Let (Sn)nzo be a simple random walk starting in 0 (i.e. So = 0) with p = 0.3 and q = 1-p = 0.7. Compute the following probabilities: (i) P(S₂ = 2, S5 = 1), (ii) P(S₂ = 2, S4 = 3, S5 = 1), (iii) P(

A graph that represents the following ellipse (x + 1)²/16 + (y - 3)²/36 = 1 is: C. graph C.

Mathematically, the standard form of the equation of an ellipse is given by mathematical expression:

x²/a² + y²/b² = 1  

Where;

This ultimately implies that, the major axis of this ellipse is the x-axis. By critically observing the graph (see attachment), we can logically deduce the following values:

a = √16 = 4.

b = √36 = 6.

In conclusion, the standard form of the equation of this ellipse is given by (x + 1)²/16 + (y - 3)²/36 = 1.

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Answer:

they are right!!!!!!!

Answer:

figure????

X

610

x=

1

Answer:

128 <33

Step-by-step explanation:

6 ÷ 3 + 32 • 4 - 2 = 128

▪︎▪︎▪︎▪︎▪︎▪︎

Answer:

2

Step-by-step explanation:

(4-5)-(13-18+2)

=-1-(13-18+2)

=-1-(-5+2)

=-1(-3)

=2

Given:

The equation is

\(4(x-3)+5=17\)

To find:

The solution of given equation.

Solution:

We have,

\(4(x-3)+5=17\)

Using distributive property, we get

\(4(x)+4(-3)+5=17\)

\(4x-12+5=17\)

\(4x-7=17\)

Adding 7 on both sides, we get

\(4x-7+7=17+7\)

\(4x=24\)

Divide both sides by 4.

\(x=\dfrac{24}{4}\)

\(x=6\)

Therefore, the value of x is 6.

The statement that describes when the plans are based on the same number of aerobic exercise sessions is B. Each plan utilizes a combination of 2 strength-training sessions and 3 aerobic exercise sessions per week.

It was illustrated that the beginner plan has 15 minutes per session of strength-training and 20 minutes per session of aerobic exercise for a total of 90 minutes of exercise in a week.

Since the combination is 2 strength-training sessions and 3 aerobic exercise sessions. This will be:

= 15(2) + 20(3)

= 30 + 60

= 90

This confirms with the number.

Also, advanced plan has 20 minutes per session of strength-training and 30 minutes of aerobic exercise for a total of 130 minutes of exercise in a week. This will be:

= 20(2) + 30(3)

= 40 + 90

= 130

This Confirms to the value above.

The correct option is B.

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Answer:

The slope would be 4/3 or 1 1/3.

Step-by-step explanation:

When you put the coordinates in the point-slope formula, you will get -5 - (-13) over 7 - 1. The answer becomes 8/6 but if simplified, it becomes 4/3 or 1 1/3 as a mixed fraction.

The system equation that models the amount of calories from fats (f) and proteins (p) that the individual can consume to reach 1,800 calories is: f + p = 1,200. The equation that relates calories from protein (p) to calories from fat (f) based on the prescribed diet is: p = 3f. Solving the system of equations, we find that the individual should consume 300 calories from fats and 900 calories from proteins.

To find the equation that models the amount of calories from fats and proteins that the individual can consume in order to reach 1,800 calories, we consider that 600 calories will come from carbohydrates. Since the total goal is 1,800 calories, the remaining calories from fats and proteins should add up to 1,800 - 600 = 1,200 calories. Therefore, the equation is f + p = 1,200.

Based on the prescribed diet, the individual is required to consume calories from protein that are three times the calories from fat. This relationship can be expressed as p = 3f, where p represents the calories from protein and f represents the calories from fat.

To solve the system of equations, we substitute the value of p from the second equation into the first equation: f + 3f = 1,200. Combining like terms, we get 4f = 1,200, and dividing both sides by 4 yields f = 300. Substituting this value back into the second equation, we find p = 3(300) = 900.

Therefore, the individual should consume 300 calories from fats and 900 calories from proteins to meet the diet requirements and achieve a total of 1,800 calories.

