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Maru [420]
3 years ago
11

For the point P(−19,18) and Q(−14,23)​, find the distance​ d(P,Q) and the coordinates of the midpoint M of the segment PQ.

Mathematics
1 answer:
Irina18 [472]3 years ago
3 0
For the point P(−19,18) and Q(−14,23)​, find the distance​ d(P,Q) and the coordinates of the midpoint M of the segment PQ.

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Suppose that the mean and s.d of the tuition fee paid by bs mathematics students in umt is 150and 30 in uds, respectively. it is
Serga [27]

From the information give,

A) the probability that the amount paid by him is greater than 105 is ≈ 0.9332

B) the probability that the amount paid by him is not between 120 to 180 ≈ 0.3173

C) the probability that the amount paid by him is between 140 and 160 ≈ 0.2611

D) the probability that the amount paid by him is less than 180 ≈ 0.8413

E) the probability that the amount paid by him is between 140 and 160 given that greater than 120 ≈ 0.3104

F) the probability that the amount paid by him is between 120 and 160orbetween 140 to 180 ≈ 0.6827. See all computations below.

<h3>What is the calculation of for the solutions indicated above?</h3>

Let X be the amount that the student paid;

Note that X ∼ N(μ,σ² ).

Since we have μ = 150, and σ = 30; The1refore


A) P(X>105)=1−P(X≤105)

= 1 - P (Z ≤ (105 -150)/30)

=  - P (Z ≤ -1.5)

P(X>105) ≈ 0.9332

B) P(X<120 or X>180)

=P(X<120)+1−P(X≤180)
= P ((Z < (120 - 150)/30) + 1 -  P ((Z < (180 - 150)/30)

=P(Z<−1)+1−P(Z≤1)

≈0.158655+0.158655

P(X<120 or X>180) ≈ 0.3173

C) P(140<X<160)

= P(X<160)−P(X≤140)

=  P ((Z < (160 - 150)/30) + 1 -  P ((Z < (140 - 150)/30)

≈P(Z<0.33333)−P(Z≤−0.33333)

≈0.63056−0.36944

P(140<X<160) ≈ 0.2611


D) P(X<180) =
P(Z< (180−150)/30

​=P(Z<1)

P(X<180) ≈0.8413

E) P(140<X<160∣X>120)

= (P((140<X<160)∩(X>120))/P(X>120)

= (P((140<X<160))/P(X>120)

= (P Z < (160 -150)/30) - P (Z ≤ (140-150)/30))/ 1- P (Z ≤ (120 - 150)/30)
≈ (P(Z<0.33333)−P(Z≤−0.33333))/ 1 - P(Z ≤ - 1)
≈ (0.63056−0.36944)/0.84134

P(140<X<160∣X>120) ≈ 0.3104

F) P(120<X<160 or 140<X<180)

= P(120<X<180)

= P(X<180)−P(X≤120)

= P ((Z < (180 -150)/30) - P ((Z ≤ (120 -150)/30)

= P(Z<1)−P(Z≤−1)

≈ 0.841345−0.158655

​P(120<X<160 or 140<X<180) ≈ 0.6827

Learn more about probability at:

brainly.com/question/24756209
#SPJ1

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