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3 years ago

Some one plz tell me how to do this

Mathematics

2 answers:

goldenfox [79]3 years ago

8 0

Subtract the 18 from the 78 then multiply both sides by -1, then divide by 5.

Eva8 [605]3 years ago

8 0

You subtract 18-5y which leaves 13 which is what your y is

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Round 100.9052 to the nearest hundredths

Bas_tet [7]

To round 100.952 to the nearest tenth consider the hundredths’ value of 100.952, which is 5 and equal or more than 5. Therefore, the tenths value of 100.952 increases by 1 to 0.

100.952 rounded to the nearest tenth = 101.0

6 0

3 years ago

you have enough tickets to play 6 different games at a amusement park. if there are 14 games, how many ways can you choose six?

Musya8 [376]

Answer:

3003 ways

Step-by-step explanation:

You can basically choose 6 games from 14 games in total. This is essential a combination problem. We want the number of ways to choose 6 things from 14 things. The general formula for combinations is:

$nCr=\frac{n!}{r!(n-r)!}$

Which tells us the number of ways to choose "r" things from a total of "n" things.

The factorial notation is:

n! = n * (n-1) * (n-2) * ....

Example: 3! = 3 * 2 * 1

Now, we know from the problem,

n = 14

r = 6

So, substituting, we get:

$nCr=\frac{n!}{r!(n-r)!}\\14C6=\frac{14!}{6!(14-6)!}\\=\frac{14!}{8!*6!}\\=\frac{14*13*12*11*10*9*8!}{6!*8!}\\=\frac{14*13*12*11*10*9}{6*5*4*3*2*1}\\=3003$

You can choose in 3003 ways

7 0

3 years ago

Bill has enough money to buy no more than 1/2 pound of cheese. How much cheese could he buy?

andriy [413]

1/2 I guess I don't think you typed in the question correctly. Not trying to be mean.

3 0

4 years ago

F-e+v=2 solve for f

Rasek [7]

$f-e+v=2\ \ \ \ |+e\\\\f+v=2+e\ \ \ |-v\\\\\boxed{f=2+e-v}$

3 0

3 years ago

Five individuals from an animal population thought to be near extinction in a certain region have been caught, tagged, and relea

Talja [164]

Answer:

a) For this case the random variable X follows a hypergometric distribution.

b) $E(X)= n\frac{M}{N}=10 \frac{5}{25}=2$

$Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}=10\frac{5}{25}\frac{25-5}{25}\frac{25-10}{25-1}=1$

c) $P(X=0)= \frac{(5C0)(25-5 C 10-0)}{25C10}=\frac{1*184756}{3268760}=0.0565$

d) $P(X=5)= \frac{(5C5)(25-5 C 10-5)}{25C10}=\frac{1*15504}{3268760}=0.00474$

Step-by-step explanation:

The hypergometric distribution is a discrete probability distribution that its useful when we have more than two distinguishable groups in a sample and the probability mass function is given by:

$P(X=k)= \frac{(MCk)(N-M C n-k)}{NCn}$

Where N is the population size, M is the number of success states in the population, n is the number of draws, k is the number of observed successes

The expected value and variance for this distribution are given by:

$E(X)= n\frac{M}{N}$

$Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}$

a. What is the distribution of X?

For this case the random variable X follows a hypergometric distribution.

b. Compute the values for E(X) and Var(X)

For this case n=10, M=5, N=25, so then we can replace into the formulas like this:

$E(X)= n\frac{M}{N}=10 \frac{5}{25}=2$

$Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}=10\frac{5}{25}\frac{25-5}{25}\frac{25-10}{25-1}=1$

c. What is the probability that none of the animals in the second sample are tagged?

So for this case we want this probability:

$P(X=0)= \frac{(5C0)(25-5 C 10-0)}{25C10}=\frac{1*184756}{3268760}=0.0565$

d. What is the probability that all of the animals in the second sample are tagged?

So for this case we want this probability:

$P(X=5)= \frac{(5C5)(25-5 C 10-5)}{25C10}=\frac{1*15504}{3268760}=0.00474$

4 0

4 years ago