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3 years ago

6

An animal shelter has 9 puppies. If the puppies are 36% of the total dog and cat population, how many dogs and cats are in the a

nimal shelter?

1 answer:

Luda [366]3 years ago

5 0

Answer:

Let total number of dog and cat population in animal shelter be x.

As per the statement:

An animal shelter has 9 puppies.

⇒ Total number of puppies = 9

It is also given that if the puppies are 36% of the total dog and cat population.

⇒ $9 = 36\% \text{of}{\text{total number of dogs and Cat}$

or

$9 = \frac{36}{100} \times x$

By cross multiply we get;

$900 = 36 x$

Divide both sides by 36 we get;

$25 = x$

Therefore, total number of dogs and cat population in animal shelter are 25

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Dmitry_Shevchenko [17]

Answer:

$\displaystyle{\int^{\infty}_0 {50e^{-0.04t}}\, dt}=1250$

Step-by-step explanation:

Given:

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now we need to solve this integral!

$T =\displaystyle{50\int^{\infty}_0 {e^{-0.04t}} \, dt}$

$T = \left|50\left(\dfrac{e^{-0.04t}}{-0.04}\right)\right|^{\infty}_0$

$T = \left|-1250e^{-0.04t}\right|^{\infty}_0$

$T = (-1250e^{-0.04(\infty)})-(-1250e^{-0.04(0)})$

when any number has a power of negative infinity it is 0. because: $a^-{\infty} = \dfrac{1}{a^{\infty}} = \dfrac{1}{\infty} = 0$ , like something being divided by a very large number!

$T = (-1250(0))-(-1250e^0)$

$T = 1250$

this is the total amount of waste

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3 years ago

Simplify -4(x - 5) + 3(a - 7)

STatiana [176]

using the distributive property:

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Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

A single card is drawn from a standard 52-card deck. Let D be the event that the card drawn is a black card and let F be the eve

tester [92]

Answer:

The indicated probability of $P(D \cup F')=\frac{25}{26}$

Step-by-step explanation:

Probability of an event E to be;

P(E) = $\frac{Number of events within E}{Total number of possible outcomes}$

As per the given condition:

Total number of possible outcomes = 52 cards.

Let the event be D and F as follows;

D : Drawn card is a black card

F : Drawn card is a 10 card.

Then,

From the given condition:

P(D) = $\frac{26}{52}$ [Out of 52 cards, 26 were black] ,

P(F) = $\frac{4}{52}$ [Out of 52 cards, there are four 10 cards]

For any two events A and B we always have;

$P(A \cup B) = P(A)+P(B)-P(A \cap B)$

Now, we have to find the indicated probability:

$P(D \cup F')=P(D)+P(F')-P(D \cap F')$ ......[1]

First find the P(F');

P(F') =1-P(F) = $1-\frac{4}{52} =\frac{52-4}{52} =\frac{48}{52}$

Also, to find $P(D \cap F')$ .

We use the formula :

For any event A and B independent variable.

$P(A \cap B) =P(A) \cdot P(B)$

then;

$P(D \cap F') =P(D) \cdot P(F') = \frac{26}{52} \cdot \frac{48}{52} =\frac{24}{52}$

Now, substitute these in [1];

$P(D \cup F')=\frac{26}{52} +\frac{48}{52} -\frac{24}{52}$ = $\frac{26+48-24}{52} =\frac{50}{52} = \frac{25}{26}$

Therefore, the probability of $P(D \cup F')=\frac{25}{26}$

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3 years ago