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3 years ago

Please help me, this is PEDMAS

Mathematics

1 answer:

rusak2 [61]3 years ago

3 0

Answer:

1). 183

2). 83

3). 353

4). -2652

5). 59

6). 510

7). -293

8). 79

9). 78

10). 87

Step-by-step explanation:

Considering order of operation, parenthesis must be simplified first.

The solutions to the given questions are shown below;

$1). \ \ 5+(7\times (3+2)^2) + 3=8 + (7\times (5)^2) = 8 + (7\times 25) = 8 + 175 = 183$

$2). \ 16+(8+ (6+2)^2) - 5=16-5 + (8+ (8)^2) = 11 + (8+64) = 11 + 72 = 83$

$3). \ (15+ 3)^2 + ((14+6)+3^2)= (18)^2 + ((20) + 9) = 324 + (29) = 353$

$4). \ ((11+ 2)-(24+2)^2)\times 2^2 = (13 - 676) \times 4 = (-663)\times 4 = -2652$

$5). \ ((4+3)^2+ 6) -5+3^2= (49 + 6) - 5 + 9= (49+6+9) - 5 = 64-5 = 59$

$6). \ (20+ 2)^2 +((15+7)+2^2)= (484) + (22+ 4)= 484 + 26 =510$

$7). \ ((12 + 4)- (15+3)^2+ 5^2 = (16 - 324) + 25 = -318 + 25 = -293$

$8). \ (6^2+ (24+ 3+5^2)) - 3^2= (36 + (52)) - 9 = (88) - 9 = 79$

$9). \ ((10-4)^2\times 2) + 2 + 2^2= ((6)^2 \times 2) + 2 + 4 = (36\times 2)+ 6 = 72 + 6 = 78$

$10). \ (7^2 + (16 + 8 + 2^2)) + 3^2 = (49 + 28) + 9 = 78 + 9 = 87$

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Option D.

Step-by-step explanation:

Suppose that we have a set of N values:

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Because here we have "the median" then we will assume that N is odd, and there is only one median, the value "k" (if N was even, the median would be the mean of the two middle values, that case is really similar to the case where N is odd, so solving only one of the cases is enough)

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The new mean will be:

$M' = \frac{(x_1 - 7) + (x_2 - 7) + ... + (x_{k}) + ... + (x_n + 7)}{N} = \frac{(x_1 + x_2 + ... + x_n) + (-7)*(n/2 - 0.5) + 7*(n/2 - 0.5)}{N} = \frac{(x_1 + x_2 + ... + x_n)}{N} = M$

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