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vodomira [7]
3 years ago
6

A bag contains 10 marbles: 4 are green, 4 are red, and 2 are blue. Heather chooses a marble at random, and without putting it ba

ck, chooses another one at random. What is the probability that both marbles she chooses are blue? Write your answer as a fraction in simplest form.
Mathematics
2 answers:
puteri [66]3 years ago
3 0
Hence there are 2 blue marbles: P = E/S = 2/10 + 1/9 = 28/90 = 0.3111111
Deffense [45]3 years ago
3 0
Probability of both marbles blue
= 2/10 × 1/9
= 1/45
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The given line passes through the points (0, −3) and (2, 3).
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If the 2 lines are parallel, their slopes are the same.

The slope of both is  
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          2 - 0

If the new line <span>passes through the point (−1, −1), then the equation in point slope form is:

y+1 = 3(x+1).</span>
4 0
3 years ago
What is equivalent to 3x-4y=6
Andreyy89

Answer:

y= (3/4) x - (3/2)

Step-by-step explanation:

3x - 4y = 6 (using algebra add the 4y to the other side & subtract the 6 to the other side >> 3x - 6 = 4y (then divide everything by 4 to get y by itself) >> 3/4x - 6/4 = y (then simplify) >> 3/4x - 3/2 = y

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: <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
The life in hours of a 75-watt light bulb is known to be normally distributed with standard deviation σ = 25 hours. A random sam
irina [24]

Answer: a) (1008.34,1019.658) b) (1009.24,1018.76)

Step-by-step explanation:

Since we have given that

n = 75

mean = 1014 hours

Standard deviation = 25 hours

At 95% two sided , z = 1.96

So, confidence interval would be

\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=1014\pm 1.96\dfrac{25}{\sqrt{75}}\\\\=1014\pm 5.658\\\\=(1014-5.658,1014+5.658)\\\\=(1008.34,1019.658)

(b) Construct a 95% lower confidence bound on the mean life.

z = 1.65

So, confidence interval would be

\bar{x}\pm z\times \dfrac{\sigma}{\sqrt{n}}\\\\=1014\pm 1.65\times \dfrac{25}{\sqrt{75}}\\\\=1014\pm 4.76\\\\=(1014-4.76,1014+4.76)\\\\=(1009.24,1018.76)

Hence, a) (1008.34,1019.658) b) (1009.24,1018.76)

4 0
3 years ago
How can a solution to a system of linear equations be determined exactly by graphing?
givi [52]
A solution something something and that’s the answer
5 0
3 years ago
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