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Natasha_Volkova [10]
3 years ago
5

each month for 6 months,kelsey completes 5 paintings. How many more paintings does she need to complete before she has completed

38 paintings?
Mathematics
2 answers:
rusak2 [61]3 years ago
4 0

<u>Answer:</u> She needs 8 more paintings to complete in total of 38 paintings.

<u>Step-by-step explanation:</u>

We are given:

Number of painting completed in 1 month = 5

Number of months = 6

Total number of paintings that she completes in 6 months = (6 × 5) = 30

Let the number of extra paintings needed be 'x'

According to the question:

\Rightarrow 30+x=38

Evaluating the value of 'x', we get:

\Rightarrow x=38-30=8

Hence, she needs 8 more paintings to complete in total of 38 paintings.

Solnce55 [7]3 years ago
3 0
5 painting every 6 months so 5x6=30
38-30=8
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The correct question statement is:

A probability experiment is conducted in which the sample space of the experiment is S = {4,5,6,7,8,9,10,11,12,13,14,15}. Let event E={7,8,9,10,11,12,13,14,15}. Assume each outcome is equally likely. List the outcomes in E^{c}. Find P(E^{c}).

Solution:

Part 1:

E^{c} means compliment of the set E. A compliment of a set can be obtained by finding the difference of the set from the universal set. The universal set is the set which contains all the possible outcomes of the events which is S in this case.

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E^{c}=S-E \\  \\ &#10;E^{c}=(4,5,6,7,8,9,10,11,12,13,14,15)-(7,8,9,10,11,12,13,14,15) \\  \\ &#10;E^{c}=(4,5,6)

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Part 2)

P(E^{c}) means probability that if we select any number from the Sample Space S, it will belong the set E compliment.

P(E^{c}) = (Number of Elements in E^{c})/Number of elements in S

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So, 

P(E^{c})= \frac{n(E^{c}) }{n(S)} \\  \\ &#10;P(E^{c})= \frac{3}{12} \\  \\ &#10;P(E^{c})= \frac{1}{4}
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