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GuDViN [60]
3 years ago
15

The equation of a linear function in point-slope form is y – y1 = m(x – x1). Harold correctly wrote the equation y = 3(x – 7) us

ing a point and the slope. Which point did Harold use?
When Harold wrote his equation, the point he used was (7, 3).
When Harold wrote his equation, the point he used was (0, 7).
When Harold wrote his equation, the point he used was (7, 0).
When Harold wrote his equation, the point he used was (3, 7).
Mathematics
2 answers:
ikadub [295]3 years ago
7 0

Answer:

When Harold wrote his equation, the point he used was (7,0).

Step-by-step explanation:

The equation of a linear function in point-slope form :

y-y_1=m(x-x_1)  --A

Where m is the slope

(x_1,y_1) are passing points

Harold's Equation : y = 3(x – 7)

On comparing Harold's Equation with A

x_1=7

y_1=0

Thus the passing point is (7,0).

Hence Option C is correct.

When Harold wrote his equation, the point he used was (7,0).

In-s [12.5K]3 years ago
6 0
I believe answer is 7,0/ hope this helps
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Simplify. (3 + 2) + (4 + 3) =
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Answer:

5 + 7

Step-by-step explanation:

3 + 2 = 5, 4 + 3 = 7

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3 years ago
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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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The Big Dipper ice cream store charges 25 cents for a cone and 75 cents for each scoop of ice cream added to the cone. What will
yulyashka [42]
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Answer: $1.75
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What is 3.75 as a mixed number in its simplest form
Fittoniya [83]
3\frac{3}{4}
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3 years ago
2. (8 points) If you have n gold coins with one a fake that you know is slightly lighter, how many weighings on the pan balance
pogonyaev

Answer:

Only 1 Time

Step-by-step explanation:

Since you know that one coin is fake and that the coin is lighter than the rest , then you would only need to weigh all the coins together 1 time to ensure that you will find the fake coin. We can determine this by doing the following.

<u>Example:</u>

Lets say we have 100 coins that weigh 1 gr each. Meaning that the total weight of the 100 coins would be (100coins*1gr = 100gr) <em><u>IF</u></em> there are no fake coins in the batch.

Knowing this we can weigh the entire batch 1 time and if the total weight is less than 100gr we know that the fake coin is in that batch.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

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2 years ago
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