Suppose you choose a marble from a bag containing 5 red marbles, 3 white marbles, and 6 blue marbles. You return the first marbl

e to the bag and then choose again. Find P(red and blue).

Mathematics

1 answer:

Aneli [31]3 years ago

5 0

$|\Omega|=14^2=196\\
|A|=5\cdot6\cdot2=60\\\\
P(A)=\dfrac{60}{196}=\dfrac{15}{49}$

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Suppose that the distance, in miles, that people are willing to commute to work is an exponential random variable with a decay p

garik1379 [7]

Answer:

m = $\frac{1}{20}$
μ = 20
σ = 20

The probability that a person is willing to commute more than 25 miles is 0.2865.

Step-by-step explanation:

Exponential probability distribution is used to define the probability distribution of the amount of time until some specific event takes place.

A random variable X follows an exponential distribution with parameter m.

The decay parameter is, m.

The probability distribution function of an Exponential distribution is:

$f(x)=me^{-mx}\ ;\ m>0, x>0$

Given: The decay parameter is, $\frac{1}{20}$

X is defined as the distance people are willing to commute in miles.

The decay parameter is m = $\frac{1}{20}$ .
The mean of the distribution is: $\mu=\frac{1}{m}=\frac{1}{\frac{1}{20}}=20$ .
The standard deviation is: $\sigma=\sqrt{variance}= \sqrt{\frac{1}{(m)^{2}} } =\frac{1}{m} =\frac{1}{\frac{1}{20}} =20$

Compute the probability that a person is willing to commute more than 25 miles as follows:

$P(X>25)=\int\limits^{\infty}_{25} {\frac{1}{20} e^{-\frac{1}{20}x}} \, dx \\=\frac{1}{20}|20e^{-\frac{1}{20}x}|^{\infty}_{25}\\=|e^{-\frac{1}{20}x}|^{\infty}_{25}\\=e^{-\frac{1}{20}\times25}\\=0.2865$

Thus, the probability that a person is willing to commute more than 25 miles is 0.2865.

7 0

3 years ago

The diagram shows a 9cm x 7cm rectangle-based pyramid. all the diagonal sides - TA, TB, TC, AND TD- are length 12cm. M is the mi

lakkis [162]

Answer:

∠TAC is approximately equal to 61.6°

Step-by-step explanation:

The given parameters for the pyramid are;

The dimension for the rectangular base are; Length = 9 cm, width = 7 cm

The length of the diagonal sides, TA, TB, TC, and TD = 12 cm each

The midpoint of the rectangular base = Point M

The diagonal AC = AM + MC

AM = MC as given M is the midpoint of the rectangular base

∴ AC = AM + MC = 2·AM

By Pythagoras' theorem, AC = √(9² + 7²) = √130

AC = √130 cm

∴ AM = AC/2 = (√130)/2 cm

Alternatively, AM = √((9/2 cm)² + (7/2 cm)²) = √(32.5) cm

∠TAC = ∠TAM

By trigonometric ratios, we have;

$cos (\theta) = \dfrac{Length \ of \ adjacent \ side \ to \ angle }{Length \ of \ hypotenuse\ side \ to \ angle}$

$\therefore cos (\angle TAM) = cos (\angle TAC) = \dfrac{\left (\dfrac{\sqrt{130} }{2} \right )}{12} = \dfrac{\sqrt{130} }{2 \times 12} = \dfrac{\sqrt{130} }{24}$