An oil company purchased an option on land in Alaska. Preliminary geologic studies assigned the following prior probabilities.
P(high-quality oil)=0.45
p(medium-quality oil)=0.15
p(no oil)=0.40
a. What is the probability of finding oil (to 2 decimals)?
b. After 200 feet of drilling on the first well, a soil test is taken. The probabilities of finding the particular type of soil identified by the test are given below.
p(soil | high-quality oil)=0.15
p(soil | medium-quality oil)= 0.75
P(soil | no oil)=0.15
Given the soil found in the test, use Bayes' theorem to compute the following revised probabilities (to 4 decimals).
What is the new probability of finding oil (to 4 decimals)?
According to the revised probabilities, what is the quality of oil that is most likely to be found?

The z-score for a backyard structure costing $2300 is -0.67.

The standard score in statistics is the amount of standard deviations that a raw score deviates from or exceeds the mean of the phenomenon being observed or measured. Standard scores are positive for raw scores above the mean and negative for raw scores below the mean.

The Z-score quantifies the deviation between a given value and the standard deviation. The Z-score, also known as the standard score, indicates how many standard deviations a given data point deviates from the mean. In essence, standard deviation represents the degree of variability present in a given data set.

The z score will be:

= (2300 - 3100) / 1200

= -0.67.

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The high costs in the California real estate market have caused families who cannot afford to buy bigger homes to consider backyard sheds as an alternative form of housing expansion. Many are using the backyard structures for home offices, art studios, and hobby areas as well as for additional storage. The mean price of a customized wooden, shingled backyard structure is $3100 (Newsweek, September 29, 2003). Assume that the standard deviation is $1200.

a. What is the z-score for a backyard structure costing $2300?

A's x- intercept is 3 and y-intercept is 5. Both are favourable.

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a) E(e^(taY)) = m(at), then the moment-generating function of U is e^(tb)m(at).

b) µ is the mean of Y and σ^2 is the variance of Y.

c) the moment-generating function of X =e^((−3µ + 4)t + (9/2)σ^2t^2) and The distribution of X is normal.

a) If Y is a random variable with moment-generating function m(t), and U is given by U = aY + b, then the moment-generating function of U is e^(tb)m(at). This can be shown by using the definition of a moment-generating function:

E(e^(tU)) = E(e^(taY + tb)) = E(e^(taY)e^(tb)) = e^(tb)E(e^(taY))

Since E(e^(taY)) = m(at), then the moment-generating function of U is e^(tb)m(at).

b) To find the mean and variance of U using the moment-generating function, we can take the first and second derivatives of e^(tb)m(at) and set t = 0. The first derivative with respect to t evaluated at t = 0 is:

(d/dt)e^(tb)m(at) = be^(tb)m(at) + ae^(tb)m'(at)

Setting t = 0, we get the mean of U:

E(U) = be^(0)m(a0) + ae^(0)m'(a0) = b + a*(m'(0)) = b + a*µ

Where µ is the mean of Y.

The second derivative of e^(tb)m(at) is:

(d^2/dt^2)e^(tb)m(at) = b^2e^(tb)m(at) + 2abe^(tb)m'(at) + a^2e^(tb)m''(at)

Setting t = 0, we get the variance of U:

Var(U) = b^2e^(0)m(a0) + 2abe^(0)m'(a0) + a^2e^(0)m''(a0) = b^2 + 2ab*(m'(0)) + a^2*(m''(0)) = b^2 + 2abµ + a^2σ^2

Where µ is the mean of Y and σ^2 is the variance of Y.

c) If the moment-generating function of a normally distributed random variable Y with mean µ and variance σ^2 is m(t) = e^(µt + (1/2)t^2σ^2), then the moment-generating function of X = −3Y + 4 is e^((−3µ + 4)t + (1/2)(−3)^2σ^2t^2) = e^((−3µ + 4)t + (9/2)σ^2t^2)

The distribution of X is normal with mean (-3µ + 4) and variance (9/2)σ^2. This can be deduced by the fact that if Y is normally distributed, then a linear transformation of Y (such as X = aY + b) is also normally distributed with mean aµ + b and variance a^2σ^2.

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The number of notepads per department used in the Southeast location is given as follows:

6.

The number of notepads per department is obtained applying the proportions in the context of this problem, considering the graph.

The number of departments of the Southeast location is given as follows:

6.

Which is given from the table given in the second image for this problem.

