What is the equation for the line that passes through the points (3, 4) and (-2, 4)?

Each question is worth 10 pts, I will do brainliest for the person who does both and follow pls help.

In-s [12.5K]

Answer:

1. 6

2.

pennies = 9

nickels = 5

quaters = 5

dimes = 10

Step-by-step explanation:

1.

18/3 = 6

as we know that there are 18 scoops, and 3 per icecream, we can divide the two to get the number of ice creams

2.

5n

#n = #q; #q = 2#d, this is a simple ratio, you just plug in the variables to get

5n = 5q = 10d = 15 coins

24 - 15 = 9 pennies

pennies = 9

nickels = 5

quaters = 5

dimes = 10

8 0

3 years ago

A coin, having probability p of landing heads, is continually flipped until at least one head and one tail have been flipped. (a

Natali [406]

Answer:

(a)

The probability that you stop at the fifth flip would be

$p^4 (1-p) + (1-p)^4 p$

(b)

The expected numbers of flips needed would be

$\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1} = 1/p$

Therefore, suppose that $p = 0.5$ , then the expected number of flips needed would be 1/0.5 = 2.

Step-by-step explanation:

(a)

Case 1

Imagine that you throw your coin and you get only heads, then you would stop when you get the first tail. So the probability that you stop at the fifth flip would be

$p^4 (1-p)$

Case 2

Imagine that you throw your coin and you get only tails, then you would stop when you get the first head. So the probability that you stop at the fifth flip would be

$(1-p)^4p$

Therefore the probability that you stop at the fifth flip would be

$p^4 (1-p) + (1-p)^4 p$

(b)

The expected numbers of flips needed would be

$\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1} = 1/p$

Therefore, suppose that $p = 0.5$ , then the expected number of flips needed would be 1/0.5 = 2.

7 0

3 years ago

Psychologist Michael Cunningham conducted a survey of university women to see whether, upon graduation, they would prefer to mar

bagirrra123 [75]

Answer:

A.The probability that exactly six of Nate's dates are women who prefer surgeons is 0.183.

B. The probability that at least 10 of Nate's dates are women who prefer surgeons is 0.0713.

C. The expected value of X is 6.75, and the standard deviation of X is 2.17.

Step-by-step explanation:

The appropiate distribution to us in this model is the binomial distribution, as there is a sample size of n=25 "trials" with probability p=0.25 of success.

With these parameters, the probability that exactly k dates are women who prefer surgeons can be calculated as:

$P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{25}{k} 0.25^{k} 0.75^{25-k}\\\\\\$

A. P(x=6)

$P(x=6) = \dbinom{25}{6} p^{6}(1-p)^{19}=177100*0.00024*0.00423=0.183\\\\\\$

B. P(x≥10)

$P(x\geq10)=1-P(x$

$P(x=0) = \dbinom{25}{0} p^{0}(1-p)^{25}=1*1*0.0008=0.0008\\\\\\P(x=1) = \dbinom{25}{1} p^{1}(1-p)^{24}=25*0.25*0.001=0.0063\\\\\\P(x=2) = \dbinom{25}{2} p^{2}(1-p)^{23}=300*0.0625*0.0013=0.0251\\\\\\P(x=3) = \dbinom{25}{3} p^{3}(1-p)^{22}=2300*0.0156*0.0018=0.0641\\\\\\P(x=4) = \dbinom{25}{4} p^{4}(1-p)^{21}=12650*0.0039*0.0024=0.1175\\\\\\P(x=5) = \dbinom{25}{5} p^{5}(1-p)^{20}=53130*0.001*0.0032=0.1645\\\\\\P(x=6) = \dbinom{25}{6} p^{6}(1-p)^{19}=177100*0.0002*0.0042=0.1828\\\\\\$

$P(x=7) = \dbinom{25}{7} p^{7}(1-p)^{18}=480700*0.000061*0.005638=0.1654\\\\\\P(x=8) = \dbinom{25}{8} p^{8}(1-p)^{17}=1081575*0.000015*0.007517=0.1241\\\\\\P(x=9) = \dbinom{25}{9} p^{9}(1-p)^{16}=2042975*0.000004*0.010023=0.0781\\\\\\$