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vovangra [49]
2 years ago
5

Chucky grabbed 121212 items in the grocery store that each had a different price and had a mean cost of about \$7.41$7.41dollar

sign, 7, point, 41. One of the items was an entire wheel of cheese that cost \$39.99$39.99dollar sign, 39, point, 99. [Show data] \$1.29dollar sign, 1, point, 29 \$1.92dollar sign, 1, point, 92 \$3.19dollar sign, 3, point, 19 \$3.79dollar sign, 3, point, 79 \$3.99dollar sign, 3, point, 99 \$4.79dollar sign, 4, point, 79 \$5.19dollar sign, 5, point, 19 \$5.29dollar sign, 5, point, 29 \$5.49dollar sign, 5, point, 49 \$6.75dollar sign, 6, point, 75 \$7.19dollar sign, 7, point, 19 \$39.99dollar sign, 39, point, 99 Chucky then decided to put the wheel of cheese back and only buy the other 111111 items. How will removing the wheel of cheese affect the mean and median?
Mathematics
2 answers:
8090 [49]2 years ago
8 0

Answer:

Both the mean and median will decrease, but the mean will decrease more than the median.

Step-by-step explanation:

Removing the wheel of cheese will decrease the median a little bit, because the median shifts from between two data points to the lower of the two data points:

       With the wheel of cheese, the median is the middle number, but there's no middle number in this data set! So, to find the median we take the mean of the two middle numbers, $4.79 and $5.19 which is $4.99.

       Without the wheel of cheese,

Removing the wheel of cheese will decrease the mean significantly, because the total cost will decrease by $39.99, and the number of items decreases by only 1.

Pepsi [2]2 years ago
4 0

Answer: B

Step-by-step explanation:

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1 year ago
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PARA ENCONTRAR CUÁNTO MIDE CAD LADO SE DIVIDE EL PERÍMETRO

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6 0
2 years ago
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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