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Orlov [11]
3 years ago
5

No one will answer this probably wasting my points XD) Spunky the terrier is chasing a chicken. Spunky runs at a rate of r feet

per second while the chicken runs at a rate of v feet per second. The chicken is originally 15 feet from Spunky and he catches up to it in 20 seconds. Which equation represents this information?
a. 20r−15=20v20r-15=20v
b. =20r+20v15=20r+20v
c. 20r=20v−1520r=20v-15
d. 20−15r=20+v
Mathematics
1 answer:
Vesna [10]3 years ago
4 0
It sucks that points have to be used in order to post questions. 

Openstudy - the company that merged into Brainly -  was much better.

Anyway, I think that the answer to this question is A. I'm not sure, though.
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What is the product?
aleksley [76]

Answer:

The first one

Step-by-step explanation:

8 0
3 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
16x^7 - 8x^3 / 4x^3
Sunny_sXe [5.5K]

Answer:

4x^4-2

Step-by-step explanation:

\dfrac{16x^7-8x^3}{4x^3}= \\\\\\\dfrac{4x^3(4x^4-2)}{4x^3}= \\\\\\4x^4-2

Hope this helps!

8 0
3 years ago
Read 2 more answers
HELP THIS IS DUE TONIGHT AT 12!!!!
Strike441 [17]

Answer:

b infinity solutions

Step-by-step explanation:

they are the same line on the graph

3 0
3 years ago
Read 2 more answers
How many distinct pairs of perfect squares differ by 35? (the pair $a, b$ is the same as the pair $b, a$.)?
Phantasy [73]
35 has 4 divisors, hence two factor pairs: 1*35 and 5*7. Each corresponds to a set of perfect squares that differ by 35

One pair is ((35±1)/2)^2 = {17^2, 18^2} = {289, 324}
The other is ((7±5)/2)^2 = {1^2, 6^2} = {1, 36}
5 0
4 years ago
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