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2 years ago

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You are planning to hire a full-time electrician who will work 40 hours per week. If you plan on giving this new hire three week

s' vacation as a benefit, how many hours will this new hire work in twelve months?

1 answer:

tresset_1 [31]2 years ago

5 0

Option b) 1960 is the total number of hours the electrician worked.

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

It is given that, the full-time electrician will work 40 hours per week.

So, we need to know the total number of hours he could work in twelve months which is one year.

Therefore, one year has a total of 365 days.

<u>To find the number of weeks in twelve months :</u>

Number of weeks = Total days in one year / 7 days of week

⇒ 365 / 7

⇒ 52.14 (approximately 52 weeks)

The number of weeks in twelve months is 52 weeks.

Now, of this 52 weeks, the three weeks are given as vacation.

The total number of weeks the electrician worked = 52 weeks - 3 weeks

⇒ 49 weeks.

The electrician worked for 49 weeks.

To calculate the total number of hours he worked = 49 weeks × 40 hours

⇒ 1960 hours.

Therefore, the total number of hours the electrician worked is 1960 hours which is option b).

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A small dealership leased 21 Suburu Outbacks on 2-year leases. When the cars were returned at the end of the lease, the mileage

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Answer:

yes 30 is the answer

Step-by-step explanation:

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2 years ago

What is the quotient (x3 – 3x2 + 3x – 2) ÷ (x2 – x + 1)?

timofeeve [1]

We need to find the quotient of the given division problem.

$\frac{x^{3}-3x^{2}+3x-2}{x^{2}-x+1}$

In order to find its quotient, we will use long division.

$x^{2}-x+1$ ) $x^{3}-3x^{2}+3x-2$

First of all, we put x in the quotient as $x^{2}$ goes into $x^{3}$ , x times.

So, we get:

$x^{2}-x+1$ ) $x^{3}-3x^{2}+3x-2$ (x

$\text{ ..................}x^{3}-x^{2}+x$

Upon subtracting, we get:

$\text{....................}-2x^{2}+2x-2$

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5 0

2 years ago

Read 2 more answers

Three couples and two single individuals have been invited to an investment seminar and have agreed to attend. Suppose the proba

Maksim231197 [3]

Answer:

(a) Probability mass function

P(X=0) = 0.0602

P(X=1) = 0.0908

P(X=2) = 0.1704

P(X=3) = 0.2055

P(X=4) = 0.1285

P(X=5) = 0.1550

P(X=6) = 0.1427

P(X=7) = 0.0390

P(X=8) = 0.0147

NOTE: the sum of the probabilities gives 1.0068 for rounding errors. It can be divided by 1.0068 to get the adjusted values.

(b) Cumulative distribution function of X

F(X=0) = 0.0602

F(X=1) = 0.1510

F(X=2) = 0.3214

F(X=3) = 0.5269

F(X=4) = 0.6554

F(X=5) = 0.8104

F(X=6) = 0.9531

F(X=7) = 0.9921

F(X=8) = 1.0068

Step-by-step explanation:

Let X be the number of people who arrive late to the seminar, we can assess that X can take values from 0 (everybody on time) to 8 (everybody late).

<u>For X=0</u>

This happens when every couple and the singles are on time (ot).

$P(X=0)=P(\#1=ot)*P(\#2=ot)*P(\#3=ot)*P(\#4=ot)*P(\#5=ot)\\\\P(X=0)=(1-0.43)^{5}=0.57^5= 0.0602$

<u>For X=1</u>

This happens when only one single arrives late. It can be #4 or #5. As the probabilities are the same (P(#4=late)=P(#5=late)), we can multiply by 2 the former probability:

$P(X=1) = P(\#4=late)+P(\#5=late)=2*P(\#4=late)\\\\P(X=1) = 2*P(\#1=ot)*P(\#2=ot)*P(\#3=ot)*P(\#4=late)*P(\#5=ot)\\\\P(X=1) = 2*0.57*0.57*0.57*0.43*0.57\\\\P(X=1) = 2*0.57^4*0.43=2*0.0454=0.0908$

<u>For X=2</u>

This happens when

1) Only one of the three couples is late, and the others cooples and singles are on time.

2) When both singles are late , and the couples are on time.

$P(X=2)=3*(P(\#1=l)*P(\#2=ot)*P(\#3=ot)*P(\#4=ot)*P(\#5=ot))+P(\#1=ot)*P(\#2=ot)*P(\#3=ot)*P(\#4=l)*P(\#5=l)\\\\P(X=2)=3*(0.43*0.57^4)+(0.43^2*0.57^3)=0.1362+0.0342=0.1704$

<u>For X=3</u>

This happens when

1) Only one couple (3 posibilities) and one single are late (2 posibilities). This means there are 3*2=6 combinations of this.

