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Semenov [28]
3 years ago
11

Martha works in a toy shop and earns $44 per day. She earns an extra $2 for each toy she sells. If Martha wants to earn at least

$80 per day, which inequality shows the minimum number of toys, n, that she should sell?
44 + 2n ≥ 80, so n ≥ 18

44 + 2n ≤ 80, so n ≤ 18

44 + 2n ≥ 80, so n ≥ 34

44 + 2n ≤ 80, so n ≤ 34
Mathematics
2 answers:
olganol [36]3 years ago
6 0

Answer:

a

Step-by-step explanation:

pantera1 [17]3 years ago
5 0
Martha's total earnings is the sum of her fixed pay pay day and the extra she gains based on the number of toys she sold. Her total earnings may be expressed as,
 
                                        44 + 2n 

Since she desires to earn $80 or more, the inequality is expressed as
 
                                         44 + 2n ≥ 80

Solving for n from the inequality,
 
                                                    n ≥18

Thus, the answer is the first choice. 

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Are your finances, buying habits, medical records, and phone calls really private? A real concern for many adults is that comput
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Answer:

a)There is a 4.88% probability that none is concerned that employers are monitoring phone calls.

b)There is a 7.89% probability that all are concerned that employers are monitoring phone calls.

c)There is a 37.23% probability that exactly two are concerned that employers are monitoring phone calls.

Step-by-step explanation:

The binomial probability is the probability of exactly x successes on n repeated trials in an experiment which has two possible outcomes (commonly called a binomial experiment).

It is given by the following formula:

P = C_{n,x}.p^{n}.(1-p)^{n-x}

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

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

And p is the probability of a success.

In this problem, a success is being concerned that employers are monitoring phone calls.

53% of adults are concerned that employers are monitoring phone calls, so p = 0.53

(a) Out of four adults, none is concerned that employers are monitoring phone calls.

Four adults, so n = 4.

Is the probability of 0 successes, so x = 0.

P = C_{n,x}.p^{n}.(1-p)^{n-x}

P = C_{4,0}.(0.53)^{0}.(0.47)^{4}

P = 0.0488

There is a 4.88% probability that none is concerned that employers are monitoring phone calls.

(b) Out of four adults, all are concerned that employers are monitoring phone calls.

Four adults, so n = 4.

Is the probability of 4 successes, so x = 4.

P = C_{n,x}.p^{n}.(1-p)^{n-x}

P = C_{4,0}.(0.53)^{4}.(0.47)^{0}

P = 0.0789

There is a 7.89% probability that all are concerned that employers are monitoring phone calls.

(c) Out of four adults, exactly two are concerned that employers are monitoring phone calls.

Four adults, so n = 4.

Is the probability of 4 successes, so x = 2.

P = C_{n,x}.p^{n}.(1-p)^{n-x}

P = C_{4,2}.(0.53)^{2}.(0.47)^{2}

P = 0.3723

There is a 37.23% probability that exactly two are concerned that employers are monitoring phone calls.

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Answer:

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