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Shalnov [3]
3 years ago
13

Consider the line y = x – 4. A line parallel to the graph of the line would have a slope of . A line perpendicular to the graph

of the line would have a slope of .
Mathematics
2 answers:
kotykmax [81]3 years ago
4 2

Answer:

A. 1

B. -1.

Step-by-step explanation:

We have been given an equation of a line y=x-4.

A. We know that slope of parallel lines is always equal,

We can see that slope of our given line is 1, therefore the slope of the line parallel to our given line would be 1.

B. We know that the product of slopes of two perpendicular lines is -1.

Let m represent slope of perpendicular to our given line, then:

m\times 1=-1

m=-1

Therefore the slope of the line perpendicular to our given line would be -1.

PSYCHO15rus [73]3 years ago
5 0

Answer:

2/3  and -3/2

Step-by-step explanation:

Sonia
2 years ago
thank you
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Free_Kalibri [48]

468 x 0.001 = 0.468

46.8 x 0.1 = 4.68

4.68 x 10^3 = 4680

0.468 x 10^2 = 46.8

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What subtracted by 26 equals -102​
Lesechka [4]

Answer:

-76

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6 0
3 years ago
Find an equation for the line that passes through the points (1, -1) and (-3, 1).
saul85 [17]

Answer:

<h3>y = - 1/2x - 1/2</h3>

Step-by-step explanation:

Equation of line is y = mx + c

where m is the slope

c is the y intercept

Slope of the line using points

(1, -1) and (-3, 1) is

m =  \frac{1 + 1}{ - 3 - 1}  =  -  \frac{2}{4}  =  -  \frac{1}{2}

Equation of the line using point

( 1 , - 1) and slope - 1/2 is

y + 1 =  -  \frac{1}{2} (x - 1) \\ y + 1 =  -  \frac{1}{2} x +  \frac{1}{2}  \\ y =   - \frac{ 1}{2} x +  \frac{1}{2}  - 1 \\ y =  -  \frac{1}{2} x -  \frac{1}{2}

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3 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
Which function has real zeros at x = 3 and x = 7?
Fynjy0 [20]
If it has real zeros at x=3 and 7, the factors of f(x) are:

(x-3)(x-7) so f(x) is

f(x)=x^2-10x+21
3 0
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