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2 years ago

Help me yall!!......

Mathematics

1 answer:

Ugo [173]2 years ago

8 0

This then allows us to see exactly how and where the subtended angle θ of a sector will fit into the sector formulas. Now we can replace the "once around" angle (that is, the 2π) for an entire circle with the measure of a sector's subtended angle θ, and this will give us the formulas for the area and arc length of that sector

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10. The expression 4 (2x-5)+7(x+2) can be written in simplest form as

ankoles [38]

Answer:

15x-6

Step-by-step explanation: 4 * 2x = 8x

4*-5 = -20

7*x = 7x

7*2= 14

Combine like terms 8x + 7x + 14-20

15x-6

4 0

2 years ago

Suppose that 4 fair coins are tossed. Let Equals The event that exactly 2 coins show tails and Equal The event that at least 2 c

faust18 [17]

Answer:

a) P ( E | F ) = 0.54545

b) P ( E | F' ) = 0

Step-by-step explanation:

Given:

- 4 Coins are tossed

- Event E exactly 2 coins shows tail

- Event F at-least two coins show tail

Find:

- Find P ( E | F )

- Find P ( E | F prime )

Solution:

- The probability of head H and tail T = 0.5, and all events are independent

So,

P ( Exactly 2 T ) = ( TTHH ) + ( THHT ) + ( THTH ) + ( HTTH ) + ( HHTT) + ( HTHT) = 6*(1/2)^4 = 0.375

P ( At-least 2 T ) = P ( Exactly 2 T ) + P ( Exactly 3 T ) + P ( Exactly 4 T) = 0.375 + ( HTTT) + (THTT) + (TTHT) + (TTTH) + ( TTTT)

= 0.375 + 5*(1/2)^4 = 0.375 + 0.3125 = 0.6875

- The probabilities for each events are:

P ( E ) = 0.375

P ( F ) = 0.6875

- The Probability to get exactly two tails given that at-least 2 tails were achieved:

P ( E | F ) = P ( E & F ) / P ( F )

P ( E | F ) = 0.375 / 0.6875

P ( E | F ) = 0.54545

- The Probability to get exactly two tails given that less than 2 tails were achieved:

P ( E | F' ) = P ( E & F' ) / P ( F )

P ( E | F' ) = 0 / 0.6875

P ( E | F' ) = 0

6 0

2 years ago

Find the protect of -3.5 (-1 1/3

STatiana [176]

Answer:

14/3 4 2/3, 4.6

Step-by-step explanation:

Hope this helps!!

5 0

2 years ago

In ΔOPQ, m∠O = 107° and m∠P = 28°. Which statement about the sides of ΔOPQ must be true?

11Alexandr11 [23.1K]

Answer:

QO < OP < QP

Step-by-step explanation:

The third angle, angle Q, must measure 45° since angles in a triangle add to 180°.

So, OQ < OP < QP.

3 0

1 year ago

The value of x in the equation log(x-1)9=2 is

mel-nik [20]

$\log_{(x-1)} 9=2$

the domain:
$x-1 >0 \ \land \ x-1 \not=1 \\
x>1 \ \land \ x \not= 2 \\
x \in (1; 2) \cup (2;+\infty)$

the equation:
$\log_{(x-1)}9=2 \\
(x-1)^2=9 \\
\sqrt{(x-1)^2}=\sqrt{9} \\
|x-1|=3 \\
x-1=3 \ \lor \ x-1=-3 \\
x=4 \ \lor \ x=-2$

4 is in the domain
-2 is not in the domain

The answer:
$x=4$

***
Also, the answer in 13 is {8}. -3 is not in the domain. Replace x with -3 and you'll see:
$x=\sqrt{5x+24} \\
-3=\sqrt{5 \times (-3)+24} \\
-3=\sqrt{-15+24} \\
-3=\sqrt{9} \\
-3=3$
It's not true so -3 isn't a solution to this equation.

4 0

3 years ago

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Help me yall!!......​

Help me yall!!......