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spin [16.1K]
3 years ago
13

Jena's income is $1600 a month, and she plans to budget 1/3 of her income for rent and 1/8 of her income for groceries. Step 2 o

f 2 : What amount of money does she intend to spend each month on these two items together? Round your answer to the nearest dollar.
Mathematics
1 answer:
WARRIOR [948]3 years ago
7 0

Answer:

Jena spends $ 733. 33 on her rent and groceries  altogether per month.

Step-by-step explanation:

Jena's total income in a month= $1600

Here , expenditure on rent = (1/3) of her income

So, now one- third of her income $1600 is  

= \frac{1}{3}  \times 1600 = 533.33

or, (1/3) of her salary is  $533. 33

Hence, Jena spends $ 533. 33 on her rent monthly.

Now, her expenditure on groceries = (1/8) of her income

So, now one- eighth of her income $1600 is  

= \frac{1}{8}  \times 1600 = 200

or, (1/8) of her salary is $ 200.

Jena spends  $ 200. on her groceries  monthly.

So, Total expenditure  = Expenditure on rent + Expenditure on groceries

= $533. 33 + $200 = $733.33

Hence, Jena spends $ 733. 33 on her rent and groceries  altogether per month.

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In a chemical plant, 24 holding tanks are used for final product storage. Four tanks are selected at random and without replacem
Damm [24]

Answer:

a) P(A) = 0,4607     or   P(A) = 46,07 %

b) P(B) = 0,7120   or 71,2 %

c) P(C) = 0,2055  or P(C) = 20,55 %

Step-by-step explanation:

We will use two concepts in solving this problem.

1.- The probability of an event (A) is for definition:

P(A) = Number of favorable events/ Total number of events FE/TE

2.- If A and B are complementary events ( the sum of them is equal to 1) then:

P(A) = 1 - P(B)

a) The total number of events is:

C ( 24,4) = 24! / 4! ( 24 - 4 )!    ⇒  C ( 24,4) = 24! / 4! * 20!

C ( 24,4) = 24*23*22*21*20! / 4! * 20!  

C ( 24,4) = 24*23*22*21/4*3*2

C ( 24,4) = 24*23*22*21/4*3*2    ⇒  C ( 24,4) =  10626

TE = 10626

Splitting the group of tanks in two 6 with h-v  and 24-6 (18) without h-v

we get that total number of favorable events is the product of:

FE = 6* C ( 18, 3)  = 6 * 18! / 3!*15!  =  18*17*16*15!/15!

FE =  4896

Then P(A) ( 1 tank in the sample contains h-v material is:

P(A) = 4896/10626

P(A) = 0,4607     or   P(A) = 46,07 %

b) P(B) will be the probability of at least 1 tank contains h-v

P(B) = 1 - P ( no one tank with h-v)

Again Total number of events is 10626

The total number of favorable events for the ocurrence of P is C (18,4)

FE = C (18,4) = 18! / 14!*4! = 18*17*16*15*14!/14!*4!

FE = 18*17*16*15/4*3*2  = 3060

Then P = 3060/10626

P = 0,2879

And the probability we are looking for is

P(B) = 1 - 0,2879

P(B) = 0,7120   or 71,2 %

c) We call P(C) the probability of finding exactly 1 tank with h-v and t-i

having 4 with t-i tanks is:

reasoning the same way but now having 4 with t-i (impurities) number of favorable events is:

FE = 6*4* C(14,2) = 24 * 14!/12!*2!

FE = 24* 14*13*12! / 12!*2

FE = 24*14*13/2    ⇒  FE = 2184

And again as the TE = 10626

P(C) = 2184/ 10626

P(C) = 0,2055  or P(C) = 20,55 %

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Answer:

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Step-by-step explanation:

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