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Using the formula, the confidence interval is: [(4)(1.3^2) / χ^2_(0.05,4), (4)(1.3^2) / χ^2_(0.95,4)]

To form a confidence interval for the variance σ^2, we can use the chi-square distribution. The formula for the confidence interval is:

[(n-1)s^2 / χ^2_(α/2,n-1), (n-1)s^2 / χ^2_(1-α/2,n-1)]

Where:

n is the sample size

s^2 is the sample variance

χ^2_(α/2,n-1) is the chi-square value for the upper α/2 percentile

χ^2_(1-α/2,n-1) is the chi-square value for the lower 1-α/2 percentile

We are given four different sets of values for x, s, and n. Let's calculate the confidence intervals for each case:

a. x = 16, s = 2.6, n = 60:

Using the formula, the confidence interval is:

[(59)(2.6^2) / χ^2_(0.05,59), (59)(2.6^2) / χ^2_(0.95,59)]

b. x = 1.4, s = 0.04, n = 17:

Using the formula, the confidence interval is:

[(16)(0.04^2) / χ^2_(0.05,16), (16)(0.04^2) / χ^2_(0.95,16)]

c. x = 160, s = 30.7, n = 23:

Using the formula, the confidence interval is:

[(22)(30.7^2) / χ^2_(0.05,22), (22)(30.7^2) / χ^2_(0.95,22)]

d. x = 8.5, s = 1.3, n = 5:

Using the formula, the confidence interval is:

[(4)(1.3^2) / χ^2_(0.05,4), (4)(1.3^2) / χ^2_(0.95,4)]

To obtain the actual confidence intervals, we need to look up the chi-square values for the given significance level α and degrees of freedom (n-1) in a chi-square distribution table.

Once we have the chi-square values, we can plug them into the confidence interval formula to calculate the lower and upper bounds of the confidence interval for each case.

Note: Since the question provides specific values for x, s, and n, the calculations for the confidence intervals cannot be completed without the corresponding chi-square values for the given significance level.

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Answer:

189.271 liters

When you convert 50 gallons to liters, you will get 189.271. Hope this helps(:

Answer:

Equation 1 x=-16

Equation 2 m=-3

Step-by-step explanation:

Equation 1

3x-x=-24-6

2x=-32

x=-32/2

x=-16

Equation 2

-2m=16-10

-2m=6

m=6/-2

m=-3

The probability that Melinda will be able to make an integer between 10 and 20 (inclusive) is 1/6.

The probability that Melinda will be able to make an integer between 10 and 20 (inclusive) with the two numbers she rolls can be calculated by determining the number of favorable outcomes and dividing it by the total number of possible outcomes.

To find the number of favorable outcomes, we need to identify the combinations of two numbers that will result in a two-digit number between 10 and 20. We can list these combinations as follows:

11, 12, 13, 14, 15, 16

Notice that we only have six favorable outcomes.

Now, let's determine the total number of possible outcomes when rolling two six-sided dice. Each die has six possible outcomes (1, 2, 3, 4, 5, 6), so the total number of outcomes is 6 multiplied by 6, which equals 36.

To calculate the probability, we divide the number of favorable outcomes (6) by the total number of possible outcomes (36):

Probability = Favorable outcomes / Total outcomes
Probability = 6 / 36
Probability = 1 / 6

Therefore, the probability that Melinda will be able to make an integer between 10 and 20 (inclusive) is 1/6.

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Answer:

Is it 12?Yes,I think it's 12.

(2,9)

this needs 20 characters

The distance is the same that was covered.

Distance is speed multiplied by time.

Mr O takes 30 minutes and Mr Gordon takes 1 hour.

So the speed of Mr O is given by:

Convert the speed in meter per second.

The speed of mr G is:

So the difference between the speeds is:

So Mr O is 2.9167 m/s faster than Mr Gordon.

Answer:

3y

Step-by-step explanation:

All three expressions have the factor 3, and all three have the factor y.  Thus, the GCF is 3y.

Let

x------> the number of multiple choice question

y------> the number of free response question

we know that

-----> equation A

-----> equation B

Substitute equation B in equation A

Find the value of x

therefore

the answer is

the number of multiple choice question are

the number of free response question are

Answer:

m = 17

Step-by-step explanation:

\(4(m - 13) - 8 = 8 \\ 4(m - 13) = 8 + 8 \\ 4(m - 13) = 16 \\ (m - 13) = \frac{16}{4} \\ m - 13 = 4 \\ m = 4 + 13 \\ \huge \red { \boxed{m = 17}}\)

The natural breaks, quantile, and equal interval classification schemes are methods used to categorize data for the purpose of creating thematic maps. Each scheme has its own approach and considerations: Natural Breaks, Quantile, Equal Interval.

Natural Breaks (Jenks): This classification scheme aims to identify natural groupings or breakpoints in the data. It seeks to minimize the variance within each group while maximizing the variance between groups. Natural breaks are determined by analyzing the distribution of the data and identifying points where significant gaps or changes occur. This method is useful for data that exhibits distinct clusters or patterns.