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The value of x in the equation,  X^3= -125 is -5.

To find x in the above polynomial equation, we need to collect like terms and equate the expression to 0. To begin we can arrange the expression this way:

x^3 + 125 = 0

This means that we should find the figure which when multiplied three times would equate to 125. This figure is -5. So, when solved, we have:

-5³ + 125 = 0

-125 +125 = 0

So, the value for x in the equation above is -5.

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Answer:

4.59 × 10^(5)

Step-by-step explanation:

Answer: 4.59 x 10^5

Step-by-step explanation:

If we move the decimal place over 5 times in the answer we get 459,000.

I hope this was able to help you :D

all the mentioned angles have been found.

What is are of triangle?

The territory included by a triangle's sides is referred to as its area. Depending on the length of the sides and the internal angles, a triangle's area changes from one triangle to another. Square units like m2, cm2, and in2 are used to express the area of a triangle.

The sum of all angles of triangle = 180

∠ICA =180-20 =160

∠EAF = 20

∠ICB =180

∠GAF = 120 -20 =100

∠CAB= 180-80=100

∠HAG=180-120 =60

∠CBA=180-120=60

∠HAC.= 20

Hence all the mentioned angles have been found.

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Answer:

Step-by-step explanation:

The value of x for the given expression will be x = 0.1535.

The mathematical expression combines numerical variables and operations denoted by addition, subtraction, multiplication, and division signs.

Mathematical symbols can be used to represent numbers (constants), variables, operations, functions, brackets, punctuation, and grouping. They can also denote the logical syntax's operation order and other properties.

Given that the expression is [tex]15^{5x}=8[/tex]. The expression will be solved as below,

[tex]15^{5x}=8[/tex]

Take log on both the sides and solve for the value of x.

5x = 3log₁₅(2)

x = (3/2)log₁₅(2)

x = 0.1535

Hence, the value of x will be 0.1535.

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The standard normal distribution is used when calculating the confidence interval for the mean using a sample size of 30 or greater with a known population standard deviation.

The Z-distribution (also known as the standard normal distribution) is a normal distribution with a mean of 0 and a standard deviation of 1. When the population standard deviation is known, the sample mean follows a normal distribution and the standard error is given by the population standard deviation divided by the square root of the sample size. So, the Z-score is given by (x - μ) / (σ / √(n)) where x is the sample mean, μ is the population mean and σ is the population standard deviation and n is the sample size.

With a sample size of 30 or greater, the central limit theorem states that the sample mean will be approximately normally distributed, regardless of the shape of the population distribution. So, a Z-score can be used to find the probability that a sample mean falls within a certain range of the population mean, which is commonly used to construct a confidence interval.

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The closest to the average(arithmetic mean) number of seeds per apple is 6.

What is average ?

An average, also known as a mean, is a measure of central tendency that represents the typical value of a set of data. To calculate the average of a set of numbers, you add them up and then divide by the number of items in the set. For example, the average of the numbers 2, 4, 6, and 8 is (2 + 4 + 6).

Average =  Total number of seeds/12

⇒ Average = 2×3+4×5+1×6+2×7+3×9/12

​⇒ Average = 73/12

⇒ Average ≈6

The closest to the average(arithmetic mean) number of seeds per apple is 6.

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One of the steps in statistical inference, to check the claim is hypothesis testing. In this method, we determine whether to reject or accept the null hypothesis.

Hypothesis testing is the procedure in statistical inference that helps to decide whether the result of the research study supports a particular hypothesis when applied to a population. This helps to test the relationship between two statistical variables.

This includes two types null and alternate hypotheses. If the results say that there is no relationship seen between the two variables, then the null hypothesis is accepted. But if the results so there is a relationship seen between two variables, then an alternate hypothesis is accepted. So to check our claim, we do hypothesis testing.

The complete question is -

The following question is from the four steps of statistical inference. If you respond with a summary of all four steps, you will receive no credit on this problem:

What do we do to check a claim?

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The papers graded exponentially at the end of 5 days are 134, and it will take approximately around 12 days to complete 6000 papers.

A function is considered to be exponentially growing if its value rises faster than any polynomial. The function e^x, serves as a good illustration.

The exponential growth is given by the following formula:

[tex]A = Pe^{rt}[/tex]

where, A is the final amount, P is the initial amount, r is the rate and t is the time.

For t = 0, the number of papers graded are, 9, hence the value of P = 9.

