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Murrr4er [49]
3 years ago
14

A rectangular piece of metal is 15 in longer than it is wide. Squares with sides 3 in long are cut from the four corners and the

flaps are folded upward to form an open box. If the volume of the box is 1350 incubed​, what were the original dimensions of the piece of​ metal? What is the original​ width? nothing in
Mathematics
1 answer:
ziro4ka [17]3 years ago
5 0

Answer:

The original dimensions of the piece of metal are

Length: 36 inches

Width: 21 inches

Step-by-step explanation:

Let

x ----> the original length pf the piece of metal

y ----> the original width pf the piece of metal

we know that

x=y+15 ----> equation A

The volume of the box is equal to

V=LWH

we have

V=1,350\ in^3

L=(x-3-3)=(x-6)\ in\\W=(y-3-3)=(y-6)\ in

H=3\ in

substitute in the formula of volume

1,350=(x-6)(y-6)(3) ----> equation B

substitute equation A in equation B

1,350=(y+15-6)(y-6)(3)

solve for y

450=(y+9)(y-6)

y^2-6y+9y-54=450\\y^2+3y-504=0

Solve the quadratic equation by graphing

using a graphing tool

The solution is y=21

Find the value of x

x=21+15=36

therefore

The original dimensions of the piece of metal are

Length: 36 inches

Width: 21 inches

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The television show CSI: Shoboygan has been successful for many years. That show recently had a share of 18, meaning that among
jeyben [28]

Answer:

(a) The value of P (None) is 0.062.

(b) The value of P(at least one) is 0.938.

(c) The value of P(at most one) is 0.253.

(d) The event is not unusual.

Step-by-step explanation:

Let <em>X</em> = number of households watching the show.

The probability of the random variable <em>x</em> is, P (X) = <em>p</em> = 0.18.

The sample selected for the survey is of size, <em>n</em> = 14

The random variable <em>X</em> follows a Binomial distribution with parameter <em>n</em> = 14 and <em>p</em> = 0.18.

The probability of a Binomial distribution is computed using the formula:

P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,...

(a)

Compute the probability that none of the households are tuned to CSI: Shoboygan as follows:

P(X=0)={14\choose 0}(0.18)^{0}(1-0.18)^{14-0}=1\times1\times0.06214=0.062

Thus, the value of P (None) is 0.062.

(b)

Compute the probability that at least one household is tuned to CSI: Shoboygan as follows:

P (X ≥ 1) = 1 - P (X < 1)

             = 1 - P (X = 0)

             =1-0.062\\=0.938

Thus, the value of P(at least one) is 0.938.

(c)

Compute the probability that at most one household is tuned to CSI: Shoboygan as follows:

P (X ≤ 1) = P (X = 0) + P (X = 1)

             ={14\choose 0}(0.18)^{0}(1-0.18)^{14-0}+{14\choose 1}(0.18)^{1}(1-0.18)^{14-1}\\=0.062+0.191\\=0.253

Thus, the value of P(at most one) is 0.253.

(d)

An event that has a very low probability of occurrence is known as an unusual event.

The probability of the event "at most one household is tuned to CSI: Shoboygan" is 0.253.

This probability value is not low.

Hence, the event is not unusual.

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