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vovikov84 [41]
3 years ago
14

Suppose that the IQs of university​ A's students can be described by a normal model with mean 140 and standard deviation 8 poi

nts. Also suppose that IQs of students from university B can be described by a normal model with mean 120 and standard deviation 11.A) Slect a student at random from university A. Find the probability that the student's IQ is at least 135 points.B) Select a student at random from each school. Find the probability that the university A students IQ is at least 5 points highter than the university B students' IQ.C) Select 3 university B students at random. Find the probability that this group's average IQ is at least 115 points.
Mathematics
1 answer:
AnnZ [28]3 years ago
3 0

Answer:

0.0266, 0.9997,0.7856

Step-by-step explanation:

Given that the IQs of university​ A's students can be described by a normal model with mean 140 and standard deviation 8 points. Also suppose that IQs of students from university B can be described by a normal model with mean 120 and standard deviation 11. Let x be the score by A students and Y the score of B.

A)P(X>135) = \\P(Z>0.625)\\=0.0266

B) Since X and Y are independent we have

X-Y is Normal with mean = 140-120 =20 and Var (x-y)=Var(x)+Var(y) = 19

P(X-Y)>5\\\\=1-0.00029\\=0.9997

C) For a group of 3, average has std deviation = \frac{11}{\sqrt{3} } \\=6.351

P(\bar y >115)\\= P(z>\frac{-5}{6.351} \\=0.7856

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Martha is late for class, so she races her 75 kg body up a 4.0 meter flight of stairs in only 2.3 seconds. What is Martha’s powe
vovangra [49]

Answer:

1.3\cdot 10^3\text{ J}

Step-by-step explanation:

Power is defined by work over time. In physics, work can be defined as the energy transferred over from an applied force over some distance. As a formula:

W=Fd, where F = force applied and d = distance that force is applied over.

*It is worth noting that F must be parallel to the distance travelled. If it is perpendicular, no work is done, and if it is at an angle, find the parallel component and use that for F.

In this case, the force applied must counter the force of gravity on Martha, which is given by F_g=mg, where m = Martha's mass and g = gravitational constant 9.8 m/s/s. Therefore, F_g=75\cdot 9.8=735\text{ N}. Since she raises her body 4.0 meters, the work done must be W=Fd=735\cdot 4=2940\text{ J}.

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