5. A dice is biased so that it lands on 1 one quarter of the time. It lands on 3 twice as often as it land on 2 and
three times as often as it lands on 4. It lands on 5 ten percent more often than it lands on 4 and ten percent
less often than it lands on 6. If I roll the dice 350 times, how often should I expect to roll a 6?

91

Step-by-step explanation:

P(1) = 0.25

P(3) = 2P(2) = 3P(4)

P(5) = P(4) + 0.1 = P(6) − 0.1

The probabilities add up to 1:

1 = P(1) + P(2) + P(3) + P(4) + P(5) + P(6)

Write each probability in terms of P(4).

1 = 0.25 + 1.5P(4) + 3P(4) + P(4) + P(4) + 0.1 + P(4) + 0.2

1 = 0.55 + 7.5P(4)

0.45 = 7.5P(4)

P(4) = 0.06

Therefore, P(6) = P(4) + 0.2 = 0.26.

The expected value is the number of trials times the probability.

E = 350 × 0.26

E = 91

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12

the rocket cannot penetrate the ground, in a contextual sense, therefore the roots of this equation should be the domain of this function. i'll let you find the roots of this quadratic equation (hint: use the quadratic formula:  

1- c

2-a

step-by-step explanation:

mark brainliest!