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pickupchik [31]
3 years ago
6

Need help ASAP! Please

Mathematics
2 answers:
iren [92.7K]3 years ago
6 0

Answer:

112

Step-by-step explanation:

20 x 20 is 40

6 x 9 is 54

6 x 3 is 18

54 + 18 is 72

72 + 40 is 112

dedylja [7]3 years ago
5 0

Answer:

472

Step-by-step explanation:

Use the order of operations PEMDAS

Parenthesis

Exponents

Multiplication

Division

Addition

Subtraction

9 + 3 = 12

20^2 = 400

6 * 12 = 72

400 + 72 = 472

hope this helped! :)

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Can 8/45 be simplified as a fraction
max2010maxim [7]

Answer:

Step-by-step explanation:

isn't it already a fraction or is it a division

if it's a division then yes

the answer is 5 and 2/8

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3 years ago
Graph this piecewise function:
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4 years ago
What's the area of the quadrilateral ​
shepuryov [24]

Answer: Use the formula (base1+base2)/2 multiplied by height(3 in)

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Answer:25.5

5 0
3 years ago
Read 2 more answers
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
3 years ago
HELP AS SOON AS POSSIBLE NEEDS ANSWERS ASAP
likoan [24]

Answer:

C. The mean will increase and ty e median will remain the same.

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