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AnnZ [28]
3 years ago
15

Kairi spent $40.18 on CDs. Each CD cost the same amount. The sale tax was$2.33. Kairi also used a coupon for $1.00 off his purch

ase. How much did each CD cost?
Mathematics
1 answer:
Arte-miy333 [17]3 years ago
8 0

Therefore the cost of each CD is =$2.30

Step-by-step explanation:

Given , Kairi spent $40 .18 no CDs. The sale tax was $2.33.Kairi also used a coupon for $1.00 0ff his purchase.

Total cost price of The CDs is = $(40+2.33-1.00)

                                                =$41.33

Therefore the cost of each CD is =$(41.33÷18)

                                                       =$2.30

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What is the common factor of the numerator and denominator in the expression <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%282x%2
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Answer:

Step-by-step explanation:

(x - 4)

What you are being asked to do is find the exact same binomial in the top as is in the bottom.

But there's a small catch. You must stipulate that x cannot equal 4. If it does, then you will get 0/0 which is undefined. You can't have that happening -- not at this level.

Any other value for x is fine.

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Which chocolate chip cookies have the greater spread in the number of chips per cookie?
Elan Coil [88]
Reduced fat had the greater spread
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3 years ago
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Given that x overbarequals3.6667​, s Subscript xequals2.0656​, y overbarequals4.2167​, s Subscript yequals1.5613​, and requalsne
Shtirlitz [24]

Answer:

y = 0.7063x+2.7790

Step-by-step explanation:

Given that

x bar = 3.6667\\s_x= 2.0656\\y bar = 4.2167\\s_7 = 1.5613\\r = -0.9344

We have slope of linear regression line is

a=r*\frac{s_y}{s_x} =0.9344(\frac{1.5613}{2.0656} )\\=0.7063

So regression line would be of the form

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Question 5 of 20
Cerrena [4.2K]

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3 years ago
A heavy rope, 30 ft long, weighs 0.4 lb/ft and hangs over the edge of a building 80 ft high. Approximate the required work by a
gizmo_the_mogwai [7]

Answer:

180 fb*lb

45 ft*lb

Step-by-step explanation:

We have that the work is equal to:

W = F * d

but when the force is constant and in this case, it is changing.

 therefore it would be:

W = \int\limits^b_ a {F(x)} \, dx

Where a = 0 and b = 30.

F (x) = 0.4 * x

Therefore, we replace and we would be left with:

W = \int\limits^b_a {0.4*x} \, dx

We integrate and we have:

W = 0.4 / 2 * x ^ 2

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W = 0.2 * (30 ^ 2) - 0.2 * (0 ^ 2)

W = 180 ft * lb

Now in the second part it is the same, but the integral would be from 0 to 15.

we replace:

W = 0.2 * (15 ^ 2) - 0.2 * (0 ^ 2)

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3 years ago
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