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ohaa [14]
2 years ago
12

A fish tank holds 1.029 yd3 of water. What is this volume in cubic meters given that 1 m = 1.093 yd? (a) 1.062 m3 (b) 0.9414 m3

(c) 1.125 m3 (d) 0.7881 m3
Mathematics
1 answer:
8_murik_8 [283]2 years ago
4 0

Answer:

d.0.7881 m^{3}

Step-by-step explanation:

Hello

i think I can help you with this, this is unit conversion problem.

Step 1

we need to know the equivalence between m3 and yd 3

1 m = 1.093\ yd\\\\1m^{3}=(1.093yd)^{3} \\ \\1m^{3}=1.3057\ yd^{3} \\\\

Step 1

using a rule of three convert 1.029 yd3 into m3 with problem data

1m^{3}=1.3057\ yd^{3}\\ xm^{3}=1.029\ yd^{3}\\\frac{1m^{3}}{1.3057\ yd^{3}} =\frac{xm^{3}}{1.029\ yd^{3}}

now, isolating x

\frac{1.029\ yd^{3}* 1m^{3}}{1.3057\ yd^{3}} =x\\x=0.7881 m^{3}

Have a great day

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(B) 0.1875

(C) Can't be computed

Step-by-step explanation:

We are given that the amount of time it takes for a student to complete a statistics quiz is uniformly distributed between 32 and 64 minutes.

Let X = Amount of time taken by student to complete a statistics quiz

So,   X ~ U(32 , 64)

The PDF of uniform distribution is given by;

    f(X) = \frac{1}{b-a} ,  a < X < b      where a = 32 and b = 64

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(B) Probability that student completes the quiz in a time between 37 and 43 minutes = P(37 <= X <= 43)  = P(X <= 43) - P(X < 37)

    P(X <= 43) = \frac{43-32}{64-32} = \frac{11}{32} = 0.34375

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(C) Probability that student complete the quiz in exactly 44.74 minutes

     = P(X = 44.74)

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