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Semmy [17]
3 years ago
6

Wren has a 30% chance of being selected as the president of the school sports club and a 80% chance of being elected editor of t

he school magazine.
What is the probability that Wren will be selected as the president of the sports club and will also be elected editor of the school magazine?
A) 11%
B) 15%
C) 24%
D) 50%
Mathematics
2 answers:
anyanavicka [17]3 years ago
4 0

Answer with Step-by-step explanation:

Wren has a 30% chance of being selected as the president of the school sports club and a 80% chance of being elected editor of the school magazine.

What is the probability that Wren will be selected as the president of the sports club and will also be elected editor of the school magazine?

A: Wren will be elected president of the school

P(A)=0.3     (30%)

B: Wren will be elected editor of school magazine

P(B)=0.8   (80%)

Both the events are independent hence,

P(A∩B)=P( Wren will be selected as the president of the sports club and will also be elected editor of the school magazine)

             = P(A)×P(B)

           = 0.3×0.8

          =0.24

          = 24%

Hence, the probability that Wren will be selected as the president of the sports club and will also be elected editor of the school magazine

C) 24%

SashulF [63]3 years ago
3 0
Multiply the two probabilities together to get the answer. 3/10 ×8/10 = 24/100 or 24 %
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For the following telescoping series, find a formula for the nth term of the sequence of partial sums {Sn}. Then evaluate limn→[
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Answer:

The following are the solution to the given points:

Step-by-step explanation:

Given value:

1) \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\2) \sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}

Solve point 1 that is \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\:

when,

k= 1 \to  s_1 = \frac{1}{1+1} - \frac{1}{1+2}\\\\

                  = \frac{1}{2} - \frac{1}{3}\\\\

k= 2 \to  s_2 = \frac{1}{2+1} - \frac{1}{2+2}\\\\

                  = \frac{1}{3} - \frac{1}{4}\\\\

k= 3 \to  s_3 = \frac{1}{3+1} - \frac{1}{3+2}\\\\

                  = \frac{1}{4} - \frac{1}{5}\\\\

k= n^  \to  s_n = \frac{1}{n+1} - \frac{1}{n+2}\\\\

Calculate the sum (S=s_1+s_2+s_3+......+s_n)

S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....\frac{1}{n+1}-\frac{1}{n+2}\\\\

   =\frac{1}{2}-\frac{1}{5}+\frac{1}{n+1}-\frac{1}{n+2}\\\\

When s_n \ \ dt_{n \to 0}

=\frac{1}{2}-\frac{1}{5}+\frac{1}{0+1}-\frac{1}{0+2}\\\\=\frac{1}{2}-\frac{1}{5}+\frac{1}{1}-\frac{1}{2}\\\\= 1 -\frac{1}{5}\\\\= \frac{5-1}{5}\\\\= \frac{4}{5}\\\\

\boxed{\text{In point 1:} \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2} =\frac{4}{5}}

In point 2: \sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}

when,

k= 1 \to  s_1 = \frac{1}{(1+6)(1+7)}\\\\

                  = \frac{1}{7 \times 8}\\\\= \frac{1}{56}

k= 2 \to  s_1 = \frac{1}{(2+6)(2+7)}\\\\

                  = \frac{1}{8 \times 9}\\\\= \frac{1}{72}

k= 3 \to  s_1 = \frac{1}{(3+6)(3+7)}\\\\

                  = \frac{1}{9 \times 10} \\\\ = \frac{1}{90}\\\\

k= n^  \to  s_n = \frac{1}{(n+6)(n+7)}\\\\

calculate the sum:S= s_1+s_2+s_3+s_n\\

S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(n+6)(n+7)}\\\\

when s_n \ \ dt_{n \to 0}

S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(0+6)(0+7)}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{6 \times 7}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{42}\\\\=\frac{45+35+28+60}{2520}\\\\=\frac{168}{2520}\\\\=0.066

\boxed{\text{In point 2:} \sum ^{\infty}_{k = 1} \frac{1}{(n+6)(n+7)} = 0.066}

8 0
2 years ago
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