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strojnjashka [21]
3 years ago
13

I need showing work for this question for finding the volume of it

Mathematics
2 answers:
Rudiy273 years ago
5 0
The formula for finding volume of a rectangular prism is
V = l * w * h
Where l is length = 6yd
w is width = 1 2/3
( 1 2/3 can be rewritten as 5/3)
h is height = 3 1/3
( 3 2/3 can be rewritten as 11/3)
we then want to plug them into our formula
V = 6 * 5/3 * 11/3
We then multiply our fractions so we can just multiply across to get
V = 30/3 * 11/3
V = 330/9
Which we can simplify and get
V = 110/3 yd^3
lesya692 [45]3 years ago
3 0

Volume of rectangular prism:
V=l•w•h
V=6•1 2/3•3 1/3
V=4
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Hewo!

Step-by-step explanation:

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3 0
2 years ago
Is -2 a solution of 2x+8=2?
Katarina [22]

Answer:2x+8=2

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Step-by-step explanation:

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3 years ago
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businessText message users receive or send an average of 62.7 text messages per day. How many text messages does a text message
KiRa [710]

Answer:

(a) The probability that a text message user receives or sends three messages per hour is 0.2180.

(b) The probability that a text message user receives or sends more than three messages per hour is 0.2667.

Step-by-step explanation:

Let <em>X</em> = number of text messages receive or send in an hour.

The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em>.

It is provided that users receive or send 62.7 text messages in 24 hours.

Then the average number of text messages received or sent in an hour is: \lambda=\frac{62.7}{24}= 2.6125.

The probability of a random variable can be computed using the formula:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!} ;\ x=0, 1, 2, 3, ...

(a)

Compute the probability that a text message user receives or sends three messages per hour as follows:

P(X=3)=\frac{e^{-2.6125}(2.6125)^{3}}{3!} =0.21798\approx0.2180

Thus, the probability that a text message user receives or sends three messages per hour is 0.2180.

(b)

Compute the probability that a text message user receives or sends more than three messages per hour as follows:

P (X > 3) = 1 - P (X ≤ 3)

              = 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)

             =1-\frac{e^{-2.6125}(2.6125)^{0}}{0!}-\frac{e^{-2.6125}(2.6125)^{1}}{1!}-\frac{e^{-2.6125}(2.6125)^{2}}{2!}-\frac{e^{-2.6125}(2.6125)^{3}}{3!}\\=1-0.0734-0.1916-0.2503-0.2180\\=0.2667

Thus, the probability that a text message user receives or sends more than three messages per hour is 0.2667.

6 0
3 years ago
Help show work step,<br> By step how to get the awnser
Archy [21]

Step-by-step explanation:

8 \frac{1}{3}  - 5 \frac{1}{2}  =  \frac{25}{3}  -  \frac{11}{2}  =  \frac{50}{6}  -  \frac{33}{6}  =  \frac{17}{6}  = 2 \frac{5}{6}

3 0
2 years ago
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