One step equations- solve for "X" or find it. Please help
1 answer:
Since these are one-step equations we only have one step.
To solve, these single steps will include the following:
[1] addition (+)
[1] subtraction (-)
[2] division (/ or ÷)
[2] multiplication (* or x)
How do we answer these? We do the opposite of what is shown. If we have addition in 4 + x = 6, we will use subtraction to solve. The numbers one (1) and two (2) above show the "sets" of which we use.
[] 3x = -12
-> -4
We have multiplication so we will divide -12 by 3 for an answer of -4
[] 14x = 280
-> 20
We have multiplication so we will divide 280 by 14 for an answer of 20
[] x + 6 = -5
-> -11
We have addition so we will subtract 6 from -5 for an answer of -11
[] x + 27 = 125
-> 98
We have addition so we will subtract 27 from 125 for an answer of 98
[] x - 324 = 760
-> 1,084
We have subtraction so we will add 324 to 760 for an answer of 1,084
[] x - 43 = 35
-> 78
We have subtraction so we will add 43 to 35 for an answer of 78
[] x/6 = -24
-> -144
We have division so we will multiply 6 by -24 for an answer of -144
[] x/13=214
-> 2,782
We have division so we will multiply 13 by 214 for an answer of 2,782
[] 26 - x = 47
-> -21
We have subtraction so we will add x to both sides to reach the equation 26 = 47 + x
Then, we have addition so we will subtract 47 from 26 to get an answer of -21
[] -x/3=-998
-> 2,994
First, we will move the negative to the bottom of the fraction, which looks like this:
Then, we have division so we will multiply -3 by -998 for an answer of 2,994
Have a nice day!
I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)
- Heather
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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 <em>X</em> = number of aircraft arrive at a certain airport during 1-hour period.
The arrival rate is, <em>λ</em>t = 8 per hour.
(a)
For <em>t</em> = 1 the average number of aircraft arrival is:
The probability distribution of a Poisson distribution is:
Compute the value of P (X = 6) as follows:
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:
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:
Thus, the probability that at least 10 small aircraft arrive during a 1-hour period is 0.2834.
(b)
For <em>t</em> = 90 minutes = 1.5 hour, the value of <em>λ</em>, the average number of aircraft arrival is:
The expected value of the number of small aircraft that arrive during a 90-min period is 12.
The standard deviation is:
The standard deviation of the number of small aircraft that arrive during a 90-min period is 3.464.
(c)
For <em>t</em> = 2.5 the value of <em>λ</em>, the average number of aircraft arrival is:
Compute the value of P (X ≥ 20) as follows:
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:
Thus, the probability that at most 10 small aircraft arrive during a 2.5-hour period is 0.0108.