To calculate the rate at which the work is done, substitute the value of P = 9, A = 27 and t = 2 in the equation:

[tex]27 = 9 e^{2r}\\3 = e^{2r}[/tex]

Taking ln on both sides of the equation:

[tex]ln (3) = ln(e^{2r})\\\\ln(3) = 2r[/tex]

1.09 = 2r

r = 0.54

To calculate the number of papers graded in five days, substitute the value of P = 9, t = 5 and r = 0.54.

[tex]A = 9 e^{(0.54)(5)}[/tex]

A = 134

Hence, the papers graded at the end of 5 days are 134.

To complete grading 6000 papers, the time taken will be:

6000 = 9e^(0.54)(t)

666.66 = e^(0.54)(t)

Taking ln on both sides of the equation:

ln (666.66) = ln(e^(0.54)(t))

6.50 = (0.54)(t)

t = 12.04

Hence, it will take approximately around 12 days to complete 6000 papers.

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An exponential model has the formula A is A0(b)t/c, so, we get 1071371.

A = A0(b)t/c,

97532 ( 455277)1/ 41446 = 1071371.

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The quartic function y = 0.5 · x⁴ - x³ - 15.5 · x² + 30 · x + 21 contains roots 4 - √2 and - 3 + √6 and intercept - 21.

Quartic functions are polynomials of grade 4, whose factor form is introduced below:

y = a · (x - r₁) · (x - r₂) · (x - r₃) · (x - r₄)

Where:

According to statement, the quartic equation has an intercept of - 21 and two real roots: 4 - √2, - 3 + √6. The two remaining roots can be found based on features exhibited by quadratic formula and that quartic functions are the product of two quadratic functions: 4 + √2, - 3 - √6

Now we proceed to expand the quartic function:

y = a · (x - 4 + √2) · (x - 4 - √2) · (x + 3 - √6) · (x + 3 + √6)

y = a · (x² - 8 · x + 14) · (x² + 6 · x + 3)

y = a · [x² · (x² - 8 · x + 14) + 6 · x · (x² - 8 · x + 14) + 3 · (x² - 8 · x + 14)]

y = a · (x⁴ - 8 · x³ + 14 · x² + 6 · x³ - 48 · x² + 84 · x + 3 · x² - 24 · x + 42)

y = a · (x⁴ - 2 · x³ - 31 · x² + 60 · x + 42)

And we find the lead coefficient:

- 21 = 42 · a

a = - 0.5

And we obtain the quartic function in standard form:

y = 0.5 · x⁴ - x³ - 15.5 · x² + 30 · x + 21

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The Average Acceleration of the car will be 14.67 ft/[tex]s^{2}[/tex].

What is Acceleration?

The rate at which an object's velocity with respect to time changes is referred to as acceleration in mechanics. Any procedure where the velocity varies is referred to as acceleration. There are only two ways to accelerate: changing your speed or changing your direction, or changing both. This is because velocity is both a speed and a direction.

Formula: a = (v(final) - v(initial))/t

In the given formula, it is given that:

v(initial) = 30mph = 44 ft/s

v(final) = 70mph = 102.67 ft/s

t(time) = 4s

Putting these values in our formula, we get:

a(acceleration) = (102.67 - 44)/4

a = 14.67 ft/[tex]s^{2}[/tex].

Therefore as per the given problem, the acceleration of the given car is 14.67 ft/[tex]s^{2}[/tex].

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The unit vector orthogonal to the plane can be found by taking the cross product of the vectors p-q and p-r. The vector should have a positive y-component if the cross product points in the positive y direction.

To find a unit vector orthogonal to the plane through the points p, q, and r, with a positive y-component, we will use the cross product of two of the vectors. The cross product of two vectors is a vector perpendicular to both of them. First, we will create vector p-q, which is a vector from point p to point q. We can do this by subtracting the coordinates of point q from the coordinates of point p. Next, we will create vector p-r, which is a vector from point p to point r. We can do this by subtracting the coordinates of point r from the coordinates of point p. Finally, we will take the cross product of the two vectors. This will give us a vector orthogonal to the plane through the three points. If the vector has a positive y-component, then it is the unit vector orthogonal to the plane that we are looking for.

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If the residual is negative, it means the observed number of wins is lower than the predicted number of wins, indicating that the team performed worse than expected based on their yards allowed.

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True, When assessing schedule feasibility, a systems analyst must consider the interaction between time and costs.

