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Bezzdna [24]
3 years ago
9

Ashley, Bob, Claire, and Daniel are among 13 students who entered a lottery to win a free vacation to Paris.

Mathematics
1 answer:
AlladinOne [14]3 years ago
7 0

Answer:

0.0014 = 0.14% probability that Ashley, Bob, Claire, and Daniel will be chosen.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order in which the students are chosen is not important, so the combinations formula is used to solve this question.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

Desired outcomes:

4 students from a set of 4(Ashley, Bob, Claire, and Daniel). So

D = C_{4,4} = \frac{4!}{4!(4-4)!} = 1

Total outcomes:

4 students from a set of 13(number of students in the lottery). So

T = C_{13,4} = \frac{13!}{4!(13-4)!} = 715

Probability:

p = \frac{D}{T} = \frac{1}{715} = 0.0014

0.0014 = 0.14% probability that Ashley, Bob, Claire, and Daniel will be chosen.

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Mae earns a weekly salary of $330 plus a commission of 6.0% on a sales gift shop.How much would she make if she sold $4300 worth
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Answer:

Total income = $588

Step-by-step explanation:

Given:

Weekly salary = $330

Commission = 6%

Total sales = $4,300

Find:

Total income

Computation:

Total commission = $4,300 x 6%

Total commission = $258

Total income = Weekly salary + Total commission

Total income = $330 + $258

Total income = $588

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Answer:

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Consider the following. f(x) = x5 − x3 + 6, −1 ≤ x ≤ 1 (a) Use a graph to find the absolute maximum and minimum values of the fu
kykrilka [37]

ANSWER

See below

EXPLANATION

Part a)

The given function is

f(x) =  {x}^{5}  -  {x}^{3}  + 6

From the graph, we can observe that, the absolute maximum occurs at (-0.7746,6.1859) and the absolute minimum occurs at (0.7746,5.8141).

b) Using calculus, we find the first derivative of the given function.

f'(x) = 5 {x}^{4} - 3 {x}^{2}

At turning point f'(x)=0.

5 {x}^{4} - 3 {x}^{2}  = 0

This implies that,

{x}^{2} (5 {x}^{2}  - 3) = 0

{x}^{2}  = 0 \: or \: 5 {x}^{2}  - 3 = 0

x =  - \frac{ \sqrt{15} }{5}   \: or \: x = 0 \:  \: or \: x =\frac{ \sqrt{15} }{5}

We plug this values into the original function to obtain the y-values of the turning points

(   -  \frac{ \sqrt{15} }{5}  , \frac{1}{125} ( 6 \sqrt{15}  +750)) \:and \:  (0, - 6) \: and\: (   \frac{ \sqrt{15} }{5}  , \frac{1}{125} ( - 6 \sqrt{15}  +750))

We now use the second derivative test to determine the absolute maximum minimum on the interval [-1,1]

f''(x) = 20 {x}^{3}  - 6x

f''( -  \frac{ \sqrt{15} }{5} ) \:   <  \: 0

Hence

(   -  \frac{ \sqrt{15} }{5}  , \frac{1}{125} ( 6 \sqrt{15}  + 750))

is a maximum point.

f''( \frac{ \sqrt{15} }{5} ) \:    >  \: 0

Hence

(     \frac{ \sqrt{15} }{5}  , \frac{1}{125} (- 6 \sqrt{15}  + 750))

is a minimum point.

f''(0) \: =\: 0

Hence (0,-6) is a point of inflexion

4 0
2 years ago
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