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2 years ago

Anyone know this? I don't.

Mathematics

1 answer:

Elza [17]2 years ago

5 0

Heyo, the answer should be A.

B is 3.407

C is 3.72

D is 3.645

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a copy shot prints 5,493 pages they use the pages to make 68 same size booklets about how many pages are in each booklet

Aliun [14]

Answer:

80 pages

Step-by-step explanation:

Given:

Number of pages printed = 5493

Number of booklets made = 68

Let each booklet use 'x' pages.

So, pages used by 68 booklets is given by unitary method and is equal to:

$=68\times x\\=68x$

Now, total number of pages are 5493.

Therefore, the pages used by the 68 booklets should be less than or equal to the total number of pages available. So,

$68x=5493\\\textrm{Dividing both sides by 68,we get:}\\x=\frac{5493}{68}=80.8\approx 80(\textrm{As number of pages can't be in decimal})$

Therefore, the number of pages in each booklet is 80.

3 0

2 years ago

Hii please help i’ll give brainliest if you give a correct answer please please hurry it’s timed

Alexxandr [17]

Answer:

The first book ur

3 0

2 years ago

What is the slope of the line?

jeyben [28]

Answer:

C, -2

Step-by-step explanation:

-4x + 7 = 2y - 3

-4x + 10 = 2y

-2y = 4x - 10

y = -2x + 5

5 0

2 years ago

Can someone pls help me this is due in 4 mins

GuDViN [60]

9514 1404 393

Answer:

5450

Step-by-step explanation:

Put 9% where r is in your formula and evaluate it. Of course, you must use the decimal equivalent.

5000(1 +9%) = 5000(1.09) = 5450

The account will have 5450 in it at the end of the year.

7 0

2 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

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

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

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

5 0

2 years ago