When assessing the viability of scheduling a project, an organisation determines the anticipated completion time. The feasibility analysis assists in identifying any potential obstacles the proposed project may encounter, such as: Internal project constraints, including those related to technology, finances, and resources

The expansion and acceptance of project management training have undergone significant changes over the past few years, and it is projected that these developments will continue and accelerate. And as project management becomes more popular, feasibility studies become more important. It may be exhilarating to begin a difficult, extensive project that will have a significant influence on your organization.

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Answer:

Question 33

a) x-int (2, 0).  y-int (0, 2/3)

b) Vertical asymptotes: x = -1 and x = 3

   Horizontal asymptote: y = 0

c) f(x) → -∞, as x → -1⁻

   f(x) → +∞, as x → -1⁺

   f(x) → -∞, as x → 3⁻

   f(x) → +∞, as x → 3⁺

d) Domain: (-∞, -1) ∪ (-1, 3) ∪ (3, ∞)

   Range: (-∞, ∞)

Question 34

a)  x-int (-1, 0) and (2,0).  y-int (0, 2/3)

b) Vertical asymptote: x = 3

c) f(x) → -∞, as x → 3⁻

   f(x) → +∞, as x → 3⁺

d) Domain: (-∞, 3) ∪ (3, ∞)

   Range: (-∞, 1] ∪ [9, ∞)

Question 35

a) No axis intercepts

b) Vertical asymptotes: x = -2 and x = 2.  Hole at (0, ¹/₄).

c) f(x) → -∞, as x → -2⁻

   f(x) → +∞, as x → -2⁺

   f(x) → +∞, as x → 2⁻

   f(x) → -∞, as x → 2⁺

d) Domain: (-∞, -2) ∪ (-2, 0) ∪ (0, 2) ∪ (2, ∞)

   Range: (-∞, 0) ∪ (¹/₄, ∞)

Step-by-step explanation:

Definitions

Given rational function:

[tex]g(x) =\dfrac{x - 2}{x^2 - 2x - 3}[/tex]

x-intercept

[tex]\begin{aligned}y=0 \implies \dfrac{x - 2}{x^2 - 2x - 3}&=0\\x - 2&=0\\x&=2\end{aligned}[/tex]

y-intercept

[tex]x=0 \implies \dfrac{0 - 2}{0^2 - 2(0) - 3}=\dfrac{2}{3}[/tex]

Vertical asymptotes

Set the denominator to zero and solve for x:

[tex]\implies x^2-2x-3=0[/tex]

[tex]\implies x^2-3x+x-3=0[/tex]

[tex]\implies x(x-3)+1(x-3)=0[/tex]

[tex]\implies (x+1)(x-3)=0[/tex]

[tex]\implies x+1=0 \implies x=-1[/tex]

[tex]\implies x-3=0 \implies x=3[/tex]

Horizontal asymptote

As the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at y = 0.  

Holes

Factor the denominator:

[tex]g(x) =\dfrac{x - 2}{(x+1)(x-3)}[/tex]

There are no holes as there is no common factor in the numerator and the denominator.

End behaviour near the vertical asymptotes

f(x) → -∞, as x → -1⁻

f(x) → +∞, as x → -1⁺

f(x) → -∞, as x → 3⁻

f(x) → +∞, as x → 3⁺

Domain

As there are vertical asymptotes at x = -1 and x = 3, the domain is restricted:

Range

The range is unrestricted:

Given rational function:

[tex]k(x)= \dfrac{x^2 - x - 2}{x - 3}[/tex]

x-intercepts

[tex]\begin{aligned}y=0 \implies \dfrac{x^2 - x - 2}{x - 3}&=0\\x^2 - x - 2&=0\\x^2-2x+x-2&=0\\x(x-2)+1(x-2)&=0\\(x+1)(x-2)&=0\\\\x+1&=0 \implies x=-1\\x-2&=0 \implies x=2\end{aligned}[/tex]

y-intercept

[tex]x=0 \implies \dfrac{(0)^2 - 0 - 2}{0 - 3}=\dfrac{2}{3}[/tex]

Vertical asymptotes

Set the denominator to zero and solve for x:

[tex]\implies x-3=0[/tex]

[tex]\implies x=3[/tex]

Holes

Factor the numerator:

[tex]k(x)= \dfrac{(x+1)(x-2)}{x - 3}[/tex]

There are no holes as there is no common factor in the numerator and the denominator.

