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Anton [14]
2 years ago
10

A six-sided fair die is rolled and a fair coin is tossed. If event M represents getting an odd number on the die and event N rep

resents landing tails on the coin, are these two events dependent or independent? The two events are _____. In the given scenario, P(M and N) =______ . HELP!
Mathematics
1 answer:
poizon [28]2 years ago
3 0
<span>Answer:

The two events are independent. In the given scenario, P(M and N) =0.25

Explanation:

1) Independent events are those that do not affect each other. In this case the number that you roll does not change the flip of the coin. So, whatever number you get with the die, the probabilities of landing tail or head when tossing the coin are the same.

2) The probability of the two simultaneous events: getting an odd number and landing a tail, since they are independent, is equal to the product of the two independent events:
P(M and N) = P(M) x P (N) (only because they are independent)

3) The probability of getting and odd number is 0.5 (half of the numbers are odd and half are even => probability = 1/2)

4) The probability of the coin lands tail is 0.5 (1/2).

5) Then the joint probability is 0.5 x 0.5 = 0.25</span>
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(a) Find a vector-parametric equation r⃗ 1(t)=⟨x(t),y(t),z(t)⟩r→1(t)=⟨x(t),y(t),z(t)⟩ for the shadow of the circular cylinder x2
motikmotik

Answer: (a) r1(t) = <2cost , 0 , 2sint>

(b) <2cost , (1 - 12sint - 10cost)/8 , 2sint>

Step-by-step explanation:

x2+z2=4

a)

Now, in the xz plane, we know that y = 0...

So, x^2 + z^2 = 4 will simply be a circle centered at (0,0)..

This can be easily parameterized as

x = 2cos(t)

z = 2sin(t)

So, the required parameterization is :

r1(t) = <2cost , 0 , 2sint>

b)

Cylinder : x^2 + z^2 = 4

Plane : 5x+8y+6z=1

Easily enough, the x^2 + z^2 = 4 can again be parameterized as

x = 2cost , z = 2sint

With this, we can find y using plane equation...

5x+8y+6z=1

5(2cost) + 8y + 6(2sint) = 1

8y = 1 - 12sint - 10cost

y = (1 - 12sint - 10cost)/8

So, the parameterization is :

<2cost , (1 - 12sint - 10cost)/8 , 2sint>

6 0
2 years ago
Use Bayes' rule to find the indicated probability. The incidence of a certain disease on the island of Tukow is 4%. A new test h
lina2011 [118]
These are the events in the question above:

<span>D - has disease 
</span>
<span>H - healthy (does not have disease) 
</span>
<span>P - tests positive </span>

<span>It is the probability that a person has the disease AND tests positive divided by the probability that the person tests positive.
</span>
Sick, + [.04*.91] = .0364 

<span>Sick, - [.04*.09] = .0036 </span>

Healthy, + [.96*.04] = 0.0384

<span>Healthy, - [.96*.96] = .9216 

</span>
.0364 / (.0364 + .0.0384) = 0.487
7 0
2 years ago
A taxi travels 2 hours to make a 220 km trip. How fast does it travel?<br>​
sertanlavr [38]
He would be driving approximately 110 km an hour
7 0
1 year ago
Read 2 more answers
The national average for mathematics SATs in 2011 was 514 and the standard deviation was approximately 40. a) Within what bounda
blondinia [14]

Answer:

0.75 = 1-\frac{1}{k^2}

If we solve for k we can do this:

\frac{1}{k^2}= 1-0.75=0.25

\frac{1}{0.25}= k^2

k^2 =4

k =\pm 2

So then we have at last 75% of the data withitn two deviations from the mean so the limits are:

Lower = \mu -2\sigma = 514- 2*40=434

Upper = \mu +2\sigma = 514 + 2*40=594

Step-by-step explanation:

We don't know the distribution for the scores. But we know the following properties:

\mu = 514 , \sigma =40

For this case we can use the Chebysev theorem who states that "At least 1 -\frac{1}{k^2} of the values lies between \mu -k\sigma and \mu +k\sigma"

And we need the boundaries on which we expect at least 75% of the scores. If we use the Chebysev rule we have this:

0.75 = 1-\frac{1}{k^2}

If we solve for k we can do this:

\frac{1}{k^2}= 1-0.75=0.25

\frac{1}{0.25}= k^2

k^2 =4

k =\pm 2

So then we have at last 75% of the data withitn two deviations from the mean so the limits are:

Lower = \mu -2\sigma = 514- 2*40=434

Upper = \mu +2\sigma = 514 + 2*40=594

4 0
2 years ago
Suppose that in one region of the country the mean amount of credit card debt perhousehold in households having credit card debt
kvv77 [185]

Answer:

The probability that the mean amount of credit card debt in a sample of 1600 such households will be within $300 of the population mean is roughly 0.907 = 90.7%.

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are 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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 15250, \sigma = 7125, n = 1600, s = \frac{7125}{\sqrt{1600}} = 178.125

The probability that the mean amount of credit card debt in a sample of 1600 such households will be within $300 of the population mean is roughly

This probability is the pvalue of Z when X = 1600 + 300 = 1900 subtracted by the pvalue of Z when X = 1600 - 300 = 1300. So

X = 1900

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

By the Central Limit Theorem

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

Z = \frac{1900 - 1600}{178.125}

Z = 1.68

Z = 1.68 has a pvalue of 0.9535.

X = 1300

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

Z = \frac{1300 - 1600}{178.125}

Z = -1.68

Z = -1.68 has a pvalue of 0.0465.

0.9535 - 0.0465 = 0.907.

The probability that the mean amount of credit card debt in a sample of 1600 such households will be within $300 of the population mean is roughly 0.907 = 90.7%.

7 0
2 years ago
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