Consider 3 trials, each having the same probability of successes. Let X denote the total number of successes in these trials.
1 answer:
Since each trial has the same probability of success,
Let, <span><span><span>Xi</span>=1</span></span> if the <span><span>i<span>th</span></span></span> trial is a success (<span>0</span> otherwise). Then, <span><span>X=<span>∑3<span>i=1</span></span><span>Xi</span></span><span>X=<span>∑<span>i=1</span>3</span><span>Xi</span></span></span>,
and <span><span>E[X]=E[<span>∑3<span>i=1</span></span><span>Xi</span>]=<span>∑3<span>i=1</span></span>E[<span>Xi</span>]=<span>∑3<span>i=1</span></span>p=3p=1.8</span><span>E[X]=E[<span>∑<span>i=1</span>3</span><span>Xi</span>]=<span>∑<span>i=1</span>3</span>E[<span>Xi</span>]=<span>∑<span>i=1</span>3</span>p=3p=1.8</span></span>
So, <span><span>p=0.6</span><span>p=0.6</span></span>, and <span><span>P{X=3}=<span>0.63</span></span><span>P{X=3}=<span>0.63</span></span></span>
I thought what I did was sound, but the textbook says the answer to (a) is <span>0.60.6</span> and (b) is <span>00</span>.
Their reasoning (for (a)) is as follows:
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Answer:
167,886,383 basketballs
Step-by-step explanation:
We are asked to find how many basketballs would take to circle around the equator. We have given the earth's radius. So, we need the formula to obtain it's perimeter. Thus:
Earth's diameter is simply radius*2. It means:
On the other hand, we have a basketball crcumference; however, we need to obtain its diameter so that we can later calculate how many basketballs fit on earth's equator by simply dividing earth's circumference by a basketball's diameter.
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We need to change units to fit in the international system.
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Finally, dividing total earth's circumference by a basketball diameter:
Answer:
Line passing through .
Graph is attached.
Step-by-step explanation:
y-intercept:
x-intercept:
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Graph is attached.
