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Aloiza [94]
3 years ago
12

The tires on Eric's car have a diameter of 18 inches. Rounded to the nearest whole number, how many rotations per second does ea

ch tire
make when Eric is driving at a rate of 88 feet per second?


A.) 2
B.) 19
C.) 37
D.) 59
Mathematics
1 answer:
Step2247 [10]3 years ago
4 0

Answer:

Step-by-step explanation:

We need to figure out how far Eric's car takes him in 1 rotation based on the diameter of his tire in feet. If the diameter of the tire is 18", then it is 1.5'. The circumference will tell us the distance 1 rotation of his tires will take him:

C = 3.1415(1.5) so

C = 4.71225 feet is how far he goes in 1 rotation. If he travels 88 feet in 1 second, we can figure the number of rotations by dividing 88 by 4.71225 to give us the unit rate (or, rotations per second his car makes). This quotient is 18.67 feet per sec, which rounds to 19, choice B.

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What is the solution of -2x = -16
Vilka [71]
Divide both sides by -2

two negatives make a positive and that's true so x = 16/2 so divide it

x = 8 because 2 x 8 = 16 and 2/16 = 8 so true.

Answer: x = 8.
7 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: <em>The probability of no arrivals in a one-minute period is 0.000045.</em>

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

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

Part d: <em>The probability of at least one arrival in a 15-second period​ is 0.9179.</em>

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}

<em>The probability of no arrivals in a one-minute period is 0.000045.</em>

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}

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

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}

<em>The probability of no arrivals in a 15-second is 0.0821.</em>

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}

<em>The probability of at least one arrival in a 15-second period​ is 0.9179.</em>

​

​

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3 years ago
PLEASE ANSWER THANK YOU SO MUCH GUYS:)
babunello [35]

Answer:

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4 0
3 years ago
Read 2 more answers
What is the square root of 16x^<br> 36?
Vsevolod [243]

Note that \sqrt[2]{x}=x^{\frac{1}{2}}

\sqrt[2]{16x^{36}}=\sqrt[2]{4^2x^{18\cdot2}}

4^{2\cdot\frac{1}{2}}x^{18\cdot2\cdot\frac{1}{2}}=4^{\frac{2}{2}}x^{\frac{18\cdot2}{2}}

\boxed{4x^{18}}

Hope this helps.

r3t40

4 0
3 years ago
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Makovka662 [10]

Answer:

x = 9

Step-by-step explanation:

Simplifying

6x + -24 = 2x + 12

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Reorder the terms:

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Move all terms containing x to the left, all other terms to the right.

Add '-2x' to each side of the equation.

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Combine like terms: 6x + -2x = 4x

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Combine like terms: 2x + -2x = 0

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Add '24' to each side of the equation.

-24 + 24 + 4x = 12 + 24

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Combine like terms: 12 + 24 = 36

4x = 36

Divide each side by '4'.

x = 9

Simplifying

x = 9

4 0
1 year ago
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