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Saturday, July 1, 2023

DSAT Math For Scoring 750-800 (Part 1)

1. What is the length of major axis of $\frac{{{x^2}}}{5} + \frac{{{y^2}}}{6} = 24$

$\begin{array}{l} A.{\rm{ }}\sqrt {30} \\ B.{\rm{ 2}}\sqrt {30} \\ C.{\rm{ 4}}\sqrt {30} \\ D.{\rm{ 24}}\\ E.{\rm{ 30}} \end{array}$

Solution

$\frac{{{x^2}}}{5} + \frac{{{y^2}}}{6} = 24 \Rightarrow \frac{{{x^2}}}{{120}} + \frac{{{y^2}}}{{144}} = 1$

$\Rightarrow \left\{ \begin{array}{l} a = \sqrt {120} \\ b = \sqrt {144} \end{array} \right. \Rightarrow 2a = 2\sqrt {120} = 4\sqrt {30}$

2. Bob averaged 40 mph going to work and 60 mph going home. What was his average speed for the trip ?

$\begin{array}{l} A.{\rm{ }}44{\rm{ }}mph\\ B.{\rm{ }}46{\rm{ }}mph\\ C.{\rm{ }}48{\rm{ }}mph\\ D.{\rm{ }}50{\rm{ }}mph\\ E.{\rm{ 52 }}mph \end{array}$

Solution

$\frac{{40 + 60}}{2} = 50{\rm{ }}mph$

3. In ∆ABC, AB=2, BC=3 and AC=4. What is the measure of the largest angle to the nearest degree ?

$\begin{array}{l} A.{\rm{ 10}}{{\rm{4}}^ \circ }\\ B.{\rm{ 10}}{{\rm{8}}^ \circ }\\ C.{\rm{ 11}}{{\rm{2}}^ \circ }\\ D.{\rm{ 11}}{{\rm{6}}^ \circ }\\ E.{\rm{ 12}}{{\rm{0}}^ \circ } \end{array}$

Solution

$\begin{array}{l} {b^2} = {a^2} + {c^2} - 2ac\cos B \Rightarrow \cos B = \frac{{{a^2} + {c^2} - {b^2}}}{{2ac}} = - \frac{1}{4}\\ \Rightarrow B = {104^ \circ }29' \end{array}$

4. If sinx=a, $0 \le x \le {30^ \circ }$ , what is sin2x in terms of a ?

$\begin{array}{l} A.{\rm{ }}a\sqrt {1 + {a^2}} \\ B.{\rm{ }}a\sqrt {1 - {a^2}} \\ C.{\rm{ 2}}a\sqrt {1 + {a^2}} \\ D.{\rm{ 2}}a\sqrt {1 - {a^2}} \\ E.{\rm{ 2}}a\sqrt {4 - {a^2}} \end{array}$

Solution

$\begin{array}{l} \cos x = \sqrt {1 - {{\sin }^2}x} = \sqrt {1 - {a^2}} \\ \Rightarrow \sin 2x = 2\sin x\cos x = 2a\sqrt {1 - {a^2}} \end{array}$

6. What are the y intercepts of $3{x^4}{y^2} + {y^2} + 2y = 4xy + 15$

$A.{\rm{ (0,5) and (0,}} - {\rm{3)}}$

$B.{\rm{ (0,}} - {\rm{5) and (0,3)}}$

$C.{\rm{ (0,5) and (0,3)}}$

$D.{\rm{ (0,}} - {\rm{3) and (0,}} - {\rm{5)}}$

$E.{\rm{ (0,3) and (0,}} - {\rm{5)}}$

Solution

$x = 0 \Rightarrow {y^2} + 2y - 15 = 0 \Rightarrow y = - 5{\rm{ }}and{\rm{ }}3$

7. The half-life of titanium-44 is 63 years. After how many years will 1 gram be left of 70 grams titanimum-44 to the nearest year ?

