How do I find the missing side
1 answer:
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That's a pretty tall order for Brainly homework. Let's start with the depressed cubic, which is simpler.
Solve
We'll put coefficients on the coefficients to avoid fractions down the road.
The key idea is called a split, which let's us turn the cubic equation in to a quadratic. We split unknown y into two pieces:
Substituting,
Expanding it out,
There a few moves we could make from here. The easiest is probably to try to solve the simultaneous equations:
which would give us a solution to the cubic.
Substituting,
By the quadratic formula (note the shortcut from the even linear term):
By the symmetry of the problem (we can interchange s and t without changing anything) when s is one solution t is the other:
We've arrived at the solution for the depressed cubic:
This is all three roots of the equation, given by the three cube roots (at least two complex), say for the left radical. The two cubes aren't really independent, we need their product to be .
That's the three roots of the depressed cubic; let's solve the general cubic by reducing it to the depressed cubic.
We want to eliminate the squared term. If substitute x = y + k we'll get a 3ky² from the cubic term and ay² from the squared term; we want these to cancel so 3k=-a.
Substitute x = y - a/3
Comparing that to
we have
which we can substitute in to the depressed cubic solution and subtract a/3 to get the three roots. I won't write that out; it's a little ugly.
Complete Question
The complete question is shown on the first uploaded image
Answer:
Yes the students are reasonably good at estimating one minute
a
Option A is correct
b
The test statistics is
Step-by-step explanation:
From the question we are told that
The set of data is
68, 82 , 38 , 62 , 41, 25 , 57 , 64, 67, 47, 61, 71, 91, 87, 64
The population mean is
The level of significance is given as
The critical value for this level of significance obtained from the normal distribution table is
The null hypothesis is
The alternative hypothesis is
Generally the sample mean is mathematically represented as
where n = 15
So
The standard deviation is mathematically represented as
substituting values
=>
The test statistics is evaluated as
substituting values
Now comparing the statistics and the critical value of the level of significance we see that the the test statistics is less than the critical value
Hence the we fail to reject the null hypothesis which mean that these times are from a population with a mean equal to 60 seconds
So we can state that yes the students are reasonably good at estimating one minute given that the sample mean is not far from the population mean
Step-by-step explanation:
I hope this is right, I used demos
