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kupik [55]
2 years ago
15

(6 x 10^8) divide (1.5 x 10^-4)

Mathematics
2 answers:
stiv31 [10]2 years ago
5 0

Answer:

Step-by-step explanation:

the answer is A OK

kherson [118]2 years ago
5 0

Answer:

Step-by-step explanation:

(6 x 10^8) =600,000,000

(1.5 x 10^-4) = 0.00015

________________________________________

600,000,000 ÷ 0.00015 = 4 ×10^12

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Kawai requires a loan of $33,000 to buy a used sport utility vehicle. His banks offers financing at 6.75% compounded monthly, fo
UkoKoshka [18]

Answer:

$49,126.87

Step-by-step explanation:

Given data

Principal= $33,000

rate= 6.75%

Time= 6 years

First, convert R as a percent to r as a decimal

r = R/100

r = 6.65/100

r = 0.0665 rate per year,

Then solve the equation for A

A = P(1 + r/n)^nt

A = 33,000.00(1 + 0.0665/12)^(12)(6)

A = 33,000.00(1 + 0.005541667)^(72)

A = $49,126.87

Hence the final amount is  $49,126.87

7 0
3 years ago
Of the following points, name all that lie on the same vertical line?
Masteriza [31]

Answer:

(-2,-6) (6,2)

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
What is the quotient (x^3-3x^2+3x-2) ÷ (x^2-x+1)​
qaws [65]

Answer:

x-2

Step-by-step explanation:

x3−3x2+3x−2/x2−x+1

=

x3−3x2+3x−2/x2−x+1

=

(x−2)(x2−x+1)/x2−x+1

=x−2

8 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>

​

​

7 0
3 years ago
A spinner has 10 numbered sections. The arrow on the spinner was spun 30 times, and the results are recorded in the table. Based
Ann [662]
I can’t understand the question can you write it more clearly?
7 0
3 years ago
Read 2 more answers
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