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zhannawk [14.2K]
2 years ago
8

Caroline has $2.55 in her purse and nickels and dimes. the number of nickels is nine less than three times the number of dimes.

find the number of each type of coin.
Mathematics
1 answer:
My name is Ann [436]2 years ago
6 0
.05N+.1d=2.55
N=3d-9
0.05(3d-9)+0.1d=2.55
0.15d-0.45+0.1d=2.55
0.25d=2.55+0.45
0.25d=3
D=12
N=3(12)-9
N=36-9
N=27
Hope this helps
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In a survey, the ratio of the number of people who preferred tea to those who preferred coffee was 9:5
kodGreya [7K]

Answer:

45 people prefered coffee, 81 people prefered tea. A total of 126 people were surveyed.

Step-by-step explanation:

Let t be the people who prefer tea, and let c be the people who prefer coffee.

The ratio between tea and coffee is 9:5, thus we have t/c=9/5. Cross products  give us 5t=9c.

Also, we have 36 more people prefer tea than coffee, so we have c+36=t

Thus we have: 5(c+36)=9c

5c+180=9c

180=4c

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c+36=t

45+36=81

5 0
2 years ago
The blue M&M was introduced in 1995. Before then, the color mix in a bag of plain M&Ms was (30% Brown, 20% Yellow, 20% R
IRISSAK [1]

Answer:

The probability that the yellow M&M came from the 1994 bag is 0.07407 or 7.407%

Step-by-step explanation:

Given

Before 1995

(Br) Brown = 30%

(Y) Yellow = 20%  =0.2

(R) Red = 20%

(G) Green =10%  =0.1

(O) Orange = 10%

(T) Tan = 10%

 

After 1995

(Br) Brown = 13%

(Y) Yellow = 14%  =0.14

(R) Red = 13%

(G) Green = 20% = 0.2

(O) Orange = 16%

(Bl) Blue = 24%

Since there are two bags, let A be the bag from 1994, and B be the bag from 1996

Then let AY imply we drew a yellow M&M from the 1994 bag

AG implies we drew a green M&M from the 1994 bag

BY implies imply we drew a yellow M&M from the 1996 bag

BG implies we drew a green M&M from the 1996 bag

P(AY) =0.2

P (BY) = 0.14

P(AG) =0.1

P(BG) =0.2

Since the draws from the 1994 and 1996 bag are independent,

therefore

P(AY n BG) = 0.2 * 0.2 = 0.04  -------(1)\\P(AG n BY) =0.1 * 0.14 =0.014   --------(2)\\

The draws can happen in either of the 2 ways in (1) and (2) above

therefore total probability E is given as

E =P( AY n BG) u P(AG n BY)\\=0.04 + 0.014 =0.O54

For the yellow one to be from 1994, it implies that the event to be chosen is

P(AYnBG) = 0.2*0.2

Since the total probability is given as E=0.054

then P((AYnBG) /E) =\frac{0.04}{0.054} = 0.07407

Concluding statement: This is the condition for the Yellow one to come from 1994 and green from 1996 provided that they obey the condition from E

4 0
2 years ago
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