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tankabanditka [31]
3 years ago
10

Roll two fair dice separately. Each die has six faces. a. List the sample space. b. Let A be the event that either a three or fo

ur is rolled first, followed by an even number. Find P(A). c. Let B be the event that the sum of the two rolls is at most seven. Find P(B). d. In words, explain what "P(A|B)" represents. Find P(A|B). e. Are A and B mutually exclusive events? Explain your answer in one to three complete sentences, including numerical justification. f. Are A and B independent events? Explain your answer in one to three complete sentences, including numerical justification.
Mathematics
1 answer:
dezoksy [38]3 years ago
3 0

Answer:

Step-by-step explanation:

a) {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6)(4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}

\Omega=6*6=36

b)A=(2*3)=6

P(A)=6/36=1/6

c) B=6+5+4+3+2+1=21

P(B)=21/36

d) P(A|B) - an event where either a 3 or 4 is rolled first and is followed by an even number and their sum goes over 7

P(A|B)=3

e) Not always, not sure how to explain, I'm not good with English Math

f) Same as above

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Solve this problem.
Katyanochek1 [597]

Answer:

Let x = the width of the smaller rectangle.

The length of the smaller rectangle is 2x - 1.

Area is A = lw

So the area of the smaller rectangle is A = (x)(2x - 1) = 2x^2 - x

Step-by-step explanation:

hope it helped

3 0
3 years ago
The density of a fluid is given by the empirical equation rho=70.5 exp(8.27 X 10^-7 P) where rho is density (lbm/ft^3) and P is
jasenka [17]

Answer:

(a) The unit of 70.5 is lbm/ft^3 and the unit of 8.27×10^-7 is in^2/lbf

(b) density = 0.1206g/cm^3

(c) rho = 0.1206exp(8.27×10^-7P)

Step-by-step explanation:

(a) The unit of 70.5 is the same as the unit of rho which is lbm/ft^3. The unit of 8.27×10^-7 is the inverse of the unit of P (lbf/in^2) because exp is found of a constant. Therefore, the unit of 8.27×10^-7 is in^2/lbf

(b) P = 9×10^6N/m^2

rho = 70.5exp(8.27×10^-7× 9×10^6) = 70.5exp7.443 = 70.5×1.71 = 120.6kg/m^3

rho = 120.6kg/m^3 × 1000g/1kg × 1m^3/10^6cm^3 = 0.1206g/cm^3

(c) Formula for rho (g/cm^3) as a function of P (N/m^2) is

rho = 0.1206exp(8.27×10^-7P) (the unit of 0.1206 is g/cm^3)

5 0
3 years ago
Calculate (A⃗ ×B⃗ )⋅C⃗ for the three vectors A⃗ with magnitude A = 5.08 and angle θA = 25.6 ∘ measured in the sense from the +x
alexgriva [62]

Answer:

(A⃗ ×B⃗ )⋅C⃗ = - 76.415

Step-by-step explanation:

First we need to calculate (A⃗ ×B⃗ ) :

(A⃗ ×B⃗ ) = A.B.sin (α).n

Where A is the magnitude of A⃗

Where B is the magnitude of B⃗

Where α is the angle between A⃗ and B⃗ = 63.9 - 25.6 = 38.3

Finally n is the vector orthogonal to A⃗ and B⃗

n magnitude is 1 and his direction is given by the right hand-rule

so n = ( 0 , 0 , 1 )

(A⃗ ×B⃗ ) = A.B.sin (α).n = 5.08 . 3.94 . sin (38.3) . (0 , 0 , 1 ) = (0,0,12.4)

C⃗ can be written as C.(0,0,-1) because of his +z - direction

C.(0,0,-1) = 6.16.(0,0,-1) = (0,0,-6.16)

(A⃗ ×B⃗ )⋅C⃗ = (0,0,12.4).(0,0,-6.16) = -76.41480787 = -76.415

8 0
3 years ago
A simple random sample of 110 analog circuits is obtained at random from an ongoing production process in which 20% of all circu
telo118 [61]

Answer:

64.56% probability that between 17 and 25 circuits in the sample are defective.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 110, p = 0.2

So

\mu = E(X) = np = 110*0.2 = 22

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{110*0.2*0.8} = 4.1952

Probability that between 17 and 25 circuits in the sample are defective.

This is the pvalue of Z when X = 25 subtrated by the pvalue of Z when X = 17. So

X = 25

Z = \frac{X - \mu}{\sigma}

Z = \frac{25 - 22}{4.1952}

Z = 0.715

Z = 0.715 has a pvalue of 0.7626.

X = 17

Z = \frac{X - \mu}{\sigma}

Z = \frac{17 - 22}{4.1952}

Z = -1.19

Z = -1.19 has a pvalue of 0.1170.

0.7626 - 0.1170 = 0.6456

64.56% probability that between 17 and 25 circuits in the sample are defective.

4 0
3 years ago
5.25 divided by 3<br> Thank You.
labwork [276]

Answer:

1.75

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
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