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3 years ago

Find each missing dimension each prism

Mathematics

1 answer:

Anika [276]3 years ago

5 0

Answer: 3

Step-by-step explanation:

Multiply 7 x 5.2

Divide 109.2/36.4

Answer 3

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150 adults complete a survey 80 are women the ratio of men and women in its simplest form

zavuch27 [327]

Answer:

Step-by-step explanation:

150 - 80 equals 70

70:80 = 7:8

please give me brainliest, I'm going to cry if you don't >:(

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2 years ago

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Pls help me how do I write 20,484,163 in expanded form I'm only in 2nd grade

mestny [16]

Answer:

twenty thousand four hundred and eight four one hundred and sixty four

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3 years ago

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You guys are life savers but does anyone know the answers to these questions? I’m so confused rn

timofeeve [1]

Answer:

The third choice

Step-by-step explanation:

We need to find the slope and y-intercept of the line and then put it into y = mx = b form. To find the slope, pick a point on the line; I will use (-2, 5); count how many units up you need to go to get to the next point on the line, which in this case it would be 3. The count how many to the right or left you would need to go, which is 1 to the left. Moving left means a negative, so it is -1. Your slope fraction would be $\frac{3}{-1}$ , since slope is rise over run. You can sub this fraction in for m in y = mx + b, which will give you a revised equation of y = -3x = b. To find the y intercept, or b, just find the point where the line crosses the y-axis, which is -1. So, the equation is now y = -3x - 1.The correct answer is third choice.

6 0

3 years ago

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

Pls send me step by step pic

Thepotemich [5.8K]

To solve, simply do this:

$(\frac{1}{2}^3)-(\frac{3}{4} )^2\\\\\frac{1}{2} *\frac{1}{2} *\frac{1}{2} *\frac{1}{2} \\\\\frac{1}{8} -(\frac{3}{4})^2\\\\\frac{1}{8}-\frac{9}{16}\\\\=\frac{-7}{16}$

Then you'll get the answer, -7/16

7 0

3 years ago