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algol13
3 years ago
14

What is the equation of the midline for the function f(x) ?

Mathematics
1 answer:
Inessa05 [86]3 years ago
5 0

let's recall the graph of sin(x), is simply a sinusoidal line waving about, but its midline is at the x-axis, namely y = 0.

this equation is simply a transformation of it, the 1/2 changes the amplitude by half, midline stays the same though, the +3, moves the whole thing upwards, a vertical shift of 3, meaning the midline went from 0 to 3, y = 3.

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Each pair of points lies on a line with the given slope. Find x.
Gwar [14]

Answer:

the answer is -2 inorder to get -1 slope

8 0
3 years ago
In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal dist
Marrrta [24]

Answer:

a) Bi [P ( X >=15 ) ] ≈ 0.9944

b) Bi [P ( X >=30 ) ] ≈ 0.3182

c)  Bi [P ( 25=< X =< 35 ) ] ≈ 0.6623

d) Bi [P ( X >40 ) ] ≈ 0.0046  

Step-by-step explanation:

Given:

- Total sample size n = 745

- The probability of success p = 0.037

- The probability of failure q = 0.963

Find:

a. 15 or more will live beyond their 90th birthday

b. 30 or more will live beyond their 90th birthday

c. between 25 and 35 will live beyond their 90th birthday

d. more than 40 will live beyond their 90th birthday

Solution:

- The condition for normal approximation to binomial distribution:                                                

                    n*p = 745*0.037 = 27.565 > 5

                    n*q = 745*0.963 = 717.435 > 5

                    Normal Approximation is valid.

a) P ( X >= 15 ) ?

 - Apply continuity correction for normal approximation:

                Bi [P ( X >=15 ) ] = N [ P ( X >= 14.5 ) ]

 - Then the parameters u mean and σ standard deviation for normal distribution are:

                u = n*p = 27.565

                σ = sqrt ( n*p*q ) = sqrt ( 745*0.037*0.963 ) = 5.1522

- The random variable has approximated normal distribution as follows:

                X~N ( 27.565 , 5.1522^2 )

- Now compute the Z - value for the corrected limit:

                N [ P ( X >= 14.5 ) ] = P ( Z >= (14.5 - 27.565) / 5.1522 )

                N [ P ( X >= 14.5 ) ] = P ( Z >= -2.5358 )

- Now use the Z-score table to evaluate the probability:

                P ( Z >= -2.5358 ) = 0.9944

                N [ P ( X >= 14.5 ) ] = P ( Z >= -2.5358 ) = 0.9944

Hence,

                Bi [P ( X >=15 ) ] ≈ 0.9944

b) P ( X >= 30 ) ?

 - Apply continuity correction for normal approximation:

                Bi [P ( X >=30 ) ] = N [ P ( X >= 29.5 ) ]

- Now compute the Z - value for the corrected limit:

                N [ P ( X >= 29.5 ) ] = P ( Z >= (29.5 - 27.565) / 5.1522 )

                N [ P ( X >= 29.5 ) ] = P ( Z >= 0.37556 )

- Now use the Z-score table to evaluate the probability:

                P ( Z >= 0.37556 ) = 0.3182

                N [ P ( X >= 29.5 ) ] = P ( Z >= 0.37556 ) = 0.3182

Hence,

                Bi [P ( X >=30 ) ] ≈ 0.3182  

c) P ( 25=< X =< 35 ) ?

 - Apply continuity correction for normal approximation:

                Bi [P ( 25=< X =< 35 ) ] = N [ P ( 24.5=< X =< 35.5 ) ]

- Now compute the Z - value for the corrected limit:

                N [ P ( 24.5=< X =< 35.5 ) ]= P ( (24.5 - 27.565) / 5.1522 =<Z =< (35.5 - 27.565) / 5.1522 )

                N [ P ( 24.5=< X =< 25.5 ) ] = P ( -0.59489 =<Z =< 1.54011 )

- Now use the Z-score table to evaluate the probability:

                P ( -0.59489 =<Z =< 1.54011 ) = 0.6623

               N [ P ( 24.5=< X =< 35.5 ) ]= P ( -0.59489 =<Z =< 1.54011 ) = 0.6623

Hence,

                Bi [P ( 25=< X =< 35 ) ] ≈ 0.6623

d) P ( X > 40 ) ?

 - Apply continuity correction for normal approximation:

                Bi [P ( X >40 ) ] = N [ P ( X > 41 ) ]

- Now compute the Z - value for the corrected limit:

                N [ P ( X > 41 ) ] = P ( Z > (41 - 27.565) / 5.1522 )

                N [ P ( X > 41 ) ] = P ( Z > 2.60762 )

- Now use the Z-score table to evaluate the probability:

               P ( Z > 2.60762 ) = 0.0046

               N [ P ( X > 41 ) ] =  P ( Z > 2.60762 ) = 0.0046

Hence,

                Bi [P ( X >40 ) ] ≈ 0.0046  

4 0
3 years ago
Find the surface area of a square pyramid with base edge 5 m and slant height 10 m
Dominik [7]

h = 10 m

s = 10.3078 m

a = 5 m

e = 10.6066 m

r = 2.5 m

V = 83.3333 m^3

L = 103.078 m^2

B = 25 m^2

A = 128.078 m^2

3 0
3 years ago
Read 2 more answers
-2(3y+5)=-4, find the value of 5y
Inessa05 [86]
Y=64 that's the answer to the problem
3 0
3 years ago
Read 2 more answers
Solve for x.<br><br> - 6 ≥ 10 - 8 x
Burka [1]

Answer:

x \geqslant 2 \:  \:  \:  \: or \:  \:  \:  \: 2 \leqslant x \\

Step-by-step explanation:

- 6 \geqslant 10 - 8x \\ - 6 - 10  \geqslant  - 8x \\  - 16x \geqslant  - 8x \\  \frac{ - 16}{ - 8}  \leqslant x \\ 2 \leqslant x

3 0
2 years ago
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