Quantile (Equal Count): The quantile classification scheme divides the data into equal-sized classes based on the number of data values. It ensures that an equal number of observations fall into each class. This approach is beneficial when the goal is to have an equal representation of data points in each category. Quantiles are useful for data that is evenly distributed and when maintaining an equal sample size in each class is important.

Equal Interval: In the equal interval classification scheme, the range of the data is divided into equal intervals, and data values are assigned to the corresponding interval. This method is straightforward and creates classes of equal width. It is useful when the range of values is important to represent accurately. However, it may not account for data distribution or variations in density.

In summary, the natural breaks scheme focuses on identifying natural groupings, the quantile scheme ensures an equal representation of data in each class, and the equal interval scheme creates classes of equal width based on the range of values. The choice of classification scheme depends on the nature of the data and the desired representation in the thematic map.

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Answer:

7

Step-by-step explanation:

5+10-13+5

15-13+5

2+5

Answer:

7

Step-by-step explanation:

Order of Operations: BPEMDAS

B - Brackets

P - Parenthesis

E - Exponents

M - Multiplication

D - Division

A - Addition

S - Subtraction

Step 1: Write expression

5 + 10 - 13 + 5

Step 2: Evaluate

Add: 15 - 13 + 5

Subtract: 2 + 5

Add: 7

Answer:

its d

Step-by-step explanation:

8% of 20 is 1.6 so 20 + 1.6

Answer:

The answer is

Step-by-step explanation:

The midpoint M of two endpoints of a line segment can be found by using the formula

\(M = ( \frac{x1 + x2}{2} , \: \frac{y1 + y2}{2} )\)

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

(6, 7) and (6,-5)

The midpoint is

\(M = ( \frac{6 + 6}{2} \: , \: \frac{7 - 5}{2} ) \\ = ( \frac{12}{2} \: , \: \frac{2}{2} )\)

We have the final answer as

Hope this helps you

If the first bus is boarded at a rate of 15 people per minute, the domain of the function is; (15 ≤ x ≤ 45)

We are told that;

Total number of Juniors at Smithers High School that are taking a class trip = 130

Number of buses taking each student to their destination = 3 buses

Capacity of each bus = 45 people

Now, if the function  f(x) describes the total number of people on the first

bus, where x is the number of minutes spent loading the bus.

Then since the maximum number of people on the first bus is 45, then it means that the domain of the function is;

(15 ≤ x ≤ 45)

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Answer:

C

Detailed explanation down below

Step-by-step explanation:

A bisector divides an angle in half, so we can say that:

7x - 27 = x + 57

7x = x + 84

6x = 84

x= 14

So now lets input this information

7 x 14 - 27 = m∠ LMO = 98 - 27 = 71°

14° + 57° = 71°

(Actually we didnt have to do this second equation because both angles should be equal.)

71° + 71° = 142°

So the answer is C : x=14° .  m∠LMN = 142°

Thus, it is evident that for any value of r, nr = n - 1r - 1n - 1r, where 1 ≤ r ≤ n.

To show that nr = n - 1r - 1n - 1r, let's consider a few cases:

Case 1: When r = 1
n1 = n - 11 - 1n - 11
n = n - 1 - 1n - 1
n = n

Case 2: When r = 2
n2 = n - 12 - 1n - 12
n - 12 = n - 1 - 1n - 1
n - 1 = n - 2

Case 3: When r = n
nn = n - 1n - 1n - 1n
n - 1n = n - 1 - 1n - 1
n - 1 = n - n

To show that nCr = n-1Cr-1 + n-1Cr, we can use the formula for nCr, which is:
nCr = n! / (r! * (n-r)!)
Now, let's plug in the values of n-1 and r-1 into the formula:
n-1Cr-1 = (n-1)! / ((r-1)! * (n-r)!)
n-1Cr = (n-1)! / (r! * (n-r-1)!)
Now, let's add these two equations together:
n-1Cr-1 + n-1Cr = (n-1)! / ((r-1)! * (n-r)!) + (n-1)! / (r! * (n-r-1)!)
= (n-1)! * (r + (n-r)) / (r! * (n-r)!)
= (n-1)! * n / (r! * (n-r)!)
= n! / (r! * (n-r)!)
= nCr
Therefore, nCr = n-1Cr-1 + n-1Cr.
I hope this helps! Let me know if you have any further questions.

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Given:

a.) The width of the rectangle is 0.7 meters less than the length.

b.) The perimeter of a rectangle is 44.6 meters.