$P(X=3)=6*(P(\#1=l)*P(\#2=ot)*P(\#3=ot)*P(\#4=l)*P(\#5=ot))\\\\P(X=3)=6*(0.43^2*0.57^3)=6*0.342=0.2055$

<u>For X=4</u>

This happens when

1) Only two couples are late. There are 3 combinations of these.

2) Only one couple and both singles are late. Only one combination of these situation.

$P(X=4)=3*(P(\#1=l)*P(\#2=l)*P(\#3=ot)*P(\#4=ot)*P(\#5=ot))+P(\#1=l)*P(\#2=ot)*P(\#3=ot)*P(\#4=l)*P(\#5=l)\\\\P(X=4)=3*(0.43^2*0.57^3)+(0.43^3*0.57^2)\\\\P(X=4)=3*0.0342+ 0.0258=0.1027+0.0258=0.1285$

<u>For X=5</u>

This happens when

1) Only two couples (3 combinations) and one single are late (2 combinations). There are 6 combinations.

$P(X=6)=6*(P(\#1=l)*P(\#2=l)*P(\#3=ot)*P(\#4=l)*P(\#5=ot))\\\\P(X=6)=6*(0.43^3*0.57^2)=6*0.0258=0.1550$

<u>For X=6</u>

This happens when

1) Only the three couples are late (1 combination)

2) Only two couples (3 combinations) and one single (2 combinations) are late

$P(X=6)=P(\#1=l)*P(\#2=l)*P(\#3=l)*P(\#4=ot)*P(\#5=ot)+6*(P(\#1=l)*P(\#2=l)*P(\#3=ot)*P(\#4=l)*P(\#5=ot))\\\\P(X=6)=(0.43^3*0.57^2)+6*(0.43^4*0.57)\\\\P(X=6)=0.0258+6*0.0195=0.0258+0.1169=0.1427$

<u>For X=7</u>

This happens when

1) Only one of the singles is on time (2 combinations)

$P(X=7)=2*P(\#1=l)*P(\#2=l)*P(\#3=l)*P(\#4=l)*P(\#5=ot)\\\\P(X=7)=2*0.43^4*0.57=0.0390$

<u>For X=8</u>

This happens when everybody is late

$P(X=8)=P(\#1=l)*P(\#2=l)*P(\#3=l)*P(\#4=l)*P(\#5=l)\\\\P(X=8) = 0.43^5=0.0147$

8 0

1 year ago

Six years ago, Kelsey opened a savings account that earns 1.2% simple interest every year. She started the account with $600 and

densk [106]

Answer:

600 times .012=7.20

7.20 times 6= 43.20

43.20+ 600= 643.20

the correct answer is b- $643.20

5 0

2 years ago

Read 2 more answers

Kelly hiked in the woods it took her 1/14 hour to walk 1/4 mile after she snacked she walked another 1/6 mile in 1/16 hour

mariarad [96]

General Idea:

The relationship between rate(R), distance(D) and time(T) given below:

$R= \frac{D}{T}$

Applying the concept:

We need to make use of the formula to find Kelly's walking rate before and after her snack

$Kelly \; Walking \; Rate \; before \; Snack:\\Distance \div Time = \frac{1}{4} \div \frac{1}{14} = \frac{1}{4} \times \frac{14}{1} = \frac{14}{4} = \frac{7}{2} = 3 \frac{1}{2} \; miles \; per \; hour\\\\Kelly \; Walking \; Rate \; after \; Snack:\\Distance \div Time = \frac{1}{6} \div \frac{1}{16} = \frac{1}{6} \times \frac{16}{1} = \frac{16}{6} = \frac{8}{3} = 2 \frac{2}{3} \; miles \; per \; hour\\\\$

Option A isn't correct because before snack Kelly walking rate is not 4/14 miles per hour.

Option B is <u>Correct,</u> Kelly walking rate after snack is 2 2/3 miles per hour.

Option C isn't correct because it doesn't took Kelly 2 hours longer to walk 1/6 mile than it did for her to walk 1/4 mile. It took 1/112 hour longer.

$\frac{1}{14} - \frac{1}{16} = \frac{1 \cdot 8}{14 \cdot 8} - \frac{1 \cdot 7}{16 \cdot 7} = \frac{8}{112} - \frac{7}{112} = \frac{1}{112}$

Option D isn't correct because 2 2/3 miles per hour is slower than 3 1/2 miles per hour.

Conclusion:

Option B is <u>Correct,</u> Kelly walking rate after snack is 2 2/3 miles per hour.

3 0

2 years ago

Read 2 more answers