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exis [7]
3 years ago
8

10 POINTS PLUS BRAINLIEST

Mathematics
2 answers:
stiks02 [169]3 years ago
6 0

Answer:

A or C

Step-by-step explanation:

Hope this help:)

jeyben [28]3 years ago
5 0

Answer:

B is correct.

Step-by-step explanation:

I eliminate A and B since it translates the pentagon left or right. D dilates, but does not make it bigger since the value is a fraction less than 1. 9/4 is correct because the value of that is greater than 1 and therefore makes the pentagon bigger.

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Find the value of k<br> 4x+y=8<br> y=kx+2<br><br> options<br><br> -4<br> 4<br> -1/4<br> 1/4
nexus9112 [7]

Answer:

4 i think im not sure

Step-by-step explanation:

Subtract yy from both sides of the equation.

4x=8−y4x=8-y

y=kx+2y=kx+2

Divide each term by 44 and simplify.

Tap for more steps...

x=2−y4x=2-y4

y=kx+2

6 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
2 years ago
A table is in the shape of a regular pentagon with a side length of 3 feet and an
ehidna [41]

Answer:

15.5 ft²

Step-by-step explanation:

The area (A)  of a regular polygon is calculated as

A = \frac{1}{2} pa ( p is the perimeter and a the apothem )

Here side of regular pentagon is 3 ft, thus

perimeter = 5 × 3 = 15, so

A = 0.5 × 15 × 2.06 ≈ 15.5 ft ( to the nearest tenth )

8 0
3 years ago
Distributive property of<br> 36+ 9y
vaieri [72.5K]
Find the greatest common factor (GCF)
in this case it would be 9

divide each term by the GCF
36 / 9 = 4
9y / 9 = y

put the two "new" terms in parenthesis and the GCF in front of it
9 (4 + y)
it could also be 9 (y + 4) because of Commutative Property of Addition
6 0
3 years ago
Read 2 more answers
Write and solve an equation using the constant of proportionality to answer each question.
yaroslaw [1]

Answer:

15 teachers

Step-by-step explanation:

Given:

number of children is c

number of teachers is t

\frac{c}{t} = \frac{14}{3} (rearrange)

t = \frac{3}{14} x C

When C = 70,

t = \frac{3}{14} x 70

t = 15 teachers

8 0
2 years ago
Read 2 more answers
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