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3 years ago

7

What is the solution to the following system of equations?

1 answer:

Alchen [17]3 years ago

3 0

Answer:

it has infinitely many solutions

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0.34 into a percent

vivado [14]

34% is the correct answer

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3 years ago

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You roll three die. what is the probability that 5 will not appear on the die?

castortr0y [4]

No of out comes =216
favorable outcome=3
<span>probability that 5 will not appear on the die=3/216</span>

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3 years ago

Dadas las funciones f(x) = x + 3, g(x)= x(cuadrada)+ 5x + 6,r(x) = x+2, s(x) = x(cuadrada) - 3x - 10

Aleksandr [31]

So the answer is D hope this would help you

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3 years ago

at the dog show, there are 4 times as many boxer as spaniels. if there are a total of 30 dogs, how many dogs are spaniel?

Lyrx [107]

Firstly, given this information there could be 25 labradors and poodles and only 4 boxers and 1 spaniel. However, assuming that there are only boxers and spaniels at the dog show the first thing to do is to divide 30 by 5 = 6 so the answer is, There are 6 spaniels.

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3 years ago

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Blababa [14]

Answer:

$P(A\textrm{ and }B)=\frac{3}{14}$

Step-by-step explanation:

Given:

$P(A)=\frac{4}{7}$

$P(B|A)=\frac{3}{8}$

We know that, conditional probability of B given that A has occurred is given as:

$P(B|A)=\frac{P(A\cap B}{P(A)}$ . Expressing this in terms of $P(A\cap B)$ , we get

$P(A\cap B)=P(B|A)\times P(A)$

Plug in the known values and solve for $P(A\cap B)$ . This gives,

$P(A\cap B)=P(B|A)\times P(A)\\P(A\cap B)=\frac{3}{8}\times \frac{4}{7}\\P(A\cap B)=\frac{12}{56}=\frac{3}{14}$

Therefore, the probability of events A and B occurring is $\frac{3}{14}$ .

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3 years ago

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