For a probability distribution the expected value is the summation of product of probabilities with their respective data values. Let x be the probability that Jackson goes gym for 2 days and y be the probability that he goes gym for 3 days.
For the given case we have following values and their probabilities:
0 : 0.1
2 : x
3 : y
So the expected value will be = 0(0.1) + 2(x) + 3(y)
Expected value is given to be 2.05. So we can write the equation as:
2x + 3y = 2.05 (Equation 1)
Also for a probability distribution, the sum of probabilities must always equal to 1. So we can set up the second equation as:
0.1 + x + y = 1
x + y = 0.9 (Equation 2)
From Equation 2 we can write the value of x to be x = 0.9 - y. Using this value in equation 1, we get:
2(0.9 - y) + 3y = 2.05
1.8 - 2y + 3y = 2.05
1.8 + y = 2.05
y = 0.25
Using the value of y in equation 2 we get value of x to be 0.65
Therefore we can conclude that:
The probability that Jackson goes to gym for 2 days is 0.65 and the probability that he goes to gym for 3 days is 0.25
⭐Solución de problemas: Cada encomiendo tiene un peso de 2 kilogramos. En fracción esto representa 1/4 de la masa total.
Y
¿por qué? Usted tiene una masa total entre los 4 encomiendos de 8 kilogramos, por lo que en orden
para expresar el peso de cada uno de ellos, tenemos
la siguiente expresión: Masa total de encomiendas (kg)/Número de encomiendas (unidad)Sustituimos:
8 kg/4 s
2 kg por encomiendaOfertamos la fracción que representa cada una en el total:
kg por encomienda/total de kg
2/8 x 1/4
Answer:
The first digit of a two digit number can be any of digits 1 - 9. It cannot be 0 though. Therefore there are 9 possible digits for the first place.
There are 5 possible digits for the second position. The two digit number has to be odd and therefore the final digit must be 1,3,5,7 or 9
Therefore for each and every one of the nine first digits there are 5 digits that the second can be.
Therefore ANSWER = 9 * 5 = 45 possible permutations.
2. The largest two digit number = 99
Subtract 57 and you get 42
ANSWER = 42
The forty two numbers are 58 ,59, 60, 61......98, 99
In Δ ABC, ∠A=120°, AB=AC=1
To draw a circumscribed circle Draw perpendicular bisectors of any of two sides.The point where these bisectors meet is the center of the circle.Mark the center as O.
Then join OA, OB, and OC.
Taking any one OA,OB and OC as radius draw the circumcircle.
Now, from O Draw OM⊥AB and ON⊥AC.
As chord AB and AC are equal,So OM and ON will also be equal.
The reason being that equal chords are equidistant from the center.
AM=MB=1/2 and AN=NC=1/2 [ perpendicular from the center to the chord bisects the chord.]
In Δ OMA and ΔONA
OM=ON [proved above]
OA is common.
MA=NA=1/2 [proved above]
ΔOMA≅ ONA [SSS]
∴ ∠OAN =∠OAM=60° [ CPCT]
In Δ OAN
OA=1
∴ OA=OB=OC=1, which is the radius of given Circumscribed circle.
