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gregori [183]
3 years ago
15

A man works 35 5/6 hours in 5 days. what is his average work day?​

Mathematics
1 answer:
valina [46]3 years ago
4 0

Answer:

7 1/6

Step-by-step explanation:

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Which number sentence shows the correct placement of the decimal point in the product, based on the two factors?
Lana71 [14]

2.52 * 27.94 = 70.4088, so not that one

94.1 * 2.5 = 235.25, so not that one

1.83 * 2.9 = 5.307, so not that one

25.07 * 5.4 = 135.378, so that one

Answer:

D. 25.07 * 5.4 = 135.378

3 0
3 years ago
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the area of a trapezoid is 780 cm squared. the shorter base is 48 cm and the height is 15 cm. Find the length of the other base
Anna35 [415]
If you multiply 48cm times 15cm,you get 720cm.Which would leave you with 60 left because 780 minus 720 equals 60.The height of the other base is 60cm.
6 0
3 years ago
What is the least common multiple of both 9 and 12
Crank
LCM is found by finding the factors of both and making sure that both are inthere
9=3 times 3
12=2 times 2 times 3
common one is 3
so we have to have at least 2 threes and 2 twos
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3 0
3 years ago
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How to solve - 10+5x=7x-4
Triss [41]
10+5x=7x-4
1. simplify 7x - 5x
10=2x-4

2. add four to both sides
10+4=2x
+ 4 4
14=2x

3. divide 14 to both sides
14=2x

x=7
8 0
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

5 0
3 years ago
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