All categories

3 years ago

Which set of points does not represent a function

1 answer:

8 0

If it is a bunch of dots on a graph, then the answer will be the one that has two sets of dots on the same vertical line (the y axis). so if there are two dots right above and below eachother, that graph is the correct answer.

You might be interested in

A person places $290 in an investment account earning an annual rate of 2.2%, compounded continuously. Using the formula V = Pe^

Tomtit [17]

Answer:

Step-by-step explanation:

Given that the principal p= $290

rate r= 2.2% 2.2/100 =0.022

time t= 18years

by applying the expression

$V = Pe^{rt}$

We have

$V = 290e^{0.022*18}\\\\V=290e^{0.396}\\\\V=290*1.48586931755\\\\V=$430.90$

Hence after 18years the money in the account will be $430.90

8 0

3 years ago

Which of the following made Soviet leader Mikhail Gorbachev MOST eager to end the Cold War arms race?

Ivanshal [37]

D. The economy of the Soviet Union was falling behind that of other major industrialized nations.

4 0

3 years ago

Tim has two spinners , spinnerA and spinner B. Each spinner can only land on red or blue . The probability that spinner A will l

Ganezh [65]

Answer:

Step-by-step explanation:

Spinner A

Probability of red = 0.5,
Probability of blue = 0.5

Spinner B

Probability of red = 0.6,
Probability of blue = 0.4

Probability of both A and B land on red:

0.5*0.6= 0.3

Number of attempts to get outcome of 84 red on both spinners is:

84/0.3= 260

Probability of both spinners land on blue:

0.5*0.4= 0.2

Estimated number of both spinners land on blue:

260*0.2= 52

4 0

3 years ago

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

I dislike school but it’s 2.2 quick check, and it’s hard.

Mademuasel [1]

Answer:umm ok

Step-by-step explanation:

3 0

3 years ago