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Luba_88 [7]
3 years ago
6

Write the standard equation of the ellipse with center at the origin, and vertices at (-4,0), (4,0), (0, -3), and (0,3).

Mathematics
1 answer:
levacccp [35]3 years ago
5 0

9514 1404 393

Answer:

  (a)  x^2/16 +y^2/9 = 1

Step-by-step explanation:

The form for the equation of an ellipse centered at the origin is ...

  (x/(semi-x-axis))^2 +(y/(semi-y-axis))^2 = 1

The vertex values tell you the semi-x-axis is 4 units, and the semi-y-axis is 3 units. Then you have ...

  (x/4)^2 +(y/3)^2 = 1

  x^2/16 +y^2/9 = 1

__

In case you don't remember that form, you can try any of the points in the equations. The equation that works will quickly become apparent.

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A mixed number has a whole number part and a fractional part. <br>O A. True <br>O B. False​
scoray [572]

Answer: True

Step-by-step explanation:

The statement that a mixed number has a whole number part and a fractional part is true.

The mixed number is made up of a whole number, and a proper fraction. A mixed number simply means a number that is between any two whole numbers. Examples of mixed numbers are 4¾, 7½ etc.

6 0
3 years ago
Identify the constant term in this expression. 25x + 9y + 20 + 2y
kvv77 [185]
20

Constant in a mathematical expression is a term whose value never changes.
The terms with x and y will vary based on the variable x and or y.
7 0
3 years ago
Q8:<br> What is the solution to the equation below?<br> ​<br> ​ 4w = 2/3 <br> W=?<br><br> ​
iVinArrow [24]

Answer:

<h2>W = 1/6</h2>

Step-by-step explanation:

Hope this helps

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5 0
3 years ago
Read 2 more answers
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
What is 3.98 X10 raise to power 10 In standard form
bonufazy [111]
39,800,000,000 is the correct answer, I believe.
3 0
3 years ago
Read 2 more answers
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