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2 years ago

The diameter of a ball is 8 in. What is the volume of the ball? Use 3.14 for pi.

Mathematics

2 answers:

Snezhnost [94]2 years ago

8 0

6 million 6 hundred and 66 thousands 6 hundred and 66

SSSSS [86.1K]2 years ago

3 0

Answer:

As said above.

Step-by-step explanation:

I also took the test, thank you for the right answer unlike this unhelfpful idiot who just wrote a bunch of sixes.

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1) Which of the following points is not on the graph of y=-3x^2+6x-4

Margaret [11]

1 is B.
2 is A
3 is B
graph the equation and check to see if the answers match any intercepting points.

6 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,

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

$\begin{array}{c}\\ = 1 - 0.082\\\\ = 0.9179\\\end{array}$

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

7 0

3 years ago

There are an infinite number of vector and parametric equations for a given plane. Why is the scalar equation of a given plane u

Sonbull [250]

Answer:

The answer is below

Step-by-step explanation:

When you refer to a normal vector you mean the form a*x + b*y + c*z = d, if that's the case then it's not unique in the nose because it gives you its normal vector. Taking into account that uniqueness only supports multiplicative constants, which means that you can multiply the equation with whatever you want, that is, it remains the same

5 0

3 years ago

Kenya is solving the equation 5 3/4 - 1/6x = 4 1/2. She gets to 1 1/4 = 1/6x. Is this a valid way to solve the equation? answer

Alona [7]

Answer:

15/2

Step-by-step explanation:

5 3/4 - 1/6x = 4 1/2

move the variable to it's own side.

5 3/4 - 4 1/2 = 1/6x

change the subtracting side to have common denominators

5 3/4 - 4 2/4 = 1/6x

1 1/4 = 1/6 x

divide both sides by 1/6, same thing as multiplying both sides by 6 (reciprocal)

30/4 = x

reduce: 15/2 = x

8 0

2 years ago

What are the domain, range, and asymptote of h(x) = (0.5)x – 9?

Rzqust [24]

Let's solve for h

hx = 0.5x - 9

Step 1: Divide both sides by x.

hx/x = 0.5x - 9/ x

Answer = h = 0.5x - 9/ x

8 0

2 years ago