Evaluate the expression 24x - 13y for x = 3 and y = 2 i need help

What's the answer for does too can anyone help

Anna007 [38]

Hello,
mes QRT=3x+5=3*14+5=47°
mes TRS=10x-7=10*14-7=133°
since
3x+5+10x-7=180
==>13x=184
==>x=14(°)

8 0

3 years ago

What is the equation of the line in slope-intercept form?

GREYUIT [131]

Y=5/2x+5 because the slope is 5/2 and the y-intercept is 5

6 0

3 years ago

The sum of 5 times a number and minus −2, plus 7 times a number

Olenka [21]

Answer:

12x + 2

Step-by-step explanation:

Let the number be represented by x.

Then five times the number = 5*x

Seven times the number = 7*x

Sum of 5 times the number minus -2 = $5*x - (-2)$ = $5x +2$

Adding seven times the number to this expression yields, $5x+2+7x$

$= (5+7)x+2$

$= 12x+2$

So the simplified expression corresponds to 12x + 2.

6 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

If 8 identical blackboards are to be divided among 4 schools, how many divisions are possible? how many if each school must rece

DaniilM [7]

Alright, so we'd use the combinations with repetition formula, so we choose from 4 schools to distribute to and distribute 8 blackboards. It's then

( 8+4-1)!/8!(4-1)!=11!/(3!*8!)=165

For at least one blackboard, we first distribute 1 to each school and then have 4 blackboards left, getting (4+4-1)!/4!(4-1)!=7!/(4!*3!)=35

8 0

3 years ago