The answer for this question is 24

Jasmine believe the lengths 7 in 12 in and 20 inch will form a triangle is Jasmine correct use mathematics or words to explain y

madreJ [45]

Answer:

7 in 12 30 inch jasmine correct is triangle

7 0

1 year ago

Determine whether the point (1, 1) is a solution to the system of equations. Explain your reasoning in complete sentences. graph

erastova [34]

The solutions to system of equations are the point of intersection of the graphs of the equations.
The point of intersection of the graphs of the given equation is (0, 2). Therefore, point (1, 1) is not a solution to the system of equation.

6 0

3 years ago

Please Answer!!!!

ExtremeBDS [4]

First mug holds the most

Solution:

Given that,

You are choosing between two mugs

The volume of cylinder is given as:

$V = \pi r^2h$

Where,

r is the radius and h is the height

One has a base that is 5.5 inches in diameter and a height of 3 inches

$radius = \frac{diameter}{2}$

Therefore,

$r = \frac{5.5}{2}\\\\r = 2.75$

Also, h = 3 inches

Thus volume of cylinder is given as:

$V = \pi \times 2.75^2 \times 3\\\\V = 3.14 \times 22.6875\\\\V = 71.23875 \approx 71.24$

Thus first mug holds 71.24 cubic inches

The other has a base of 4.5 inches in diameter and a height of 4 inches

$Radius = \frac{4.5}{2}\\\\r = 2.25$

h = 4 inches

Therefore,

$V = 3.14 \times 2.25^2 \times 4\\\\V = 3.14 \times 20.25\\\\V = 63.585$

Thus the second mug holds 63.585 cubic inches

On comparing, volume of both mugs,

Volume of first mug > volume of second mug

First mug holds the most

8 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

8+4(s-4t) simplify the expression. HELP, 20 POINTS

ella [17]

8+ 4s -16t= 4 (2+s-4t)

5 0

3 years ago