)^5.d^2+\frac{6!}{(4)!2!} (3c)^4.d^4+\frac{6!}{(6-3)!3!} (3c)^3.d^6+\frac{6!}{(2)!4!} (3c)^2.d^8+\frac{6!}{(1)!5!} (3c).d^{10}+\frac{6!}{(0)!6!} (3c)^0.d^{12}

⇒(3c+d^2)^6=(3c)^6.d^0+\frac{720}{120} (3c)^5.d^2+\frac{720}{48} (3c)^4.d^4+\frac{720}{36} (3c)^3.d^6+\frac{720}{48} (3c)^2.d^8+\frac{720}{120} (3c).d^{10}+.d^{12}

⇒(3c+d^2)^6=729c^6 + 1,458c^5d^2 + 1,215c^4d^4 + 540c^3d^6 + 135c^2d^8 + 18cd^{10} + d^{12}

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3 years ago
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