$\begin{array}{l} A.{\rm{ 368}}\\ B.{\rm{ 386}}\\ C.{\rm{ 638}}\\ D.{\rm{ 683}}\\ E.{\rm{ 836}} \end{array}$

Solution

8. Bob invested $100,000 and 20 years later it was $500,000. What was the average percent increase per year ?

$\begin{array}{l} A.{\rm{ 7}}{\rm{.5\% }}\\ B.{\rm{ 7}}{\rm{.7\% }}\\ C.{\rm{ 7}}{\rm{.9\% }}\\ D.{\rm{ 8}}{\rm{.1\% }}\\ E.{\rm{ 8}}{\rm{.3\% }} \end{array}$

Solution

9. ${x^4} - 5{x^2} - 36 = 0$ . What are the solutions over complex number ?

$\begin{array}{l} A.{\rm{ 3, }} - {\rm{2, 2i, }} - 3i\\ B.{\rm{ 3, }} - {\rm{2, 3i, }} - 2i\\ C.{\rm{ 2, }} - 3{\rm{, 2i, }} - 3i \end{array}$

$\begin{array}{l} D.{\rm{ 3, }} - 3{\rm{, 2i, }} - 2i\\ E.{\rm{ 2, }} - {\rm{2, 3i, }} - 3i \end{array}$

Solution

10. tan θ=3a, $0 \le \theta \le \frac{\pi }{2}$ , what is cscθ ?

$\begin{array}{l} A.{\rm{ }}\frac{{\sqrt {9{a^2} + 1} }}{{3a}}\\ B.{\rm{ }}\frac{{\sqrt {9{a^2} + 2} }}{{2a}}\\ C.{\rm{ }}\frac{{\sqrt {9{a^2} + 3} }}{a} \end{array}$

$\begin{array}{l} D.{\rm{ }}\frac{{\sqrt {9{a^2} - 2} }}{{2a}}\\ E.{\rm{ }}\frac{{\sqrt {9{a^2} - 1} }}{{3a}} \end{array}$

Solution

[Probablity and Statistics] Sampling and Sampling Distribution

Đặng Pete8:54 AMNo comments

I. Sample Distribution of Sample Mean

Prob1. Assume the prices of meals at a restaurant are normally distributed with an average of $35 and a standard deviation of $5. I have $33 in my pocket.

1. What is the probability I have enough money for my meal ?

2. I’m out to dinner with friends and we’re splitting the check evenly by 10.

a. What is the expected price per meal now ?

b. What is the probability I have enough moneny now ?

c. Why did the probability change ?

Solution:

$1.{\rm{ }}P(X \le 33) = P(Z \le \frac{{33 - 35}}{5}) = P(Z \le - 0.4) = 0.3446$

$2a.{\rm{ }}E(\overline X ) = \mu = 35\$ ;{\rm{ }}\overline X \sim N(\mu ,\frac{\sigma }{{\sqrt n }})$

$2b.{\rm{ }}P(\overline X \le 33) = P(Z \le \frac{{33 - 35}}{{(5/\sqrt {10} }}) = P(Z \le - 1.26) = 0.1038$

$2c.{\rm{ }}We're{\rm{ }}averaging{\rm{ }}10{\rm{ }}values,{\rm{ }}the{\rm{ }}std{\rm{ }}errror{\rm{ }}is{\rm{ }}less{\rm{ }}and{\rm{ }}values{\rm{ }}convergence{\rm{ }}toward{\rm{ }}\$ 35$

Prob2. You manage a fast-food franchise in LA. Last month the mean waiting time at the drive-through windown for branches in LA was 3.8 minutes and the standard deviation was 0.95 minutes. You can assume the waiting times are normally distributed.

a. Let X denote the waiting time of a random customer. What is the probability that X is between 3 and 5 minutes ?

b. What is the probability the average waiting time of the next 10 customers is greater than 3 minutes ?

c. Suppose your restaurant has 10 customers in line for picking up lunch. What is the total amount of time required so that with probability 90% that all 10 orders can be completed during that period ?

