Jimmy has 42 minutes to get ready for

The patriot diner sells 2 cheeseburgers and one soda for $11.00 and 3 hamburgersand 2 sodas for $18.00. What is the cost of a ch

kvasek [131]

Answer: the cost of a cheeseburger is $4

Step-by-step explanation:

Let x represent the cost of one cheeseburgers.

Let y represent the cost of one Soda.

The patriot diner sells 2 cheeseburgers and one soda for $11.00. It means that

2x + y = 11 - - - - - - - - - - - 1

She also sells 3 cheeseburgers and 2 sodas for $18.00. It means that

3x + 2y = 18 - - - - - - - - - - -2

Multiplying equation 1 by 3 and equation 2 by 2, it becomes

6x + 3y = 33

6x + 4y = 36

Subtracting, it becomes.

- y = - 3

y = 3

Substituting y = 3 into equation 1, it becomes

2x + 3 = 11

2x = 11 - 3 = 8

x = 8/2 = 4

6 0

3 years ago

There are 50 chairs in the auditorium. 15 of the chairs are red and the rest are blue. What is the ratio of the number of blue c

navik [9.2K]

Answer: 7:3

Step-by-step explanation:

15 chairs are red

35 are blue

35:15

Simplifies to

7:3

7 0

3 years ago

In January, Music Marathon priced all the compact discs (CD’s) at $20. In February, these same CD’s were discounted 50% and in M

skad [1K]

Answer:I think it would be $9

Step-by-step explanation:

50% of 20 is $10

10% of $10 is $1

Subtract 1 from 10, which gives you $9

3 0

3 years ago

I cant figure out 17 :3 and 68 :12 are equivalent

sleet_krkn [62]

Answer:

17:3 is 68:12 simplified

15 slices will be eaten in 10 minutes

Step-by-step explanation:

first off 17:3 is 68:12 simplified (both 68 and 12 go into 4 resulting in 17 and 3)

you have to divide 10 by 2 which will give you 5

and then multiply by 3

to get 15

5 0

3 years ago

Read 2 more answers

An article reports that 1 in 500 people carry the defective gene that causes inherited colon cancer. In a sample of 2000 individ

Tema [17]

Answer:

a) P=0.558

b) P=0.021

Step-by-step explanation:

We can model this random variable as a Poisson distribution with parameter λ=1/500*2000=4.

The approximate distribution of the number who carry this gene in a sample of 2000 individuals is:

$P(x=k)=\frac{\lambda^ke^{-\lambda}}{k!} =\frac{4^ke^{-4}}{k!}$

a) We can calculate that the approximate probability that between 4 and 9 (inclusive) as:

$P(4\leq x\leq 9)=\sum_{k=4}^9P(k)\\\\\\ P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\P(9)=4^{9} \cdot e^{-4}/9!=262144*0.0183/362880=0.013\\\\\\$

$P(4\leq x\leq 9)=\sum_{k=4}^9P(k)=0.195+0.156+0.104+0.060+0.030+0.013=0.558$

b) The approximate probability that at least 9 carry the gene is:

$P(x\geq9)=1-P(x\leq 8)\\\\\\$

$P(0)=4^{0} \cdot e^{-4}/0!=1*0.0183/1=0.018\\\\P(1)=4^{1} \cdot e^{-4}/1!=4*0.0183/1=0.073\\\\P(2)=4^{2} \cdot e^{-4}/2!=16*0.0183/2=0.147\\\\P(3)=4^{3} \cdot e^{-4}/3!=64*0.0183/6=0.195\\\\P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\$

$P(x\geq9)=1-P(x\leq 8)\\\\P(x\geq9)=1-(0.018+0.073+0.147+0.195+0.195+0.156+0.104+0.060+0.030)\\\\P(x\geq9)=1-0.979=0.021$

8 0

3 years ago