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shtirl [24]
3 years ago
8

Chris has a blue coat and a

Mathematics
1 answer:
Tju [1.3M]3 years ago
3 0

Answer: P = 0.125 = 1/8

Step-by-step explanation:

We know that he has a blue coat and a black coat.

If he dresses at random, then the probability of getting the blue coat is equal to the quotient between the number of blue coats (1) and the total number of coats (2).

Then the probability is:

p = 1/2

We also know that he has blue pants and brown pants, the probability of getting at random the blue pants is calculated in the same way than above, then:

q = 1/2

And for the shirt he has a blue shirt and a red one, the probability of randomly selecting the blue one is calculated in the same way than above, then:

k = 1/2

Now, the joint probability (he selects all blue clothes) is equal to the product of the individual probabilities:

P = p*q*k = (1/2)*(1/2)*(1/2) = 1/8 = 0.125

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Can u guys PLEASE answer this question ASAP. Two wooden boxes, both with dimensions 80 cm, 1 m and 25 cm, are placed on the grou
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Answer:

34000cm or 340m

Step-by-step explanation:

We have to find total surface area.

dimensions of 1 wooden box =

l = 1m = 100cm

b = 80cm

h = 25cm

∴ In the case of 2 =

l = 100cm

b= 80cm

h  = 25 + 25 = 50cm

T.S.A = 2(lb + lh + bh)

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3 years ago
Using pemdas what’s the correct solution
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parenthesis first

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change to an improper fraction (6*3+5)/6

4 - 2/3 ( 23/6)divided by 3/4

4 - 46/18 divide by3/4

copy dot flip

4 - 46/18 * 4/3

4 - 23/9 * 4/3

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Kyla has the number of cherries that Denise has. If Kyla has 46 cherries, how many cherries do they have altogether?​
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Answer:

92

Step-by-step explanation:

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5 0
1 year ago
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Solve the following differential equations or initial value problems. In part (a), leave your answer in implicit form. For parts
shepuryov [24]

Answer:

(a) (y^5)/5 + y^4 = (t^3)/3 + 7t + C

(b) y = arctan(t(lnt - 1) + C)

(c) y = -1/ln|0.09(t + 1)²/t|

Step-by-step explanation:

(a) dy/dt = (t^2 + 7)/(y^4 - 4y^3)

Separate the variables

(y^4 - 4y^3)dy = (t^2 + 7)dt

Integrate both sides

(y^5)/5 + y^4 = (t^3)/3 + 7t + C

(b) dy/dt = (cos²y)lnt

Separate the variables

dy/cos²y = lnt dt

Integrate both sides

tany = t(lnt - 1) + C

y = arctan(t(lnt - 1) + C)

(c) (t² + t) dy/dt + y² = ty², y(1) = -1

(t² + t) dy/dt = ty² - y²

(t² + t) dy/dt = y²(t - 1)

(t² + t)/(t - 1)dy/dt = y²

Separating the variables

(t - 1)dt/(t² + t) = dy/y²

tdt/(t² + t) - dt/(t² + t) = dy/y²

dt/(t + 1) - dt/(t(t + 1)) = dy/y²

dt/(t + 1) - dt/t + dt/(t + 1) = dy/y²

Integrate both sides

ln(t + 1) - lnt + ln(t + 1) + lnC = -1/y

2ln(t + 1) - lnt + lnC = -1/y

ln|C(t + 1)²/t| = -1/y

y = -1/ln|C(t + 1)²/t|

Apply y(1) = -1

-1 = ln|C(1 + 1)²/1|

-1 = ln(4C)

4C = e^(-1)

C = (1/4)e^(-1) ≈ 0.09

y = -1/ln|0.09(t + 1)²/t|

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