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elena55 [62]
3 years ago
13

At work, Elon takes a break 1/5​​ of each hour he works. He also takes a lunch break for 1/2​​ of an hour each day he works. Elo

n’s break time totals 12 1/2​​ hours at the end of 5​ days. Which equation could be used to determine the number of hours, h​ , Elon works during one 5​-day work week?
Mathematics
1 answer:
Anton [14]3 years ago
8 0

Answer:

h=50\ hours

Step-by-step explanation:

Let

h -----> the number of hours, Elon works during one 5​-day work week

we know that

12\frac{1}{2}\ h=\frac{12*2+1}{2}=\frac{25}{2}\ h

Elon’s break time must be equal to 1/2​​ of an hour multiplied by 5 days plus  1/5​​ multiplied by the total hours h

so

\frac{25}{2}=5(\frac{1}{2})+\frac{1}{5}h

Solve for h

Multiply by 10 both sides to remove the fraction

125=25+2h

2h=125-25

2h=100

h=50\ hours

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Suppose that in a senior college class of 500 students itis found that 210 smoke, 258 drink alcoholic beverages, 216 eatbetween
Katarina [22]

Answer:

a)Smoke but doesn't drink alcoholic beverages

P=0.176  or 17.6%

b)eats between meals and drinks alcoholic beverages but doesn't smoke

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c.) neither smokes nor eats between meals

P=0.342  or 34.2%

Step-by-step explanation:

Using a Venn diagram with three circles, one for each bad health habit. Let us a for the students that only smoke, b for only drink alcoholic beverages, and c for only eat between meals. Following this logic, d represents smoke and drink alcoholic beverages but not eat between meals,  e drink alcoholic beverages and eat between meals but not smoke,f eat between meals and smoke but not drink alcoholic beverages. Finally g for having all the three, this means g=52.

To find f, subtract 52 (the three problems) from 97 because this last number represents all the students that smoke and eat between meals, including the students that have the three bad habits. The same goes for 'd' and 'e'.

f=97-52=45

e=83-52=31

d=122-52=70

To find a, subtract e, f, and g from the total of smokers. This is because e, f, and g represent smoke and at least another bad habit and a represents only smoking.

a=210-d-f-g=210-70-45-52=43

The same goes for b and c.

b=258-d-e-g=258-70-31-52=105

c=216-e-f.-g=216-31-45-52=88

Adding all letters and subtract from the total to see if there is any healthy student:

500-a+b+c+d+e+f+g =500-43+105+88+70+31+45+52=500- 434=66

a)Smoke but doesn't drink alcoholic beverages

This will be 'a' (only smokes) and 'f'  ( smokes and eats between meals but doesn't drink) divided by the total of students.

P=(43+45)/500=0.176

b)eats between meals and drinks alcoholic beverages but doesn't smoke

This probability is 'e' divided by the total of students.

P=31/500=0.062

c.) neither smokes nor eats between meals

This will be 'b' (only drinks) plus the healthy students (66) divided by the total of students.

P=(105+66)/500=0.342

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3 years ago
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