Answer:
A
Step-by-step explanation:
4 x $0.85= $3.40
$9-$3.40= $5.60
We know that
<span>The angular velocity is 10π rad/sec
step 1
convert to rad/min
1 min-----60 sec
X min----> 1 sec
X=1/60 min
V=</span>10π rad/(1/60)------> V=600 π rad/min
the answer is the option
<span>D)The angular velocity is 600π rad/min.</span>
Answer:
A salad costs $2.50
A sandwich costs $4.50
A drink costs $1.25
Step-by-step explanation:
Let x represent the salad, y represent the sandwich, and z represent the drink.
Since three salads, two sandwiches, and one drink cost $17.75:
3x + 2y + z = 17.75 (1)
Since one salad, one sandwich, and three drinks cost $10.75:
x + y + 3z = 10.75 (2)
Since a salad costs twice as much as a drink:
x = 2z (3)
Multiply the equation 2 by -2 then, sum the equation 1 and equation 2:
-2x - 2y - 6z = -21.50
3x + 2y + z = 17.75
→ x - 5z = -3.75
Replace the x with 2z using equation 3:
2z - 5z = -3.75
-3z = -3.75
z = 1.25
x = 2z → x = 2.50
x + y + 3z = 10.75 → 2.50 + y + 3.75 = 10.75 → y + 6.25 = 10.75 → y = 4.50
Complete Question
The Brown's Ferry incident of 1975 focused national attention on the ever-present danger of fires breaking out in nuclear power plants. The Nuclear Regulatory Commission has estimated that with present technology there will be on average, one fire for every 10 years for a reactor. Suppose that a certain state has two reactors on line in 2020 and they behave independently of one another. Assuming the incident of fires for individual reactors can be described by a Poisson distribution, what is the probability that by 2030 at least two fires will have occurred at these reactors?
Answer:
The value is 
Step-by-step explanation:
From the question we are told that
The rate at which fire breaks out every 10 years is
Generally the probability distribution function for Poisson distribution is mathematically represented as

Here x represent the number of state which is 2 i.e 
Generally the probability that by 2030 at least two fires will have occurred at these reactors is mathematically represented as

=> ![P(x_1 + x_2 \ge 2 ) = 1 - [P(x_1 + x_2 = 0 ) + P( x_1 + x_2 = 1 )]](https://tex.z-dn.net/?f=P%28x_1%20%2B%20x_2%20%5Cge%202%20%29%20%3D%20%201%20-%20%5BP%28x_1%20%2B%20x_2%20%3D%200%20%29%20%2B%20P%28%20x_1%20%2B%20x_2%20%3D%201%20%29%5D)
=> ![P(x_1 + x_2 \ge 2 ) = 1 - [ P(x_1 = 0 , x_2 = 0 ) + P( x_1 = 0 , x_2 = 1 ) + P(x_1 , x_2 = 0)]](https://tex.z-dn.net/?f=P%28x_1%20%2B%20x_2%20%5Cge%202%20%29%20%3D%20%201%20-%20%5B%20P%28x_1%20%20%3D%200%20%2C%20%20x_2%20%3D%200%20%29%20%2B%20P%28%20x_1%20%3D%200%20%2C%20x_2%20%3D%201%20%29%20%2B%20P%28x_1%20%2C%20x_2%20%3D%200%29%5D)
=> 
=> ![P(x_1 + x_2 \ge 2 ) = 1 - \{ [ \frac{1^0}{ 0! } * e^{-1}] * [[ \frac{1^0}{ 0! } * e^{-1}]] )+ ( [ \frac{1^1}{1! } * e^{-1}] * [[ \frac{1^1}{ 1! } * e^{-1}]] ) + ( [ \frac{1^1}{ 1! } * e^{-1}] * [[ \frac{1^0}{ 0! } * e^{-1}]]) \}](https://tex.z-dn.net/?f=P%28x_1%20%2B%20x_2%20%5Cge%202%20%29%20%3D%20%201%20-%20%5C%7B%20%5B%20%5Cfrac%7B1%5E0%7D%7B%200%21%20%7D%20%2A%20e%5E%7B-1%7D%5D%20%2A%20%5B%5B%20%5Cfrac%7B1%5E0%7D%7B%200%21%20%7D%20%2A%20e%5E%7B-1%7D%5D%5D%20%29%2B%20%28%20%5B%20%5Cfrac%7B1%5E1%7D%7B1%21%20%7D%20%2A%20e%5E%7B-1%7D%5D%20%2A%20%5B%5B%20%5Cfrac%7B1%5E1%7D%7B%201%21%20%7D%20%2A%20e%5E%7B-1%7D%5D%5D%20%29%20%2B%20%28%20%5B%20%5Cfrac%7B1%5E1%7D%7B%201%21%20%7D%20%2A%20e%5E%7B-1%7D%5D%20%2A%20%5B%5B%20%5Cfrac%7B1%5E0%7D%7B%200%21%20%7D%20%2A%20e%5E%7B-1%7D%5D%5D%29%20%5C%7D)
=> ![P(x_1 + x_2 \ge 2 )= 1- [[0.3678 * 0.3679] + [0.3678 * 0.3679] + [0.3678 * 0.3679] ]](https://tex.z-dn.net/?f=P%28x_1%20%2B%20x_2%20%5Cge%202%20%29%3D%201-%20%5B%5B0.3678%20%20%2A%200.3679%5D%20%2B%20%5B0.3678%20%20%2A%200.3679%5D%20%2B%20%5B0.3678%20%20%2A%200.3679%5D%20%20%5D)
