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3 years ago

Find the quotient and remainder.

Mathematics

2 answers:

Damm [24]3 years ago

8 0

Answer:

quotient 2.8 , reminder 0

Vitek1552 [10]3 years ago

4 0

Answer:

14 divided by 5

The quotient is 2, and the remainder is 4.

Let me know if this helps!

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X+y+z=2<br> 5x+5y+5z=3<br> 4x+y-3z=-6

aniked [119]

Answer:

Step-by-step explanation:

Examples: 1+2, 1/3+1/4, 2^3 * 2^2

(x+1)(x+2) (Simplify Example), 2x^2+2y a x=5, y=3 (Evaluate Example)

y=x^2+1 (Graph Example), 4x+2=2(x+6) (Solve Example)

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3 years ago

A student S is suspected of cheating on exam, due to evidence E of cheating being present. Suppose that in the case of a cheatin

Kay [80]

Answer:

Step-by-step explanation:

Suppose we think of an alphabet X to be the Event of the evidence.

Also, if Y be the Event of cheating; &

Y' be the Event of not involved in cheating

From the given information:

$P(\dfrac{X}{Y}) = 60\% = 0.6$

$P(\dfrac{X}{Y'}) = 0.01\% = 0.0001$

$P(Y) = 0.01$

Thus, $P(Y') \ will\ be = 1 - P(Y)$

P(Y') = 1 - 0.01

P(Y') = 0.99

The probability of cheating & the evidence is present is = P(YX)

$P(YX) = P(\dfrac{X}{Y}) \ P(Y)$

$P(YX) =0.6 \times 0.01$

$P(YX) =0.006$

The probabilities of not involved in cheating & the evidence are present is:

$P(Y'X) = P(Y') \times P(\dfrac{X}{Y'})$

$P(Y'X) = 0.99 \times 0.0001 \\ \\ P(Y'X) = 0.000099$

(b)

The required probability that the evidence is present is:

P(YX or Y'X) = 0.006 + 0.000099

P(YX or Y'X) = 0.006099

(c)

The required probability that (S) cheat provided the evidence being present is:

Using Bayes Theorem

$P(\dfrac{Y}{X}) = \dfrac{P(YX)}{P(Y)}$

$P(\dfrac{Y}{X}) = \dfrac{P(0.006)}{P(0.006099)}$

$P(\dfrac{Y}{X}) = 0.9838$

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Answer:

A. 10 weeks

Step-by-step explanation:

Oliver Marie

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week 2 320 240

week 3 330 260

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2 years ago

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20 the answer is 20 because you plug in 4 for n.

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