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nadya68 [22]
3 years ago
6

The random variable KK has a geometric distribution with mean 16. Which of the following is closest to the standard deviation of

random variable KK ?
A. 0.0625
B. 0.9375
C. 4
D. 15.49
E. 240
Mathematics
1 answer:
asambeis [7]3 years ago
5 0

Answer:

E 240

Step-by-step explanation:

Geometric Distribution mean = (1- p) / p

16 = (1 - p) / p

16p = 1 - p  ;  17p = 1 ; p = 1/17 = 0.058  

Standard Deviation = (1 - p) / p^2

= (1 - 0.058) / (0.058)^2

0.942 / 0.003364

= 280 , closest to 240

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Three angles are in the ratio 2 : 3 : 5 The smallest angle is 50 degrees Work out the sizes of the other two angles
kvasek [131]

Answer:

Step-by-step explanation:

Let angles are 2x,3x,5x

So given 2x=50

X=25.

3x=3×25=75.

5x=5×25=125.

7 0
3 years ago
Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be
Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of t=0.6482

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

h(t)=-16t^2 +v_{o}t+s       Eqn(1).

Where:

h(t) : is the height range as a function of time

v_{o}   : is the initial velocity

s     : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

h(t)=-16t^2 + v_{o}t + 40        Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

<u>Ben:</u>

We know that v_{o}=60ft/s

Thus Eqn. (2) becomes:

h(t)=-16t^2+60t+40        Eqn.(3)

<u>Josh:</u>

We know that v_{o}=48ft/s

Thus Eqn. (2) becomes:

h(t)=-16t^2+48t+40       Eqn. (4).

<em><u>Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at </u></em>h(t)=0<em><u>. Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one. </u></em>

Thus we have for Ben, Eqn. (3) gives:

h(t)=0-16t^2+60t+40=0

Using the quadratic expression to find the roots of the quadratic we have:

t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec

Since time can only be positive we reject the t_{2} solution and we keep that Ben's book took t=4.3276 seconds to reach the ground.

Similarly solving for Josh we obtain

t_{1}=3.6794sec\\t_{2}=-0.6794sec

Thus again we reject the negative and keep the positive solution, so Josh's book took t=3.6794 seconds to reach the ground.

Then we can find the difference between Ben and Josh times as

t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482

So to answer the original question:

<em>Whose textbook reaches the ground first and by how many seconds?</em>

  • Josh's textbook reached the ground first
  • Josh's textbook reached the ground first by a difference of t=0.6482

3 0
3 years ago
Find the LCD 4/7 and 3/5
blsea [12.9K]
I would multiply 4/7 by 5 and 3/5 by 7 to get 20/35 and 21/35. The Least Common factor is 35
8 0
3 years ago
50 POINTS!!!
san4es73 [151]

Step-by-step explanation:

<u>Step 1:  Determine an ordered pair</u>

A solution of an equation just means that the point lies on the line.  We can find any y-value when we plug in a specific x-value.  For example, if we want to know what ordered pair lies at x=1, we just plug in y = -1/2(1) and solve for y which gives us -1/2.  This gives us an ordered pair of (1, -1/2).  We can continue to do this for any x value.

We can also reverse the order and plug in the y-value and get the x-value in order to accomplish the same goal but it's a bit harder.

Hope this helps!

6 0
2 years ago
Read 2 more answers
I suck at math so I'm gunna ask a bunch of questions on this work
Nastasia [14]
It would be c plz mark me brainliest
3 0
3 years ago
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