Answer:
Part A:
C= cash Sue earns
H = hours Sue works
C is dependent upon H because C only changes when H is changed.
Part B:
(0,0)
(1,6)
Part C:
C=6H
Part (a)
There are 7 red out of 7+3 = 10 total
<h3>Answer: 7/10</h3>
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Part (b)
We have 3 green out of 10 total
<h3>Answer: 3/10</h3>
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Part (c)
3/10 is the probability of getting green on any selection. This is because we put the first selection back (or it is replaced with an identical copy)
So (3/10)*(3/10) = 9/100 is the probability of getting two green in a row.
<h3>Answer: 9/100</h3>
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Part (d)
Similar to part (c) we have 7/10 as the probability of getting red on each independent selection.
(7/10)*(7/10) = 49/100
<h3>Answer: 49/100</h3>
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Part (e)
7/10 is the probability of getting red and 3/10 is the probability of getting green. Each selection is independent of any others.
(7/10)*(3/10) = 21/100
<h3>Answer: 21/100</h3>
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Part (f)
We have the exact same set up as part (e). Notice how (7/10)*(3/10) is the same as (3/10)*(7/10).
<h3>Answer: 21/100</h3>
Yeah it is D i hope this helps u
Answer: 1) D 2) B
<u>Step-by-step explanation:</u>
1) The denominator cannot equal zero, Factor the denominator to find the zeros. n³ - 4n² + 3n = 0
n(n - 3)(n - 1) = 0
n=0 n-3=0 n-1=0
n=0 n=3 n=1
2) In order to be a rational expression, it must be in the form 
where both a and b are integers.
Since 
is irrational and not an integer, then the expressions containing an irrational term <em>after simplified</em> cannot result in an integer.
Therefore, options <em>(i)</em> and <em>(iv)</em> are not rational expressions.
Option <em>(iii)</em> contains b as an exponent. Since there is no information about b, it could be a fraction, which means it could be an irrational number.
Answer:
d. She should reject the null hypothesis because p < 0.10.
Step-by-step explanation:
We have a t statistic, so let's solve for the P-value on our calculators. (tcdf on a TI-84 calculator is 2nd->VARS->6.)
tcdf(left bound, right bound, degrees of freedom)
- Our left bound is t=1.457.
- Our right bound is infinity, because we're interested in the hypothesis µ>40 mg/dL. We use 999 to represent infinity in the calculator.
- Our degrees of freedom is n-1 = 15-1 = 14.
tcdf(1.457,999,14) = .084
.084 < P-value of .10, so we reject the null hypothesis.
