Both the heads and tails will have a probability of 0.5 with a fair coin. ... TO find probability that foe the 7th toss head appears exactly 4 times.
Answer:
a) 47.55
b) 58
c) 47.88
Step-by-step explanation:
Given that the size of the orders is uniformly distributed over the interval
$25 ( a ) to $80 ( b )
<u>a) Determine the value for the first order size generated based on 0.41</u>
parameter for normal distribution is given as ; a = 25, b = 80
size/value of order = a + random number ( b - a )
= 25 + 0.41 ( 80 - 25 )
= 47.55
<u>b) Value of the last order generated based on random number (0.6)</u>
= a + random number ( b - a )
= 25 + 0.6 ( 80 - 25 )
= 25 + 33 = 58
<u>c) Average order size </u>
= ∑ order 1 + order 2 + ----- + order 10 ) / 10
= (47.55 + ...... + 58 ) / 10
= 478.8 / 10 = 47.88
Answer:
Step-by-step explanation:
Okay so you are gonna substitute x for 5 so the equations will look like y= (5)2+(5)2
So 10+10 = 20
So y=20
Aplicando multiplicación cruzada, tiene-se que el valor de w es w = 9.
- Cuando una proporción es dada, con una igualdade de duas proporciones, puede-se aplicar multiplicación cruzada entre ellas.
En este problema, la ecuación que relaciona las proporciones es dada por:
Aplicando multiplicación cruzada:
El valor de w es w = 9.
Un problema similar es dado en brainly.com/question/24615636