Solution:

$a.{\rm{ }}X \sim N(\mu ,\sigma );{\rm{ }}\mu = 3.8;{\rm{ }}\sigma = 0.95$

$P(3 \le X \le 5) = P(X \le 5) - P(X \le 3) = P(Z \le \frac{{5 - 3.8}}{{0.95}}) - P(Z \le \frac{{3 - 3.8}}{{0.95}}) = ?$

$b.{\rm{ }}\overline X \sim N({\mu _{\overline x }},{\sigma _{\overline x }});$ ${\rm{ }}{\mu _{\overline x }} = \mu = 3.8;{\rm{ }}{\sigma _{\overline x }} = \frac{\sigma }{{\sqrt n }} = \frac{{0.95}}{{\sqrt {10} }}$

${\rm{ }}P(\overline X > 3) = P(Z > \frac{{3 - 3.8}}{{(0.95/\sqrt {10} }}) = P(Z > ?) = 1 - P(Z < ?) = ?$

$c.{\rm{ }}P(\overline X \le t) = 0.9 \Rightarrow P(\overline X \le t) = P(Z \le \frac{{t - 3.8}}{{(0.95/\sqrt {10} )}}) = P(Z \le \frac{{t - 3.8}}{{0.3}})$

$\Rightarrow P(Z \le \frac{{t - 3.8}}{{0.3}}) = 0.9 = P(Z \le 1.28) \Rightarrow \frac{{t - 3.8}}{{0.3}} = 1.28 \Rightarrow t = 4.184$

II. Sample Distribution of Sample Proportion

Prob3. A study found that 55% of British firms experienced a cyber-attack in the past year.

a. What are the expected value and the standard error of the sample proportion derived from a random sample of 100 firms ?

b. What are the expected value and the standard error of the sample proportion derived from a random sample of 200 firms ?

c. Comment on the value of the standard error as the sample size gets larger.

Solution:

$\begin{array}{l} a.{\rm{ }}p = 0.55;n = 100;E(\overline p ) = p = 0.55\\ se(\overline p ) = \sqrt {\frac{{p(1 - p)}}{n}} = \sqrt {\frac{{0.55(1 - 0.55)}}{{100}}} = 0.0497 \end{array}$

$\begin{array}{l} b.{\rm{ }}p = 0.55;n = 200;E(\overline p ) = p = 0.55\\ se(\overline p ) = \sqrt {\frac{{p(1 - p)}}{n}} = \sqrt {\frac{{0.55(1 - 0.55)}}{{200}}} = 0.0352 \end{array}$

$c.{\rm{ As the sample size gets larger, the standard error gets smaller}}{\rm{.}}$

[Geneneral Math 3-4] Chapter 14E: Using crashing to reduce the completion time on of a project

Đặng Pete12:25 PMNo comments

Hôm nay Gia Sư Toán Cao Cấp và Xác Suất Thống Kê sẽ giới thiệu đến các bạn dạng bài tập Crashing trong môn General Math 3-4

1. Altering completion times

- The minimum time it takes to complete a project depends upon the times it takes to complete the individual activites of the project, and upon the predecessors each of the activities have.

- Critical path analysis can be completed to find the overall minimum completion time.

- Sometime, The managers of a project might arrange for one or more activities within the project to be completed in a shorter time than originally planned. Changing the conditions of an acitivites within a project, and recalculating the minimum completion time for the project, is called crashing.

2. A simple crashing example.

- A simple activity network is shown in the diagram on the below. The forwards and backwards scanning processes have been completed and the critical path is shown in red on the diagram.

- The minimum time for completion is currently 13 hours. In order to reduce this overall time, The manager of the project should try to complete one, or more, of the activities in a shorter time than normal. Reducing the time taken to complete activity A,B or C would not achieve this goal however. These activites are not on the critical path and so they already slack time. Reducing their completion time will not shorten the overall time taken to complete the project.

- Activity D and E, on the other hand, lie on the critical path. Reducing the duration of these activities will reduce the overall time for the project. If activities D was reduced in time to 4 hours instead, the project will be completed in 11, not 13, hours.

