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siniylev [52]
3 years ago
7

Can someone explain this.

Mathematics
1 answer:
san4es73 [151]3 years ago
5 0

Answer:

9/4 yds fabric per 1 ft of ribbon

Step-by-step explanation:

3/4 yd fabric is required for every 2/3 ft of ribbon.  Let's put this info into an equation of ratios to figure out the number of yards of fabric required for every ONE ft of ribbon:

  3/4 yd fabric        2/3 ft ribbon

----------------------- = ---------------------

            x                    1 ft ribbon

Use cross mult. to solve this.  We obtain 3/4 = (2/3)x.

Mult. both sides by (3/2) yields (3/2)(3/4) = (3/2)(2/3)x = x.

This comes out to 9/4 yds fabric per 1 ft of ribbon.

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8 0
3 years ago
Let Y1 and Y2 have the joint probability density function given by:
Ann [662]

Answer:

a) k=6

b) P(Y1 ≤ 3/4, Y2 ≥ 1/2) =  9/16

Step-by-step explanation:

a) if

f (y1, y2) = k(1 − y2), 0 ≤ y1 ≤ y2 ≤ 1,  0, elsewhere

for f to be a probability density function , has to comply with the requirement that the sum of the probability of all the posible states is 1 , then

P(all possible values) = ∫∫f (y1, y2) dy1*dy2 = 1

then integrated between

y1 ≤ y2 ≤ 1 and 0 ≤ y1 ≤ 1

∫∫f (y1, y2) dy1*dy2 =  ∫∫k(1 − y2) dy1*dy2 = k  ∫ [(1-1²/2)- (y1-y1²/2)] dy1 = k  ∫ (1/2-y1+y1²/2) dy1) = k[ (1/2* 1 - 1²/2 +1/2*1³/3)-  (1/2* 0 - 0²/2 +1/2*0³/3)] = k*(1/6)

then

k/6 = 1 → k=6

b)

P(Y1 ≤ 3/4, Y2 ≥ 1/2) = P (0 ≤Y1 ≤ 3/4, 1/2 ≤Y2 ≤ 1) = p

then

p = ∫∫f (y1, y2) dy1*dy2 = 6*∫∫(1 − y2) dy1*dy2 = 6*∫(1 − y2) *dy2 ∫dy1 =

6*[(1-1²/2)-((1/2) - (1/2)²/2)]*[3/4-0] = 6*(1/8)*(3/4)=  9/16

therefore

P(Y1 ≤ 3/4, Y2 ≥ 1/2) =  9/16

8 0
3 years ago
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Multiply 18 x 8 to find the cost for 8 months. $18 x 8 = $144.

Now add $25 to $144, which equals $169

5 0
3 years ago
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