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Lesechka [4]
3 years ago
5

Please could someone help me with these questions sorry about the scribbling on one page I got stuck

Mathematics
1 answer:
Sloan [31]3 years ago
8 0
Question 4
So the polygon has 10 sides, to calculate how much all the interior angles are summed up, you have to see how many triangles you can fit into the polygon (this works for any polygon), in this case it is 8 (see file attached). Therefore you do 8 x 180 = 1440, so the overall angle is 1440º. Now you do 1440 : 10, because there are 10 equal angles, and that is 144º. 
Every angle has 144º

question 5
Bearings are always measured clockwise and you always start north. You are at point A and want to go to point B, how much do you have to turn (clockwise!!)? 
345º

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The Porter family budget in the new city is shown in the graph. Of the seven expenses shown, which of the following is fixed?
barxatty [35]

Answer:

Housing

Step-by-step explanation:

5 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
What’s the variance
creativ13 [48]

A is the right answer

8 0
3 years ago
un frasco de jugo tiene una capacidad de 3/8 de litro ¿cuantos frascos se pueden llenar con cuatro litros y medio de jugo?
tangare [24]
Tenemos que

la capacidad del frasco es igual a 3/8 de litro
cuatro litros y medio es igual a decir 4 1/2 litros----> (4*2+1)/2----> 9/2 litros

realizamos una regla de tres
 si un frasco tiene capacidad para--------------> 3/8 litros
 X frascos-----------------------------------------> para 9/2 litros
X=(9/2)/(3/8)--------> 72/6-----> 12 frascos

la respuesta es
12 frascos
6 0
3 years ago
Help please I don’t get it
AlexFokin [52]

Answer:

3/8

Step-by-step explanation:

Rearrange, find common denominator.

x =  \frac{7}{8}  -  \frac{1}{2}  \\ x =  \frac{7}{8}  -  \frac{4}{8}  \\ x =  \frac{3}{8}

8 0
3 years ago
Read 2 more answers
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