3. Crashing with Cost

Example 1:

Solution:

Example 2:

Solution:

Example 3:

Solution:

Example 4:

Solution:

Example 5:

Solution:

Exercise 1: The activity network for a project is shown in the diagram below. The duration for each activity is in hours.

a. List all four paths from the Start to the Finish of the project, With their respective completion times.

b. Identify the critical path and the minimum completion time for the project.

c. If Activity E is reduced by 3 hours, identify the new minimum completion time for this project.

Solution:

1a. A-D

B-E-F

B-E-G-I

C-H-I

1b. B-E-G-I: 21 hours

1c. 18

Exercise 2: The directed network below shows the sequence of 8 activites that are needed to complete a project. The time, in days, that is takes to complete each activity is also shown.
a. Write down the critical path for this project.
b. What is the minimum completion time for the project ? Activity B can be reduced by a maximum of 3 days at a cost of $100 per day.
c. What is the new minimum completion time for the project ?
d. What is the minimum cost that will achieve the greatest reduction in time taken to complete the project ?
Solution:
2a. A-B-F-G
2b. 21 days.
2c. 20 days
2d. $100
Exercise 3: The activity network for a project is shown in the diagram on the below. The duration for each activity is in hours.
a. Identify the critical path for this project.
b. What is the maximum number of hours that the completion time for activity E can be reduced by without changing the minimum completion time of the project ?
c. What is the maximum number of hours that the completion time for activity H can be reduced without changing the minimum completion time of the project ?
d. Every activity can be reduced in duration by a maximum of 2 hours. If every activity was reduced by the maxmimum amount possible, what is the new minimum completion time for the project ?
Solution:
3a. B-E-H-J
3b. 2 hours
3c. 6 hours
3d. 14 hours
Exercise 4: The activity network for a project is shown in the diagram below. The duration for each activity is in hours.
a. How many activities could be delayed by 4 hours without altering the minimum completion for the project ?
b. If the project is to be crashed by reducing the completion time of one activity only, what is the minimum time, in hours, that the project can be completed in ?
c. Activity G can be reduced in time at a cost of $200 per hour. Activity J can be reduced in time at a cost of $150 per hour, What is the cost of reducing the completion time of this project as much as possible ?
Solution:

4a. 4
4b. 17 hours
4c. $1200

Kết Quả Học Viên

Đặng Pete11:38 AMNo comments

1. Học Viên Nguyễn Sơn Hoàn - Đại Học Kinh Tế: Khóa Toán Cao Cấp học cấp tốc, từ không biết gì đạt được mục tiêu qua môn để ra trường.

2. Học Viên Phạm Duy Khanh - Đại Học Sư Phạm Kỹ Thuật: Khóa Giải Tích 3 cấp tốc, từ không biết gì đạt được mục tiêu qua môn để ra trường.

3. Học Viên Đức Hiền: Khóa Đại Số Tuyến Tính.

4. Học Viên Minh Trọng: Khóa Đại Số Tuyến Tính.

Confidence Interval for Difference in Means

Đặng Pete10:23 AMNo comments

Hôm nay Gia Sư Toán Cao Cấp và Xác Suất Thống Kê sẽ giới thiệu đến các bạn dạng bài tập tìm khoảng tin cậy cho hiệu hai trung bình cũng như là Code Python.

1. Determine the sample means: Calculate the means of the two independent groups or samples.

2. Calculate the standard deviation(s): Calculate the standard deviation(s) for each group or sample. This measure of variability helps determine the precision of the difference in means estimate.

3. Determine the sample sizes: Identify the number of observations or participants in each group or sample. The sample size affects the standard error and the width of the confidence interval.

4. Choose the confidence level: Determine the desired level of confidence, typically expressed as a percentage (e.g., 95% confidence level).

5. Calculate the standard error: Use the formula for the standard error of the difference in means. This involves dividing the standard deviation of each group by the square root of their respective sample sizes.

