<em>The</em><em> </em><em>right</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em>5</em><em>5</em><em>3</em><em>8</em><em>.</em><em>9</em><em>6</em><em> </em><em>units</em><em>^</em><em>2</em>
<em>please</em><em> </em><em>see</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em><em> for</em><em> </em><em>full</em><em> solution</em><em> </em>
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<em>Good</em><em> </em><em>luck</em><em> </em><em>on</em><em> </em><em>your</em><em> </em><em>assignment</em>
Answer:
Slopem=74
As a decimal:
m = 1.75
Step-by-step explanation:
When we multiply a whole number with a fraction, we can think of the whole numbers as "over one"...
5 3 5 15
3 * --- = --- * ---- = ----
6 1 6 6
We can further simplify the fraction, by dividing numerator and denominator by 3...
15 15 / 3 5
--- = ------- = ----
6 6 / 3 2
Since this is an improper fraction, we need to turn it into a mixed number...
5 2 + 2 + 1 2 2 1 1 1
---- = -------------- = --- + ---- + --- = 1 + 1 + ---- = 2 --- or 2.5
2 2 2 2 2 2 2
Answer: (a) 0.006
(b) 0.027
Step-by-step explanation:
Given : P(AA) = 0.3 and P(AAA) = 0.70
Let event that a bulb is defective be denoted by D and not defective be D';
Conditional probabilities given are :
P(D/AA) = 0.02 and P(D/AAA) = 0.03
Thus P(D'/AA) = 1 - 0.02 = 0.98
and P(D'/AAA) = 1 - 0.03 = 0.97
(a) P(bulb from AA and defective) = P ( AA and D)
= P(AA) x P(D/AA)
= 0.3 x 0.02 = 0.006
(b) P(Defective) = P(from AA and defective) + P( from AAA and defective)
= P(AA) x P(D/AA) + P(AAA) x P(D/AAA)
= 0.3(0.02) + 0.70(0.03)
= 0.027