End behaviour near the vertical asymptote

f(x) → -∞, as x → 3⁻

f(x) → +∞, as x → 3⁺

Domain

As there is vertical asymptote at x = 3, the domain is rest:

Range

The extreme points are (1, 1) and (5, 9).  Therefore, the range is restricted:

Given rational function:

[tex]f(x)= -\dfrac{x}{x^3 - 4x}[/tex]

x-intercepts

As the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at y = 0.  Therefore, there are no x-intercepts.

y-intercept

Substitute x = 0 and solve for y:

[tex]x=0 \implies -\dfrac{0}{0(x+2)(x-2)}=\dfrac{0}{0}=\textsf{unde\:\!fined}[/tex]

Therefore, there is no y-intercept.

Factor the denominator:

[tex]f(x)= -\dfrac{x}{x(x+2)(x-2)}[/tex]

Cancel the common factor x:

[tex]f(x)= -\dfrac{1}{(x+2)(x-2)}[/tex]

Vertical asymptotes

Set the denominator to zero and solve for x:

[tex]\implies (x+2)(x-2)=0[/tex]

[tex]\implies x+2=0 \implies x=-2[/tex]

[tex]\implies x-2=0 \implies x=2[/tex]

Holes

Factor the denominator:

[tex]f(x)= -\dfrac{x}{x(x+2)(x-2)}[/tex]

As the numerator and denominator have a common factor of x, the function is not defined when x = 0. Therefore, there is a hole when x = 0.

To find the y-value of the hole, cancel the common factor x and input x = 0 into the resulting function:

[tex]x=0 \implies -\dfrac{1}{(0+2)(0-2)}=\dfrac{1}{4}[/tex]

Therefore, there is a hole at (0, ¹/₄).

End behaviour near the vertical asymptotes

f(x) → -∞, as x → -2⁻

f(x) → +∞, as x → -2⁺

f(x) → +∞, as x → 2⁻

f(x) → -∞, as x → 2⁺

Domain

As there are vertical asymptotes at x = -2 and x = 2, and a hole at (0, ¹/₄), the domain is restricted:

Range

As there is a hole at (0, ¹/₄) and a horizontal asymptote at y = 0, the range is restricted:

EASY just say THANK YOU!!!

Answer:

1. 3 ones

2. 3

3. 3 tens

4. 3

5. 3 hundreds

6. 3

7. 3 halves

8. 3 (not 3/2)

9. 3 fourths

10. 3

11. 3

12. 2

13. 1/2

14. 3

15. 5/2

16. 3/4

17. 3

18. 3/2

19. 3/4

20. 7/2

21. 2/3

22. 4/11

NOTE: Read the step-by-step explanation on how we arrive on these final answers. Don't just copy-paste it and try to understand the concept so that next time you already know how to solve this type of expressions.

Step-by-step explanation:

This is a simple mathematic problem which requires basic strategies to be solved. People should already know on how to calculate these simple mathematic tasks. Most especially to the first 7 items, we will just use division to get the final answer (numerator/denominator).

[tex]\frac{9}{3}[/tex] = 3 or

9 ÷ 3 = 3

Similar, with the rest of items that includes fractions. We will just reciprocate the denominator and multiply it to the numerator to get the final answer.

Example was the item 8:

= [tex]\frac{9}{2}[/tex] ÷ [tex]\frac{3}{2}[/tex]

= [tex]\frac{9}{2}[/tex] x [tex]\frac{2}{3}[/tex]

= [tex]\frac{18}{6}[/tex]

= 3

Answer:

  0.70

Step-by-step explanation:

You want p(A+B) given that p(A) = 0.60, p(B) = 0.25, and p(A·B) = 0.15.

The relevant formula is ...

  p(A+B) = p(A) +p(B) -p(A·B)

  p(A+B) = 0.60 +0.25 -0.15

  p(A+B) = 0.70

__

Additional comment

In the attached Venn diagram, the sum of the numbers inside the circle labeled A is p(A) = 0.45 +0.15 = 0.60. Likewise, the sum of the numbers inside circle B is p(B) = 0.15 +0.10 = 0.25.

As the attachment shows, the event A includes the event 'A and B'. Likewise, event B includes the event 'A and B'. Simply adding the probabilities of A and B will cause p(A and B) to be counted twice.

The change in the function per unit over a specific period of time is the average rate of change. Between x is 5 and x is 7, the average rate of change is -4.

