Could somebody help me on this one. Due in 3 min
1 answer:
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<h3>Answer: </h3><h3>See the attached image below for the table</h3>
In column 1 we have the x values 0, 4, 8, 12
In column 2, we have the y values 1, 2, 3, 4
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Explanation:
We can start with any x value we want. I find that 0 is easiest to work with, so why not start at x = 0. For the sake of not repeating, I'll use y = 1 as the start value for y.
The first row of the table is x = 0 and y = 1.
A slope of 1/4 means rise/run = 1/4. So rise = 1 and run = 4.
A rise of 1 indicates y goes up by 1. A run of 4 means x increases by 4.
So put together, slope = 1/4 means each time x goes up by 4, y goes up by 1.
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If we started at x = 0, then going up by 4 leads to x = 4. If y starts at y = 1, then going up by 1 leads to y = 2.
The second row of the table is x = 4 and y = 2
We'll repeat the pattern of "add 4 to x, add 1 to y" getting the third row having x = 8 and y = 3
We can repeat as many times as we want. Doing another "up 1, over to the right 4" movement lands us on (x,y) = (12, 4)
The completed table is shown below in the attached image.
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If you were to use the slope formula with any two rows of the table, then you should get m = 1/4
Let's do the first two rows
m = (y2-y1)/(x2-x1)
m = (2-1)/(4-0)
m = 1/4
I'll let you check the other pairs of rows.
Answer:
Step-by-step explanation:
We are given quadratic equations as
and it can be factored as
now, we can multiply factor term
now, we can compare
so, we get
we are given that
'a' and 'b' are integers
so, we can find all possible factors
so, we can find k
At :
we can plug values
At :
we can plug values
At :
we can plug values
At :
we can plug values
So, values of k are
Answer:
(a) 0.6579
(b) 0.2961
(c) 0.3108
(d) 240
Step-by-step explanation:
The random variable <em>X</em> can be defined as the number of particles in a suspension.
The concentration of particles in a suspension is 50 per ml.
Then in 5 mL volume of the suspension the number of particles will be,
5 × 50 = 250.
The random variable <em>X</em> thus follows a Poisson distribution with parameter, <em>λ</em> = 250.
The Poisson distribution with parameter λ, can be approximated by the Normal distribution, when λ is large say λ > 10.
The mean of the approximated distribution of X is:
μ = λ = 250
The standard deviation of the approximated distribution of X is:
σ = √λ = √250 = 15.8114
Thus,
(a)
Compute the probability that the number of particles withdrawn will be between 235 and 265 as follows:
Thus, the value of P (235 < <em>X</em> < 265) = 0.6579.
(b)
Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:
Thus, the value of .
(c)
A 10 mL sample is withdrawn.
Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:
Thus, the value of .
(d)
Let the sample size be <em>n</em>.
The value of <em>z</em> for this probability is,
<em>z</em> = 1.96
Compute the value of <em>n</em> as follows:
Thus, the sample selected must be of size 240.