Answer:
a) P(X∩Y) = 0.2
b)
= 0.16
c) P = 0.47
Step-by-step explanation:
Let's call X the event that the motorist must stop at the first signal and Y the event that the motorist must stop at the second signal.
So, P(X) = 0.36, P(Y) = 0.51 and P(X∪Y) = 0.67
Then, the probability P(X∩Y) that the motorist must stop at both signal can be calculated as:
P(X∩Y) = P(X) + P(Y) - P(X∪Y)
P(X∩Y) = 0.36 + 0.51 - 0.67
P(X∩Y) = 0.2
On the other hand, the probability
that he must stop at the first signal but not at the second one can be calculated as:
= P(X) - P(X∩Y)
= 0.36 - 0.2 = 0.16
At the same way, the probability
that he must stop at the second signal but not at the first one can be calculated as:
= P(Y) - P(X∩Y)
= 0.51 - 0.2 = 0.31
So, the probability that he must stop at exactly one signal is:

Answer:
B.
Step-by-step explanation:
Parentheses mean multiplication in this case. So it would be 2 x -3 OR -3 + -3. When multiplying a negative and a positive the answer will ALWAYS be negative. So you would move to the left on a number line.
43.5 because if it were anything below that it would've been rounded down to 43 instead of up to 44.
Answer:
See answer below
Step-by-step explanation:
The possible zeroes are p/q where p is factors of the constant and q is factors of the coefficient of the largest degree.
This means possible zeroes are ±15/4, ±5/4, ±3/4, ±1/4, ±15/2, ±5/2, ±3/2, ±1/2, ±15, ±5, ±3, ±1.
<em>The</em><em> </em><em>right</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em>x^</em><em>2</em><em>-</em><em>1</em><em>.</em>
<em>EXPLANATION</em><em>:</em>
<em>To</em><em> </em><em>be</em><em> </em><em>a</em><em> </em><em>polynomial</em><em>,</em><em> </em><em>the</em><em> </em><em>power</em><em> </em><em>of</em><em> </em><em>each</em><em> </em><em>term</em><em> </em><em>must</em><em> </em><em>be</em><em> </em><em>a</em><em> </em><em>whole</em><em> </em><em>number</em><em>.</em>
<em>Hope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>be</em><em> </em><em>helpful</em><em> </em><em>to</em><em> </em><em>you</em><em>.</em><em>.</em><em>.</em>
<em>Good</em><em> </em><em>luck</em><em> </em><em>on</em><em> </em><em>your</em><em> </em><em>assignment</em>
<em>~</em><em>p</em><em>r</em><em>a</em><em>g</em><em>y</em><em>a</em>