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SVETLANKA909090 [29]
3 years ago
11

Help me please I’m having trouble with it

Mathematics
1 answer:
gavmur [86]3 years ago
4 0

Given:

Measure of arcs 50°, 115° and 85°

To find:

The measure of the numbered angle 5.

Solution:

Let the missing arc measure be A.

The arc measure of a full circle is 360°

m(ar A) + 50° + 115° + 85° = 360°

m(ar A) + 250° = 360°

Subtract 250 from both sides.

m(ar A) = 110°

<em>If two chords intersects inside a circle, then the measure of the angle formed is half of the sum of intercepted arcs.</em>

$\Rightarrow m\angle 5 =\frac{1}{2} (110^\circ+115^\circ)

$\Rightarrow m\angle 5 =\frac{1}{2} (225^\circ)

$\Rightarrow m\angle 5 =112.5^\circ

The measure of the numbered angle is 112.5°.

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Please help! also please explain how to solve
iris [78.8K]

Answer:

C

Step-by-step explanation:

The fast way: Testing the options!

Teacher loving way:

An= d×n + A0

d= A4-A3=A5-A4=A6-A5...= 6

A3= 0 = 6×3 + A0

A0= -18

Then the equation is: An= 6n - 18

4 0
2 years ago
An alarming number of U.S. adults are either overweight or obese. The distinction between overweight and obese is made on the ba
madreJ [45]

Answer:

(A) The probability that a randomly selected adult is either overweight or obese is 0.688.

(B) The probability that a randomly selected adult is neither overweight nor obese is 0.312.

(C) The events "overweight" and "obese" exhaustive.

(D) The events "overweight" and "obese" mutually exclusive.

Step-by-step explanation:

Denote the events as follows:

<em>X</em> = a person is overweight

<em>Y</em> = a person is obese.

The information provided is:

A person is overweight if they have BMI 25 or more but below 30.

A person is obese if they have BMI 30 or more.

P (X) = 0.331

P (Y) = 0.357

(A)

The events of a person being overweight or obese cannot occur together.

Since if a person is overweight they have (25 ≤ BMI < 30) and if they are obese they have BMI ≥ 30.

So, P (X ∩ Y) = 0.

Compute the probability that a randomly selected adult is either overweight or obese as follows:

P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\=0.331+0.357-0\\=0.688

Thus, the probability that a randomly selected adult is either overweight or obese is 0.688.

(B)

Commute the probability that a randomly selected adult is neither overweight nor obese as follows:

P(X^{c}\cup Y^{c})=1-P(X\cup Y)\\=1-0.688\\=0.312

Thus, the probability that a randomly selected adult is neither overweight nor obese is 0.312.

(C)

If two events cannot occur together, but they form a sample space when combined are known as exhaustive events.

For example, flip of coin. On a flip of a coin, the flip turns as either Heads or Tails but never both. But together the event of getting a Heads and Tails form a sample space of a single flip of a coin.

In this case also, together the event of a person being overweight or obese forms a sample space of people who are heavier in general.

Thus, the events "overweight" and "obese" exhaustive.

(D)

Mutually exclusive events are those events that cannot occur at the same time.

The events of a person being overweight and obese are mutually exclusive.

5 0
3 years ago
Campbell Middle school has 912 students.How many computers are in this schools computer lab?
Aleksandr [31]
You can't really determine the amount of computers from this question, they're obviously not going to have one computer for everyone so half third, or even a fourth of the school population
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3 years ago
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What is 2x^2-8x<br> factorised*
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Answer:

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Step-by-step explanation:

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HELP!!!!
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Answer:

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Step-by-step explanation:

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