All categories

3 years ago

Look at this table of values. What is the y-intercept?

Mathematics

2 answers:

ratelena [41]3 years ago

6 0

Answer:

I think y=3

Step-by-step explanation:

Elanso [62]3 years ago

5 0

Answer:

Step-by-step explanation:

Y=3

You might be interested in

1. WRUF 103.7 FM is giving away tickets to a concert. Tim, the general manager of the

Eva8 [605]

Answer:

Kindly check explanation

Step-by-step explanation:

Given that :

Upper level seat = $25

Mid level seat = $40

Number of tickets to give away is at least 25

Budget constraint = $1000

part A: write a system of two inequalities that describe this situation

Number of tickets constraint:

Upper level + mid level ≥ 25

u + m ≥ 25

Cost constraint :

$25u + $40m ≤ $1000

Part B:

Give away 10 upper level seat tickets and 15 mid level seat tickets

8 0

3 years ago

A restaurant advertises their lunch menu has 200 different combinations if you choose a soup, salad, and sandwich. There are 5 s

BARSIC [14]

Let me see if 57.3 is correct I’m not sure tho does it give you options?

7 0

4 years ago

Andrew bought 3 pairs of shoe shoes for 165 dollars and what if he bought 11 shoes how much money

LuckyWell [14K]

3 pairs cost $165
165/3 = 55
1 pair cost $55
Eleven pairs would cost 11(55) = $605

4 0

3 years ago

Fill in the blanks. Factor x 10=9x20=factor

sertanlavr [38]

The answer is x=1.00

4 0

3 years ago

A rare form of malignant tumor occurs in 11 children in a million, so its probability is 0.000011. Four cases of this tumor occ

Shkiper50 [21]

Answer:

a) The mean number of cases is 0.14608 cases.

b) The probability that the number of cases is exactly 0 or 1 is 0.990.

c) The probability of more than one case is 0.010

d) No, because the probability of more than one case is very small

Step-by-step explanation:

We can model this problem with a Poisson distribution, with parameter:

$\lambda=r*t=0.000011*13,280=0.14608$

a) The mean amount of cases is equal to the parameter λ=0.14608.

b) The probability of having 0 or 1 cases is:

$P(k=0)=\frac{\lambda^0 e^{-\lambda}}{0!}=\frac{1*0.864}{1} =0.864\\\\ P(k=1)=\frac{\lambda^1 e^{-\lambda}}{0!}=\frac{0.14608*0.864}{1} =0.126\\\\P(k\leq1)=0.864+0.126=0.990$

c) The probability of more than one case is:

$P(k>1)=1-P(k\leq 1)=1-0.990=0.010$

d) The cluster of 4 cases can not be due to pure chance, as it is a very high proportion of cases according to the average rate. Just having more than one case has a probability of 1%.

7 0

3 years ago