c. 3C5P(S)?P(A15
b. 5C3P(S)3P(F2
d. 5C3P(S)3P(F)5
The probability of 3 successes in 5 trials is represented by '5C3P(S)³P(F)²'. This equation reflects the combination of possible successful events, multiplied by the probability of success raised to the number of successes, and the probability of failure raised to the number of failures.
The correct answer is d. 5C3P(S)³P(F)².
Probability of success in binomial problems is given by the equation P(S) = nCr * (p)ⁿ * (q)ⁿ-r where n is the number of trials, r is the number of successful trials, p is the probability of success, and q is the probability of failure (1-p). In this case, there are 5 trials and 3 successes, so your p to the power of 3 (P(S)³) represents the probability of success and the your q to the power 2 (P(F)²) represents the probability of failure. The 5C3 in front is a combination which shows how many ways 3 successes can occur in 5 trials.
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Answer: option B: 5C3P(S)3P(F)2
Step-by-step explanation:
correct on Edge 2020
The linear equation is in the slope-intercept form Y = mx + b.
The equation in slope intercept form is y=-3x-14
One method for expressing a linear equation is in the slope-intercept form (the equation of a line). Y = mx + b, where m is the slope and b is the y-intercept, is the formula for the slope-intercept form (the point where the line crosses the y-axis). The equation in slope intercept form is y = mx + c.
Given slope is m=-3
Let the given point be (-2,-8).
The equation of the line when is
substitut ethe values in the above equation, we get
i.e., (y - (-8)) = -3(x - (-2))
simplifying the above equation, we get
⇒ y + 8 = -3x - 6
⇒ y = -3x - 14
Therefore, the required equation in the slope intercept form is y=-3x-14
To learn more about slope intercept form refer to:
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Answer:
Step-by-step explanation:
y + 3 = -3(x + 8)
y + 3 = -3x - 24
y = -3x - 27
2y - 4x =
10
5
10
Answer: -10
Step-by-step explanation:
Answer:
2x-y+2= 0
Step-by-step explanation:
(-5,-2)=(x1, y1)
(3, 14) =(x2, y2)
now,
slope(m)= y2-y1 / x2-x1
=14+2 / 3+5
= 16 /8
=2 units
passing point = (-5,-2) = x1,y1
now,
eqn of the line can be represented as,
y -y1 = m (x-x1)
y +2 = 2 (x +5)
y+2 = 2x +10
y = 2x+10-8
0 = 2x+2-y
2x-y+2 =0 is the required eqn
Answer:
licker 90000000000000000