Perform the operation and write the answer in the simplest form....
2 answers:
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Answer:
See explanation below
Step-by-step explanation:
<u>First we will solve the radical equation</u> (which I guess was problem 1),
Let's start by simplifying it:
Now we will solve the equation by squaring both sides of the equation:
So the calculation for x was that x = -10
However, this does not produce a solution to the equation: When we plug this value into the radical equation we get:
This happens because <u>when we first squared both sides of the equation in the first part of the problem we missed one value for x </u>(remember that all roots have 2 answers, a positive one and a negative one) while squares are always positive.
When we squared the root, we missed one value for x and that is why the calculation does not produce a solution to the equation.
Answer:
The answers are a) 1/3 and b) 1/3
Step-by-step explanation:
we will consider t to be the arrival time variable in minutes. We will have it run from 0 min to 30 min omitting the 7 hours , which won't change the results since the PDF we are about to calculate is a constant.
So the PDF of the arrival time is a constant and the since the area under this PDF distribution should be equal to 1 so, Let the height of the constant distribution equal to c, so c*30 (which is equal to the total probability) would be the area under the distribution, but this area should be equal to 1, So that gives us c=1/30 which is the value of the constant PDF for all corresponding arrival times.
part a) of the question asks for the probability that the passengers wait less than 5 minute. The passengers would have to wait for the bus 5 min or late if they arrive between the times (10 - 15 )min and (25 - 30) min. So we will have to integrate the PDF corresponding to these times and then we will will just have to add the probabilities calculated as give below,
Now part b) of the question asks for the probability that the passenger waits for more than 10 minutes. Which can be calculated by noting that that can only happen if the passenger arrives between the times (0 - 5) min and (15 - 20) min. So we will have to integrate the PDF corresponding to these times and then we will will just have to add the probabilities calculated as give below,
The answer to part b) is 1/3.
Answer:
The dimension of the box is l×w×h = (2.274 × 2.274 × 1.934) ft³
Step-by-step explanation:
From the given information;
Let a be the cost of the box
Let b be one side of the square base ; &
h to be the height of the box
We know that the volume of the box = 10 cubic feet
Then;
a²h = 10
h = 10/a²
The base = (0.65)a²
The top = (0.2)a²
The side = (0.2) a × 25/a²
= 5/a
For the four sides of the box now ;
= (0.2) 4a × 25/a²
= 0.8 × 25/a
= 20 /a
The total cost of the box is:
b = 0.65a² + 0.2a² + 20 /a
b = 0.85 a² + 20 /a
Taking differential of b with respect to a ;we have:
db/da = 1.7a - 1/a²(20) = 0
1.7 a³ - 20 = 0
1.7 a³ = 20
a³ = 20/1.7
a³ = 11.77
a =
a = 2.274 ft
Thus; the cost for the base of the box = (0.65)a²
the cost for the base of the box =(0.65) × ( 2.274)²
the cost for the base of the box = 3.362
The top of the box = (0.2)a²
The top of the box = (0.2)× ( 2.274)²
The top of the box = 1.034
The four sides of the box = 20 /a
The four sides of the box = 20/2.274
The four sides of the box = 8.795
the total cost = b = 0.85 a² + 20 /a
the total cost = 0.85 (2.274)² + 20 /2.274
the total cost = 4.395 + 8.795
the total cost = 13.19
Recall that:
the volume of the box = 10 cubic feet
Then;
a²h = 10
h = 10/a²
h = 10/ 2.274²
h 1.934
The dimension of the box is l×w×h = (2.274 × 2.274 × 1.934) ft³