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Harlamova29_29 [7]
3 years ago
8

Solve this exercise using the fact that the sum of the measures of the three angles of a triangle is 180.

Mathematics
1 answer:
den301095 [7]3 years ago
7 0

Answer:

X=70

Step-by-step explanation:

70 + 75 + 35 =180

x + x+5 + x-35

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Tom always keeps his 4 gardens clean. He takes 20 minutes to clean a single garden. How much time will it take to clean all four
Nataly [62]


If each garden takes 20 minutes to clean, it will take him (20 x 4) to clean the gardens.

20 x 4 = 100

80 minutes

or 1 hour and 10 minutes.

5 0
3 years ago
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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
We have a right triangle that has am 8 meter and 21 meter leg .Find the hypotenuse
Serjik [45]

Answer:

hypotenuse is 22.47 m

Step-by-step explanation:

The length of both legs of a right angle triangle are 8m and 21 m

We need to find the hypotenuse

To find hypotenuse we use Pythagorean theorem

Hypotenuse is AC  and other two legs are AB  and BC

AC^2 = AB^2 + BC^2

Hypotenuse ^2 = 8^2 + 21^2

hypotenuse = \sqrt{8^2 + 21^2}

= \sqrt{64 + 441}

= \sqrt{505}

= 22.47

So the length of the hypotenuse is 22.47 m

4 0
3 years ago
Ir-11 2-25
trapecia [35]

Answer:

what's ir but looks like a calculator will help  

u need a little better explanation

Step-by-step explanation:

5 0
3 years ago
A wire is bent to form a square of side 22 cm. If the wire is re bent to form a circle, its radius is with steps
ExtremeBDS [4]

Step-by-step explanation:

the circumference of a circle is

C = 2×pi×r

with r being the radius.

the square side length is 22 cm. so, the circumference or perimeter of the square (and therefore the length of the wire) is

4×22 = 88 cm.

now, it is the same wire of the same length that is now forming a circle.

so, the circumference of the square is also the circumference of the circle.

therefore,

88 = 2×pi×r

44 = pi×r

r = 44/pi = 14.00563499... cm

so, rounded, radius = 14 cm.

5 0
3 years ago
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