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3 years ago

A cube has an edge length of 10 centimeters. What is its volume, in cubic centimeters?

Mathematics

2 answers:

Grace [21]3 years ago

8 0

Answer:

1000

Step-by-step explanation:

10 x 10 x 10 = 1000

Morgarella [4.7K]3 years ago

3 0

Answer:

1000 cm³

Step-by-step explanation:

The volume (V) of a cube is calculated as

V = s³ ( s is the edge length ) , then

V = 10³ = 1000 cm³

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Write the equation for the parabola that has x− intercepts (1.2,0) and (4,0) and y− intercept (0,12).

zzz [600]

Equation of a parabola is written in the form of f(x)=ax²+bx+c.
The equation passes through points (4,0), (1.2,0) and (0,12), therefore;
replacing the points in the equation y = ax² +bx+c
we get 0 = a(4)²+b(4) +c for (4,0)
0 = a (1.2)²+ b(1.2) +c for (1.2,0)
12 = a(0)² +b(0) +c for (0,12)
simplifying the equations we get
16a + 4b + c = 0
1.44a +1.2b + c = 0
+c = 12
thus the first two equations will be
16a + 4b = -12
1.44 a + 1.2b = -12 solving simultaneously
the value of a = 5/2 and b =-13
Thus, the equation of the parabola will be given by;
y= 5/2x² - 13x + 12 or y = 2.5x² - 13x + 12

3 0

3 years ago

Read 2 more answers

Find the value of this expression if x=9 x2 + 6 / x - 6

GREYUIT [131]

Answer:

87/3

Step-by-step explanation:

Here we are given the expression:

$\frac{x^{2}+6}{x-6}$

Now we have to find the value of this expression when x equals 9.

So plugging the value of x as -9 in the given expression:

$\frac{x^{2}+6}{x-6}$

$=\frac{(9)^{2}+6}{9-6}$

=87/3

So on evaluating we get the value of expression as 87/3.

3 0

3 years ago

Read 2 more answers

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

Elenna [48]

Answer:

Part a: The probability of no arrivals in a one-minute period is 0.000045.

Part b: The probability of three or fewer passengers arrive in a one-minute period is 0.0103.

Part c: The probability of no arrivals in a 15-second is 0.0821.

Part d: The probability of at least one arrival in a 15-second period is 0.9179.

Step-by-step explanation:

Airline passengers are arriving at an airport independently. The mean arrival rate is 10 passengers per minute. Consider the random variable X to represent the number of passengers arriving per minute. The random variable X follows a Poisson distribution. That is,

$X \sim {\rm{Poisson}}\left( {\lambda = 10} \right)$

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

Substitute the value of λ=10 in the formula as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{{\left( {10} \right)}^x}}}{{x!}}$

Part a:

The probability that there are no arrivals in one minute is calculated by substituting x = 0 in the formula as,

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}}\\\\ = {e^{ - 10}}\\\\ = 0.000045\\\end{array}$

The probability of no arrivals in a one-minute period is 0.000045.

Part b:

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

The probability of the arrival of three or fewer passengers in one minute is calculated by substituting \lambda = 10λ=10 and x = 0,1,2,3x=0,1,2,3 in the formula as,

$\begin{array}{c}\\P\left( {X \le 3} \right) = \sum\limits_{x = 0}^3 {\frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}} \\\\ = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^1}}}{{1!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^2}}}{{2!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^3}}}{{3!}}\\\\ = 0.000045 + 0.00045 + 0.00227 + 0.00756\\\\ = 0.0103\\\end{array}$

The probability of three or fewer passengers arrive in a one-minute period is 0.0103.

Part c:

Consider the random variable Y to denote the passengers arriving in 15 seconds. This means that the random variable Y can be defined as $\frac{X}{4}$

$\begin{array}{c}\\E\left( Y \right) = E\left( {\frac{X}{4}} \right)\\\\ = \frac{1}{4} \times 10\\\\ = 2.5\\\end{array}$

That is,

$Y\sim {\rm{Poisson}}\left( {\lambda = 2.5} \right)$

So, the probability mass function of Y is,

$P\left( {Y = y} \right) = \frac{{{e^{ - \lambda }}{\lambda ^y}}}{{y!}};x = 0,1,2, \ldots$

The probability that there are no arrivals in the 15-second period can be calculated by substituting the value of (λ=2.5) and y as 0 as:

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = {e^{ - 2.5}}\\\\ = 0.0821\\\end{array}$

The probability of no arrivals in a 15-second is 0.0821.

Part d:

The probability that there is at least one arrival in a 15-second period is calculated as,

$\begin{array}{c}\\P\left( {X \ge 1} \right) = 1 - P\left( {X < 1} \right)\\\\ = 1 - P\left( {X = 0} \right)\\\\ = 1 - \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = 1 - {e^{ - 2.5}}\\\end{array}$