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Gnesinka [82]
3 years ago
6

Billy's plant grows at least 2 inches a month.

Mathematics
2 answers:
Viktor [21]3 years ago
8 0

Answer:

y=2x+k

Step-by-step explanation:

y=2x+k

Here we are given that the plant growth is fixed by 2 inches for every month.

hence ,

if we assume that the plant had zero height at beginning , when the time was also zero, then in the next month that is 1st month the height of the plant will be 2 inches, similarly in the 2nd month the height will be 4 inches , for the third month the height of the plant will be 6 inches and so on.

So, if we plot them on a graph , taking time in month on x axis and height of the plant on y axis , the pattern of growth as mentioned above will give us few coordinates like,

(0,0) , (1,2) , (2,4) , (3,6) and so on..

hence from here we can determine equation which refers the above situation by using two point form of an equation . The two point form is given as

\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}

where

(x_1,y_1) and (x_2,y_2) are (1,2) and (2,4) respectively

substituting them in the above formula we get

\frac{y-2}{x-1}=\frac{4-2}{2-1}

\frac{y-2}{x-1}=2

y-2=2(x-1)

y-2=2x-2

y=2x

Hence we get out scenario , in equation as y=2x

Note:

this is the situation when the height of the plant was considered 0 at time 0 . however , if the initial height of the plant was k ( assumption )

The equation would have been like this

y=2x+k

This is because in both the cases the slope of the line would be same , as slope represents the rate of increase of y with respect to x . in this case it would be the increased in the height of the plant with respect to the months Where k represent the value of y at x= 0 , or the height of the plant at time 0 month.

Please refer to the image attached with this.

emmainna [20.7K]3 years ago
5 0
We can assume that this growth rate can be expressed as y = mx + b as long as the slope is constant, because a quadratic or exponential formula wouldn't make much sense in this situation. If it is growing at least 2 inches a month, the slope (m) will be 2. As x (month) increases by 1 month, y (plant's height) increases by 2 inches. You could make it more, but as long as you plot the points in such a way that the slope for these points are 2 or higher, you will get the answer correct. If you have any other questions or need to clarify more about what you wanted let me know and I'll help.
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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
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