First, let's draw the rectangle to better understand the problem.

Recall: The formula in getting the perimeter of the rectangle.

But it stated that the width of the rectangle is 0.7 meters less than the length.

Thus,

Width = Length - 0.7

Let's now find its Length,

Therefore, the length of the rectangle is 11.5 meters.

Let's now determine the width,

Width = Length - 0.7

= 11.5 - 0.7

Width = 10.8 meters

Therefore, the width of the rectangle is 10.8 meters.

The 4th degree Taylor polynomial for \( \ln(x) \) centered at \( x = 1 \) is:

\[ T_4(x) = (x - 1) - \frac{1}{2}(x - 1)^2 + \frac{1}{3}(x - 1)^3 - \frac{1}{4}(x - 1)^4 \]

The 4th degree Taylor polynomial for \( \ln(x) \) centered at \( x = 1 \) can be found by taking the derivatives of \( \ln(x) \) and evaluating them at \( x = 1 \). The general form of the Taylor polynomial is:

\[ T_n(x) = f(c) + f'(c)(x - c) + \frac{f''(c)}{2!}(x - c)^2 + \frac{f'''(c)}{3!}(x - c)^3 + \frac{f^{(4)}(c)}{4!}(x - c)^4 \]

where \( c \) is the center of the approximation. In this case, \( c = 1 \). Let's go through the steps to find the 4th degree Taylor polynomial:

Step 1: Find the derivatives of \( \ln(x) \).

\[ f(x) = \ln(x) \]

\[ f'(x) = \frac{1}{x} \]

\[ f''(x) = -\frac{1}{x^2} \]

\[ f'''(x) = \frac{2}{x^3} \]

\[ f^{(4)}(x) = -\frac{6}{x^4} \]

Step 2: Evaluate the derivatives at \( x = 1 \).

\[ f(1) = \ln(1) = 0 \]

\[ f'(1) = \frac{1}{1} = 1 \]

\[ f''(1) = -\frac{1}{1^2} = -1 \]

\[ f'''(1) = \frac{2}{1^3} = 2 \]

\[ f^{(4)}(1) = -\frac{6}{1^4} = -6 \]

Step 3: Plug the values into the general form of the Taylor polynomial.

\[ T_4(x) = 0 + 1(x - 1) + \frac{-1}{2!}(x - 1)^2 + \frac{2}{3!}(x - 1)^3 + \frac{-6}{4!}(x - 1)^4 \]

Simplifying, we get:

\[ T_4(x) = (x - 1) - \frac{1}{2}(x - 1)^2 + \frac{1}{3}(x - 1)^3 - \frac{1}{4}(x - 1)^4 \]

Therefore, the 4th degree Taylor polynomial for \( \ln(x) \) centered at \( x = 1 \) is:

\[ T_4(x) = (x - 1) - \frac{1}{2}(x - 1)^2 + \frac{1}{3}(x - 1)^3 - \frac{1}{4}(x - 1)^4 \]

This polynomial provides an approximation of \( \ln(x) \) near \( x = 1 \) up to the 4th degree term.

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The deterministic finite automaton (DFA) M3 for the language L3, where strings have more than 3 'c' symbols and do not end in 'c', is designed with states q0, q1, q2, q3, and q4. Transitions and loops are created to determine acceptance.



To build a deterministic finite automaton (DFA) for the language L3 = {x ∈ {a, b, c}* | x has strictly more than 3 'c' symbols and does not end in 'c'}, we can design the following DFA M3:1. Start with an initial state q0.

2. Create a loop on q0 for 'a', 'b', and 'c' inputs, leading back to q0.

3. From q0, transition to a state q1 on input 'c'. This represents the first 'c' encountered.

4. From q1, create a loop for 'a' and 'b' inputs, leading back to q1.

5. Transition from q1 to q2 on input 'c'. This represents the second 'c' encountered.

6. Similarly, create a loop on q2 for 'a' and 'b' inputs, leading back to q2.

7. Transition from q2 to q3 on input 'c'. This represents the third 'c' encountered.

8. From q3, create a loop for 'a', 'b', and 'c' inputs, leading back to q3.

9. Create a final accepting state q4 and transition from q3 to q4 on 'a' or 'b' inputs.

In this DFA, any string that ends with 'c' or has less than three 'c' symbols will not reach the accepting state q4, hence not belonging to L3.

Therefore, The deterministic finite automaton (DFA) M3 for the language L3, where strings have more than 3 'c' symbols and do not end in 'c', is designed with states q0, q1, q2, q3, and q4. Transitions and loops are created to determine acceptance.

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