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stellarik [79]
3 years ago
12

THE NUMBER 1-10 ARE IN A BAG, YOU TAKE 2 OUT WITHOUT REPLACEMENT. WHAT IS THE PROBABILITY OF DRAWING THE NUMBER 7 THE FIRST TIME

AND THEN DRAWING AN EVEN NUMBER.
Mathematics
1 answer:
ehidna [41]3 years ago
6 0

Answer:

1 / 18

Step-by-step explanation:

PROBABILITY OF DRAWING THE NUMBER 7 THE FIRST TIME = Number of favorable outcome / Total number of outcomes

Favorable outcome = 1

Total number of outcome (1,2,3,4,5,6,7,8,9,10) = 10

PROBABILITY OF DRAWING THE NUMBER 7 THE FIRST TIME  = 1 / 10

PROBABILITY OF DRAWING AN EVEN NUMBER THE SECOND TIME = Number of favorable outcome / Total number of outcomes

Favorable outcome (2,4,6,8,10) = 5

Total number of outcomes (1,2,3,4,5,6,8,9,10) = 9

PROBABILITY OF DRAWING AN EVEN NUMBER THE SECOND TIME = 5/9

PROBABILITY OF DRAWING THE NUMBER 7 THE FIRST TIME AND THEN DRAWING AN EVEN NUMBER = 1 / 10  * 5 / 9 = 1 / 18

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5. Write two expressions for the area of the big rectangle. ​
4vir4ik [10]

Answer:

Two expressions for the area are;

i) x/3 + 2·y + 6

ii) (1/3) × (x + 6·y + 18)

Step-by-step explanation:

The given dimension of the rectangle are;

The height of the rectangle, h = 1/3

The width of the smallest rectangle, w₁ = x

The width of the med sized rectangle, w₂ = 6·y

The width the large rectangle, w₃ = 18

The area, 'A', of the entire rectangular figure, the big rectangle, can be expressed as follows;

A = A₁ + A₂ + A₃

Where;

A₁ = The area of the smallest rectangle

A₂ = The area of the mid sized rectangle

A₃ = The area of the large rectangle

∴ A = (1/3) × x + (1/3) × 6·y + (1/3) × 18 = x/3 + 2·y + 6

The area of the big rectangle. 'A', can also be found as follows;

A = (1/3) × (x + 6·y + 18)

Therefore, two expressions for the area of the big rectangle are;

x/3 + 2·y + 6 and (1/3) × (x + 6·y + 18).

3 0
3 years ago
Given: <br> PQ<br> ⊥<br> QR<br> , PR=20,<br> SR=11, QS=5<br> Find: The value of PS.
dlinn [17]

Answer:

The value of the side PS is 26 approx.

Step-by-step explanation:

In this question we have two right triangles. Triangle PQR and Triangle PQS.

Where S is some point on the line segment QR.

Given:

PR = 20

SR = 11

QS = 5

We know that QR = QS + SR

QR = 11 + 5

QR = 16

Now triangle PQR has one unknown side PQ which in its base.

Finding PQ:

Using Pythagoras theorem for the right angled triangle PQR.

PR² = PQ² + QR²

PQ = √(PR² - QR²)

PQ = √(20²+16²)

PQ = √656

PQ = 4√41

Now for right angled triangle PQS, PS is unknown which is actually the hypotenuse of the right angled triangle.

Finding PS:

Using Pythagoras theorem, we have:

PS² = PQ² + QS²

PS² = 656 + 25

PS² = 681

PS = 26.09

PS = 26

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3 years ago
Determine the complement and/or supplement of each angle. If it is not possible, explain.
Alecsey [184]

Answer:

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  • supplement: 122.8°

Step-by-step explanation:

The sum of an angle A and its complement C is 90°:

  A + C = 90°

  C = 90° -A . . . . . subtract A from both sides.

That is, the complement of an angle is found by subtracting the angle from 90°.

__

The sum of an angle and its supplement is 180°. This means the supplement of an angle is found by subtracting the angle from 180°. You may notice the supplement is 90° more than the complement.

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  S = 180° -A = 90° +(90° -A)

__

For the given angle, the complement is ...

  C = 90° -57.2° = 32.8°

And the supplement is ...

  S = 180° -57.2° = 122.8°

_____

<em>Additional comment</em>

We generally like angle measures to be positive (as with all measures in geometry). Hence, we might say that the complement of an angle greater than 90° does not exist. YMMV

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