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sertanlavr [38]
2 years ago
6

Each pepperoni slice is 78 of an inch long and costs $0.03. Exactly enough pepperoni slices are laid end to end to have a total

length of 56 inches. What is the total cost of these pepperoni slices?
Mathematics
1 answer:
Leokris [45]2 years ago
8 0

The total cost of these pepperoni slices is $1.92

<h3>What is the total cost of these pepperoni slices?</h3>

Each slice measures 7/8 of an inch. Here we have N slices, such that the measure of the N slices is 56 inches, then we have:

N*(7/8 in) = 56 in

Now we can find the value of N, the number of slices used:

N = (8/7)*56 = 64

And each slice costs $0.03, then the cost of the 64 slices is:

C = $0.03*64  = $1.92

The total cost of these pepperoni slices is $1.92

If you want to learn more about costs:

brainly.com/question/14315509

#SPJ1

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aivan3 [116]

Answer:

x=-3

Step-by-step explanation:

-3x-2y=9

-3x-2(0)=9

-3x-0=9

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Combine like terms.

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Hope this helps!
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Select the equation of the line parallel to the equation y = -2x - 7 that passes through the point (3, 1). y = -2x + 7 y = -2x +
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Y - y₁ = m(x - x₁)
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in a final exam you have four multliple choice questions left to do. each questions has five suggested answers and only one of t
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Answer:

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The probability that you get one questions correct is 0.4096

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Step-by-step explanation:

These probability can be describe with a Binomial Distribution. These distribution can be used when we have n identical and independent situations in which there is a probability p or probability of success and a probability q or probability of fail. Additionally q is equal to 1 - p. The probability of x for a situation in which we can apply binomial distribution is:

P( x,n,p) = n Cx * p^{x } * q^ {n-x}

Where x is the variable that says the number of success in the n situations

And nCx is calculate as:

nC x = \frac{ n!}{ x! (n-x)! }

From the question we can identify that:

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Then the probability of get zero questions correct of 4 questions is:

P( 0, 4, 0.2) = 4 C0 * p^{0 } * q^ {4-0}= 0.4096

The probability of get one question correct of 4 questions is:

P( 1, 4, 0.2) = 4 C1 * p^{1 } * q^ {4-1}= 0.4096

The probability of get three questions correct of 4 questions is:

P( 3, 4, 0.2) = 4 C3 * p^{3 } * q^ {4-3}= 0.0256

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