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Alecsey [184]
2 years ago
14

Let R be the event that a randomly chosen person lives in the city of Raleigh. Let O be the event that a randomly chosen person

is over 50 years old. Place the correct event in each response box below to show: Given that the person lives in Raleigh, the probability that a randomly chosen person is over 50 years old.
Mathematics
1 answer:
Anit [1.1K]2 years ago
7 0

Answer:

P(O|R)

Step-by-step explanation:

The conditional probability notation of two events A and B can be written as either P(A|B) or P(B|A).

The '|' sign is read as 'given'. So, P(A|B) is read as the probability of event A given event B which implies that it is the probability that event A will occur given that event B has already occurred.

In the question,

Event R = Person lives in the city of Raleigh

Event O = Person is over 50 years old

The statement says, 'given that the person lives in Raleigh' which means that event R has already occurred and we need to find the probability of event O (the randomly chosen person is over 50 years old).

Hence, this statement can be given in conditional probability notation as

P(O|R)

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Match each shape on the left with its area formula on the right.
e-lub [12.9K]

Step-by-step explanation:

A=1/2(B1+B2)h is a trapezoid

A=1/2bh is a triangle

A=bh is a parallelogram

And A=Pir2 is a circle

5 0
3 years ago
A number, X, rounded to 1 decimal place is 3,7<br> Write down the error interval for x,
sp2606 [1]

Answer:

The error interval for x is:

                 [3.65,3.74]

Step-by-step explanation:

The number after rounding off is obtained as:

                         3.7

We know that any of the number below on rounding off the number to the first decimal place will result in 3.7:

  3.65     3.66    3.67    3.68    3.69    3.70    3.71   3.72    3.73    3.74

( Because if we have to round off a number present in decimals to n place then if  there is a  number greater than or equal to 5 at n+1 place then it will result to the one higher digit  at nth place on rounding off and won't change the digit if it less than 5 )

        Hence, the error interval is:

            [3.65,3.74]

6 0
2 years ago
Solve the system of equations: <br> X+y+z=5<br> -3x-4y+4z=15<br> 2x-y-4z=-8
kolezko [41]

Answer:

Solution By Gauss jordan elimination method

x = 3, y = 2 and z = 4

6 0
2 years ago
Please help all of my piont help me<br><br>pleeeeeeeeeeeeeeeeeeeeeeeeeaaaaaaaaaaaaaaaaaasse
KengaRu [80]

Answer:

<h2>3) -82.</h2><h2>4)-70.</h2><h2>5) -15.</h2><h2>6)-14.</h2><h2>7)5</h2><h2>8)15</h2><h2>9)16</h2><h2>10 )16</h2><h2>11)64</h2><h2>12)21.</h2><h2>13)1500 in the middle (not above not below)</h2><h2>14)450 m</h2>
7 0
3 years ago
Read 2 more answers
NO LINKS OR FILES!
Archy [21]

(a) If the particle's position (measured with some unit) at time <em>t</em> is given by <em>s(t)</em>, where

s(t) = \dfrac{5t}{t^2+11}\,\mathrm{units}

then the velocity at time <em>t</em>, <em>v(t)</em>, is given by the derivative of <em>s(t)</em>,

v(t) = \dfrac{\mathrm ds}{\mathrm dt} = \dfrac{5(t^2+11)-5t(2t)}{(t^2+11)^2} = \boxed{\dfrac{-5t^2+55}{(t^2+11)^2}\,\dfrac{\rm units}{\rm s}}

(b) The velocity after 3 seconds is

v(3) = \dfrac{-5\cdot3^2+55}{(3^2+11)^2} = \dfrac{1}{40}\dfrac{\rm units}{\rm s} = \boxed{0.025\dfrac{\rm units}{\rm s}}

(c) The particle is at rest when its velocity is zero:

\dfrac{-5t^2+55}{(t^2+11)^2} = 0 \implies -5t^2+55 = 0 \implies t^2 = 11 \implies t=\pm\sqrt{11}\,\mathrm s \imples t \approx \boxed{3.317\,\mathrm s}

(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:

\dfrac{-5t^2+55}{(t^2+11)^2} > 0 \implies -5t^2+55>0 \implies -5t^2>-55 \implies t^2 < 11 \implies |t|

In interval notation, this happens for <em>t</em> in the interval (0, √11) or approximately (0, 3.317) s.

(e) The total distance traveled is given by the definite integral,

\displaystyle \int_0^8 |v(t)|\,\mathrm dt

By definition of absolute value, we have

|v(t)| = \begin{cases}v(t) & \text{if }v(t)\ge0 \\ -v(t) & \text{if }v(t)

In part (d), we've shown that <em>v(t)</em> > 0 when -√11 < <em>t</em> < √11, so we split up the integral at <em>t</em> = √11 as

\displaystyle \int_0^8 |v(t)|\,\mathrm dt = \int_0^{\sqrt{11}}v(t)\,\mathrm dt - \int_{\sqrt{11}}^8 v(t)\,\mathrm dt

and by the fundamental theorem of calculus, since we know <em>v(t)</em> is the derivative of <em>s(t)</em>, this reduces to

s(\sqrt{11})-s(0) - s(8) + s(\sqrt{11)) = 2s(\sqrt{11})-s(0)-s(8) = \dfrac5{\sqrt{11}}-0 - \dfrac8{15} \approx 0.974\,\mathrm{units}

7 0
2 years ago
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