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3 years ago

What is the approximate volume?

Mathematics

2 answers:

True [87]3 years ago

6 0

Answer:

B) 6,300 m³

Step-by-step explanation:

The volume of approximate = L×W×H/3

30×30×21/3=

answer: 6,300

$**************$

hope it helps...

have a great day!!

liubo4ka [24]3 years ago

3 0

Answer:

Step-by-step explanation:

1) find the base area

base area = side^2 = 30^2 = 900 m^2

2) find the Volume

V = (base area x height)/3 = (900 x 21)/3 = 6300 m^3

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Subtraction using ten facts

Nonamiya [84]

? what u r saying but think u r saying subtraction facts of 10. Right.
10-0=10
10-1=9
10-2=8
10-3=7
10-4=6
10-5=5
10-6=4
10-7=3
10-8=2
10-9=1
That is all the basic subtraction 10 facts i know hope it helps.
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3 0

3 years ago

Find the arc length of the semicircle.

mafiozo [28]

The arc length of the semicircle is 15.7

<h3>Calculating Arc length </h3>

From the question, we are to determine the arc length of the semicircle

Arc length can be determined by using the formula,

Arc length = θ/360° × 2πr

Where θ is the angle subtended by the arc

and r is the radius of the circle

In the given diagram,

θ = 180°

and r = 10/2

r = 5

Thus,

The arc length of the semicircle = 180°/360° ×2×3.14×5

The arc length of the semicircle = 1/2×2×3.14×5

The arc length of the semicircle = 15.7

Hence, the arc length of the semicircle is 15.7

Learn more on Calculating Arc length here: brainly.com/question/16552139

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8 0

2 years ago

Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The m

marshall27 [118]

Answer:

1. 0.0000454

2. 0.01034

3. 0.0821

4. 0.918

Step-by-step explanation:

Let X be the random variable denoting the number of passengers arriving in a minute. Since the mean arrival rate is given to be 10,

$X \sim Poi(\lambda = 10)$

1. Requires us to compute

$P(X = 0) = e^{-10} \frac{10^0}{0!} = 0.0000454$

2. We need to compute $P(X \leq 3) = P(X =0) + P(X =1) + P(X =2) + P(X =3)$

$P(X =1) = e^{-10} \frac{10^1}{1!} = 0.000454$

$P(X =2) = e^{-10} \frac{10^2}{2!} = 0.00227$

$P(X =3) = e^{-10} \frac{10^3}{3!} = 0.00757$

$P(X \leq 3) =0.0000454+ 0.000454 + 0.00227 + 0.00757 = 0.01034$

3. The expected no. of arrivals in a 15 second period is = $10 \times \frac{1}{4} = 2.5$ . So if Y be the random variable denoting number of passengers arriving in 15 seconds,

$Y \sim Poi(2.5)$

$P(Y=0) = e^{-2.5} \frac{2.5^0}{0!} = 0.0821$

4. Here we use the fact that Y can take values $0,1, \dotsc$ . So, the event that "Y is either 0 or $\geq$ 1" is a sure event ( i.e it has probability 1 ).