Theoretically, each letter should have the same probability of occurring since there are 15 of each. There are 5 letters that can be drawn, so there is a total of 75 balls, and each letter has a probability of being drawn.
This means one would expect a theoretical frequency of , i.e. any given letter should get drawn 250 times, which means the answer is D.
Answer:
Step-by-step explanation:
From the step given step 6
(b² — 4ac) / 4a² = (x + b/2a)²
Mistake in question it is (x + b/2a)²
Step 7 given
±√(b² —4ac) /2a = x + b/2a
Mistake in question it is over 2a and not 1a.
1. The step taken from step 6 to step 7 is taking square roots of both sides
(b² — 4ac) / 4a² = (x + b/2a)²
Taking square of both sodes
√(b²—4ac) / √4a² = √(x + b/2a)²
√(b²—4ac) / 2a = x + b/2a
This is the required step 7.
Now, subtracting b/2a from both sides
√(b²—4ac) / 2a - b/2a= x + b/2a -b/2a
√(b²—4ac) / 2a — b/2a = x
(√(b²—4ac) — b)/2a = x
x = [—b ± √(b²—4ac)] / 2a
So, this is the required formula method
The discriminant is D = b²—4ac.
Answer:
Is plausible that the successive throws are independent
Step-by-step explanation:
1) Table with info given
The observed values are given by the following table
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First shot Made Second shot missed Total
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Made 152 33 185
Missed 37 8 45
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Total 189 41 230
2) Calculations and test
We are interested on check independence and for this we need to conduct a chi square test, the next step would be find the expected value:
Null hypothesis: Independence between two successive free throws
Alternative hypothesis: No Independence between two successive free throws
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First shot Made Second shot missed
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Made 189(185)/230=152.0217 41(185)/230=32.9783
Missed 189(45)/230=36.9783 41(45)/230=8.0217
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On this case all the expected values are higher than 5 and the sample size 230 is enough to apply the chi squared test.
3) Calculate the chi square statistic
The statistic for this case is given by:
Where O represent the observed values and E the expected values. Replacing the values that we got we have this
Now with the calculated value we can find the degrees of freedom
on this case r means the number of rows and c the number of columns.
Now we can calculate the p value
On this case the pvalue is a very large value and that indicates that we can fail to reject the null hypothesis of independence. So is plausible that the successive throws are independent.