Six friends are going to sit together in a roller coaster car that has six seats in how many different arrangements can they sit

Members of a college chemistry department agree to contribute equal amounts of money to make up a scholarship fund of $280. Then

Zarrin [17]

Answer:

7 members

Step-by-step explanation:

Lets say that the department has n members. If each member contributes with an equal amount x and make a total of $280, we know that:

n*x = 280

Where each one colaborated with:

n*x/n = 280/n

Or what is the same that: (as the n on the left eliminate themselves)

x = 280/n (*)

Now the department hires 3 more members, so we now have n+3 members. And we know that the pass from contributing x to contribute $30 less, it is, x-30 (I omit the $ symbol for simplicity). So, know we have:

x-30 = 280/(n+3)

We can replace x here by the formulation of x we did in the (*) equation:

( 280/n) - 30 = 280/(n+3)

using common denominator:

( 280/n) - 30 n/n = 280/(n+3)

( 280 - 30 n)/n = 280/(n+3)

Cross multiplying the denominators:

( 280 - 30 n)*(n+3) = 280*(n)

280n - 30n^2 + 840 - 90n = 280n

Subtracting 280 n in both sides:

30n^2 + 840 - 90n = 0

Dividing both sides by 30:

-n^2 - 3n + 28 = 0

Factorizing this term (you can notice the factor immediately or solving with baskara resolution, I omit the baskara for simplicity):

-(n + 7)(n - 4) =0

So, n is equal to 4 and -7. As we are dealing with people we can not have -7 people, so we keep only n=4.

So, previous the new hiring there were 4 members, now there are 7.

The used to collaborate with 280/4 = $70, now the collaborate with 280/7=$40, and we see the collaboration decreased in $30.

7 0

3 years ago

Substitutions y=x+4 y=3x+0

cluponka [151]

Answer:

x = 2

Step-by-step explanation:

since both equal y, you can set them equal to each other

x + 4 = 3x + 0

subtract x on both sides

4 = 2x

divide by 2 on both sides

2 = x

♡♡ hope this helped ♡♡

4 0

3 years ago

Read 2 more answers

Share £40 in the ratio 1:4 between Tim and Sam

lesantik [10]

Answer:

$\£8:\£32$

Step-by-step explanation:

$\frac{40}{1+4}$

$\frac{40}{5}$

$=8$

$1:4$

$1 \times 8:4 \times 8$

$8:32$

7 0

3 years ago

Read 2 more answers

For what value is a is (a,9) a solution of the equation y=2x+1

yanalaym [24]

Answer:

a = 4

Step-by-step explanation:

y = 2x + 1 ......(1)

Going through y - y1 = m(x - x1)

m is slope and it is 2

y1 = 9 and x1 = a

y - 9 = 2(x - a)

y - 9 = 2x - 2a

y = 2x - 2a + 9 ........(2)

Equating (1) and (2)

2x + 1 = 2x - 2a + 9

Collecting like terms

2x - 2x + 2a = 9 - 1

2a = 8

a = 8/2

a = 4

7 0

3 years ago

Suppose small aircraft arrive at a certain airport according to a Poisson process with rate a 5 8 per hour, so that the number o

timurjin [86]

Answer:

(a) P (X = 6) = 0.12214, P (X ≥ 6) = 0.8088, P (X ≥ 10) = 0.2834.

(b) The expected value of the number of small aircraft that arrive during a 90-min period is 12 and standard deviation is 3.464.

(c) P (X ≥ 20) = 0.5298 and P (X ≤ 10) = 0.0108.

Step-by-step explanation:

Let the random variable X = number of aircraft arrive at a certain airport during 1-hour period.

The arrival rate is, λt = 8 per hour.

(a)

For t = 1 the average number of aircraft arrival is:

$\lambda t=8\times 1=8$

The probability distribution of a Poisson distribution is:

$P(X=x)=\frac{e^{-8}(8)^{x}}{x!}$

Compute the value of P (X = 6) as follows:

$P(X=6)=\frac{e^{-8}(8)^{6}}{6!}\\=\frac{0.00034\times262144}{720}\\ =0.12214$

Thus, the probability that exactly 6 small aircraft arrive during a 1-hour period is 0.12214.

Compute the value of P (X ≥ 6) as follows:

$P(X\geq 6)=1-P(X$

Thus, the probability that at least 6 small aircraft arrive during a 1-hour period is 0.8088.

Compute the value of P (X ≥ 10) as follows:

$P(X\geq 10)=1-P(X$

Thus, the probability that at least 10 small aircraft arrive during a 1-hour period is 0.2834.

(b)

For t = 90 minutes = 1.5 hour, the value of λ, the average number of aircraft arrival is:

$\lambda t=8\times 1.5=12$

The expected value of the number of small aircraft that arrive during a 90-min period is 12.

The standard deviation is:

$SD=\sqrt{\lambda t}=\sqrt{12}=3.464$

The standard deviation of the number of small aircraft that arrive during a 90-min period is 3.464.

(c)

For t = 2.5 the value of λ, the average number of aircraft arrival is:

$\lambda t=8\times 2.5=20$

Compute the value of P (X ≥ 20) as follows:

$P(X\geq 20)=1-P(X$

Thus, the probability that at least 20 small aircraft arrive during a 2.5-hour period is 0.5298.

Compute the value of P (X ≤ 10) as follows:

$P(X\leq 10)=\sum\limits^{10}_{x=0}(\frac{e^{-20}(20)^{x}}{x!})\\=0.01081\\\approx0.0108$

Thus, the probability that at most 10 small aircraft arrive during a 2.5-hour period is 0.0108.

8 0

3 years ago