6. Find the critical value: Identify the critical value (often denoted as z) based on the chosen confidence level. This value is obtained from a standard normal distribution or a t-distribution.

7. Compute the margin of error: Multiply the critical value by the standard error to determine the margin of error. The error represents the extent to which the confidence interval is likely to vary from the true difference in means.

8. Calculate the lower and upper limits: Subtract the margin of error the to obtain the lower of the confidence interval. Add the margin of error to the sample mean difference to obtain the upper limit.

9. Interpret the confidence interval: Interpret the confidence interval by stating that you are (e.g., 95% confident) that the true difference in means lies within the calculated interval. Discuss whether the interval includes zero, as this indicates statistical significance or non-significance.

10. Draw conclusions: Based on the confidence interval, make conclusions about the difference in means. If the interval includes zero, there is no significant difference between the two groups, while excluding zero indicates a significant difference.

11. Communicate the results: Effectively communicate the confidence interval results and conclusions using appropriate language and visual representations, such as tables, figures, or graphs.

[Calculus - Series and Sequences] Integral Test For Convergence

Đặng Pete3:42 PMNo comments

Solution:

Confidence Interval for Mean

Đặng Pete11:40 AMNo comments

Hôm nay Gia Sư Toán Cao Cấp và Xác Suất Thống Kê sẽ giới thiệu đến các bạn dạng bài tập tìm khoảng tin cậy cho trung bình cũng như là Code Python.

1. Find 90% percent confidence interval for the mean of a normal distribution with xichma=3 given sample (3.3, -0.3, -0.6, -0.9) . What would be the confidence interval if xichma were unknown ?

Solution

a. Code Python:

b. Code Python:

import numpy as np

import scipy.stats as st

# define sample data

gfg_data = [3.3, -0.3, -0.6, -0.9]

# create 90% confidence interval

st.t.interval(alpha=0.90, df=len(gfg_data)-1,

loc=np.mean(gfg_data),

scale=st.sem(gfg_data))

Result: (-1.937561291069545, 2.6875612910695437)

2. The breaking strength in pounds of five speciments of manila rope of diameter 3/16 inch were found to be 660, 460, 540, 580, 550

a. Estimate the mean breaking strength by a 95 percent confidence interval assuming normality ?

b. Estimate the point at which only 5 percent of such speciments would be expected to break.

c. Estimate xichma^2 by a 90 percent confidence interval; also xich

d. Plot an 81 percent confidence region for the joint estimation of nuy and xich^2; for nuy and xich

Nhận dạy: Xác Suất Thống kê, Zalo: 0979 027 945 or Inbox Facebook

Identifying Float Times and The Critical Path (General Math - Year 12)

Đặng Pete10:56 AMNo comments

1. Float Times

2. Critical Path

EX1:

Solution

Find EST:

0-Vertex: 0

1-Vertex: 4

2-Vertex: 3

Update 0-Vertex: 12 so finallisation is 12

3-Vertex: 23

4- Vertex: 15

5-Vertex: 32

6-Vertex: 42 and 24, so finallisation is 42.

Find LST

6-Vertex: 42

5-Vertex: 32

4-Vertex: (33,27) chose minimum

3-Vertex: 23

1-Vertex: 12

2-Vertex: (15,13,3)

EX2:

Solution:

EX3:

Solution:

Nhận Dạy Calculus AB-BC

Kết Quả Học Viên

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Saturday, July 1, 2023

DSAT Math For Scoring 750-800 (Part 1)

Thursday, June 29, 2023

[Probablity and Statistics] Sampling and Sampling Distribution

Wednesday, June 28, 2023

[Geneneral Math 3-4] Chapter 14E: Using crashing to reduce the completion time on of a project

Kết Quả Học Viên

Confidence Interval for Difference in Means

Tuesday, June 27, 2023

[Calculus - Series and Sequences] Integral Test For Convergence

Monday, June 26, 2023

Confidence Interval for Mean

Saturday, June 24, 2023

Identifying Float Times and The Critical Path (General Math - Year 12)

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