The probability that an individual is left-handed is 0.11. in a class of 29 students, what is the probability of finding five le
1 answer:
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Answer:
x ≥ 7
Step-by-step explanation:
|x - 7| = x - 7
A. For each absolute, find the intervals
x - 7 ≥ 0 x - 7 < 0
x ≥ 7 x < 7
If x ≥ 7, |x - 7| = x - 7 > 0.
If x < 7, |x - 7| = x - 7 < 0. No solution.
B. Solve for x < 7
Rewrite |x - 7| = x - 7 as
+(x - 7) = x - 7
x - 7 = x - 7
-x + 14 = x
14 = 2x
x = 7
7 ≮7. No solution
C. Solve for x ≥ 7
Rewrite |x - 7| = x - 7 as
+(x - 7) = x - 7
x - 7 = x - 7
True for all x.
D. Merge overlapping intervals
No solution or x ≥ 7
⇒ x ≥ 7
The diagram below shows that the graphs of y = |x - 7| (blue) and of y = x - 7 (dashed red) coincide only when x ≥ 7.
Answer:
Step-by-step explanation:
Probability distribution of this type is a vicariate distribution because it specifies two random variable X and Y. We represent the probability X takes x and Y takes y by
f(x,y) = P((X=x,Y=y) ,
and given that the random variables are independent the joint pmf isbgiven by:
f(x,y) = fX(x) .fY(y) and this gives the table (see attachment)
(b) the required probability is given by considering X=0,1 and Y =0,1
f(x,y) = sum{(P(X ≤ 1, Y ≤ 1) }
= [0.1 +0.2 ] × [0.1 + 0.3]
= 0.3 × 0.4
= 0.12
Now we find
fX(x) = sum{[f(x.y)]} for y =0, 1 and similarly for fY(y)= sum{[f(x,y)]} for x=0, 1
fX(x) = sum{[f(x,y)]}= 0.1+0.3
= 0.4
fY(y) = sum{[f(x.y)]} = 0.1 + 0.2
= 0.3
Since fY(y) × fX(x) = 0.4 × 0.3
= 0.12
Hence f(x,y) = fX(x) .fY(y), the events are independent
(c) the required event is give by: [P(X<=1, PY<=1)]
= P(X<=1) . P(Y<=1)
= [sum{[f(x.y]} over Y=0,1] × [sum{[f(x.y)]}, over X= 0, 1
= 0.4 × 0.3
= 0.12
(d) the required event is given by the idea that: either The South has no bed and the North has or the the North has no bed and the South has. Let this event be A
A =sum[(P(X = 1<= x<= 4, Y =0)] + sum[P(X =0 Y= 1 <= x <= 3)]
A = [0.02 + 0.03 + 0.02 + 0.02] + [0.03 + 0.04 + 0.02]
A = [0.09] + [ 0.09]
A = 0.18
Pls note the sum of the vertical column X=2 is 0.3
Answer:
The point estimate for population mean is 8.129 Mpa.
Step-by-step explanation:
We are given the following in the question:
Data on flexural strength(MPa) for concrete beams of a certain type:
11.8, 7.7, 6.5, 6.8, 9.7, 6.8, 7.3, 7.9, 9.7, 8.7, 8.1, 8.5, 6.3, 7.0, 7.3, 7.4, 5.3, 9.0, 8.1, 11.3, 6.3, 7.2, 7.7, 7.8, 11.6, 10.7, 7.0
a) Point estimate of the mean value of strength for the conceptual population of all beams manufactured
We use the sample mean, as the point estimate for population mean.
Formula:
Thus, the point estimate for population mean is 8.129 Mpa.