The y = f graph

The points on f(x) that connect x = 5 to x = 7 are in the attachment. A green line serves as the line segment's representation.

The following values can be used to calculate the line segment's average rate of change.

If x = 5, then y = 8.

Y Equals 0 when x = 7

Consequently, the typical rate of change (m) is

(Delta*y/(Delta*x)) = m

Where:

Delta*y equals y 2 - y 1.

Sigma*y = 0 to 8

Delta*y equals -8

Delta*x equals x 2 - x 1.

x = 7 - 5 for delta.

x*Delta = 2

We therefore have:

(Delta*y/(Delta*x)) = m

m = - 8/2 m = - 4

Hence, the average rate of change from x = 5 to 7 is -4

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The value of x which holds true for the given system of equations as required is; x = 4 / 5.

It follows from the task content that the value of x which holds true for the given system of equations be determined.

Since the given system is;

2x - 3y = 7........(i)

-4x + y = -5........(ii)

The value of x which holds true for the given system of equations can be determined by making y the subject of the formula in (ii) and substitution into (i) as follows;

y = 4x - 5.

Therefore;

2x - 3 (4x - 5) = 7

2x - 12x + 15 = 7

-10x = -8

x = 4/5

Ultimately, the value of x which holds true for the given system of equations is; x = 4 / 5.

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2 x – 3 y = 7 –4 x + y = –5 Wh

Answer:

The distance between -19 and 12 is 31.

Step-by-step explanation:

a) The total interest that Charlie, who borrows $700 at a simple interest rate of 4% per annum, will pay after 4 years is $112.

b) The total repayment amount after 4 years, including interest, is $812.

The simple interest system is one that does not add interest to the principal before calculating subsequent years' interest.

The simple interest system is different from compound interest.

The simple interest formula is given as Principal x Time x Rate.

Principal loan = $700

Loan period = 4 years

Annual interest rate = 4%

Interest system = Simple Interest

Simple interest after 4 years = $112 ($700 x 4% x 4)

Amount to be repaid = Principal + Interest

= $812 ($700 + $112)

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Answer:i think it is yes! this )--89+98

Step-by-step explanation:

The equation of the word sentence, "5 is one-fourth of a number c" is: 5 = 1/4c.

To write a word sentence as an equation, you need to identify the variables and the operations being described in the sentence, and then use mathematical symbols to represent them.

For example, we are given the sentence "5 is one-fourth of a number c,", in this case, the variable is "c" and the operation is "one-fourth of," which is represented by the fraction 1/4 or the multiplication by 1/4.

Therefore, the equation would be:

5 = 1/4c.

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The average of given set of integers is -17.

In maths, the average value in a set of numbers is the middle value, calculated by dividing the total of all the values by the number of values. When we need to find the average of a set of data, we add up all the values and then divide this total by the number of values.

The given set of integers are -20, -54, -93, 37 and 45.

Here, average = Sum of all the observations/Number of observations

= (-20-54-93+37+45)/5

= -85/5

= -17

Therefore, the average of given set of integers is -17.

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The probability that the first MP3 song heard is the fifth song played is:

[tex]\frac{C^9^0_4.C^1^0_1}{C^1^0^0_5}[/tex] =  [tex]\frac{2,555,190. 10}{75,287,520}[/tex] ≈ 0.3394

The probability that there are exactly two MP3 songs in the first five songs played is:

[tex]\frac{C^1^0_2.C^9^0_3}{C^1^0^0_5}[/tex] [tex]= \frac{45.117,480}{75,287,520}[/tex] ≈ 0.0702

Now, According to the question:

The 1st MP3 player songs heard is the 5th song played means that first 4 songs are not MP3 songs (there are 100 - 10 = 90 such songs) and the 5th song is MP3 song (there are 10 such songs). Hence, the probability that the first MP3 song heard is the fifth song played is:

[tex]\frac{C^9^0_4.C^1^0_1}{C^1^0^0_5}[/tex] =  [tex]\frac{2,555,190. 10}{75,287,520}[/tex] ≈ 0.3394

There are exactly two MP3 songs in the first five songs played means there are 2 MP3 songs and 3 not MP3 songs (the number when these songs are played do not matter). Hence, the probability that there are exactly two MP3 songs in the first five songs played is:

[tex]\frac{C^1^0_2.C^9^0_3}{C^1^0^0_5}[/tex] [tex]= \frac{45.117,480}{75,287,520}[/tex] ≈ 0.0702

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