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Gnesinka [82]
3 years ago
14

Bella pays 7 payments of $5 each to a game store. She returns none game and receives $20 back. What is the change to the amount

of money she has.
Mathematics
2 answers:
N76 [4]3 years ago
6 0
15$ 7 payments of 5$ = 35$ if she only gets 20$ back she lost 15$
nika2105 [10]3 years ago
3 0

Answer:

Bella has $15 less

Step-by-step explanation:

Bella made 7 payments of $5 each. So, total payment made by her was 7\times$5 = $35. As such, Bella had $35 initially. She played all games and received $20 back. This means that she lost $ 15 that is $35 - $20 = $15, in the game.

Therefore, she now has $15 less than the amount of $35 that she had initially. So, $15 is the the change to the amount she has now.

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8 0
3 years ago
Please answer this correctly
RUDIKE [14]

Answer:

11\dfrac{3}{8}\text{cm}

Step-by-step explanation:

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Hope this helps!

3 0
3 years ago
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 %

5 0
3 years ago
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6 0
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soldi70 [24.7K]

104 and 105 are consecutive numbers that add up to 209. Maybe this answers your question? Let me know otherwise.

8 0
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