Answer:
(A) 0.11
(B) 0.0526
(C) Related
(D) 0.28
Step-by-step explanation:
The data provided is:
DC = event that a randomly selected driver is using a cell phone
TA = event that a randomly selected driver has a traffic accident
(A)
From the provided data:
P (DC) = 0.11
(B)
From the provided data:
P (TA) = 0.0526
(C)
To determine whether the events DC and TA are dependent, we need to show that:
[tex]P(DC\cap TA)\neq P(DC)\times P(TA)[/tex]
The value of P (DC ∩ TA) is,
[tex]P(DC\cap TA)=P(DC|TA)\time P(TA)[/tex]
[tex]=0.28\times 0.0526\\=0.014728[/tex]
Now compute the value of P (DC) × P (TA) as follows:
[tex]P (DC) \times P (TA)=0.11\times 0.0526=0.005786[/tex]
So, [tex]P(DC\cap TA)\neq P(DC)\times P(TA)[/tex]
Thus, cell phone use while driving and traffic accidents are related.
(D)
The probability that the driver was distracted by a cell phone given that the driver has an accident is:
P (DC | TA) = 0.28
Use the two highlighted points to find the
equation of a trend line in slope-intercept
form.
Answer: y=(4/3)x+2/3
Step-by-step explanation:
Slope-intercept form is expressed as y=mx+b
First, find the slope (m):
m= rise/run or vertical/horizontal or y/x (found between the highlighted points)
m = 4/3
Second, find b:
Use one of the highlighted points for (x, y)
2=4/3(1)+b
6/3=4/3+b
2/3=b
b=2/3
Plug it into the equation:
You get y=(4/3)x+2/3 :)
Simply. Who ever answers this will be marked Brainlist.
Answer:
Step-by-step explanation:
Hello,
[tex]r^3s^{-2}\cdot 8r^{-3}s^4\cdot 4rs^5\\\\=r^{3-3+1}s^{-2+4+5}\cdot 8\cdot 4\\\\\boxed{=32\cdot r\cdot s^7}[/tex]
Hope this helps.
Do not hesitate if you need further explanation.
Thank you
one third multiplied by the sum of a and b
Answer:
1/3(a+b)
hope it helps :>
find the perimeter of a square of sides 10.5cm
Answer:
Perimeter = 42 cm
Step-by-step explanation:
A square has all equal sides so you would just add 10.5 + 10.5 + 10.5 + 10.5 to get 42 cm.
Answer:
42 cm
Step-by-step explanation:
Side of square = 10.5 cm (given)
Perimeter of square = Side X 4
= 10.5 X 4
= 42 cm
HOPE THIS HELPED YOU !
:)
Findℒ{f(t)}by first using a trigonometric identity. (Write your answer as a function of s.)f(t) = 12 cost −π6
Answer:
[tex]L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ][/tex]
Step-by-step explanation:
Given that:
[tex]f(t) = 12 cos (t- \dfrac{\pi}{6})[/tex]
recall that:
cos (A-B) = cos AcosB + sin A sin B
∴
[tex]f(t) = 12 [cos\ t \ cos \dfrac{\pi}{6}+ sin \ t \ sin \dfrac{\pi}{6}][/tex]
[tex]f(t) = 12 [cos \ t \ \dfrac{3}{2}+ sin \ t \ sin \dfrac{1}{2}][/tex]
[tex]f(t) = 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t)[/tex]
[tex]L(f(t)) = L ( 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t) ][/tex]
[tex]L(f(t)) = 6 \sqrt{3} \ L [cos \ (t) ] + 6\ L [ sin \ (t) ][/tex]
[tex]L(f(t)) = 6 \sqrt{3} \dfrac{S}{S^2 + 1^2}+ 6 \dfrac{1}{S^2 +1^2}[/tex]
[tex]L(f(t)) = \dfrac{6 \sqrt{3} +6 }{S^2+1}[/tex]
[tex]L(f(t)) = \dfrac{6( \sqrt{3} \ S +1 }{S^2+1}[/tex]
[tex]L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ][/tex]
Question 1: A triangle has sides with lengths 5, 6, and 7. Is the triangle right, acute, or obtuse?
A)Right
B)Obtuse
C)Can't be determined
D) Acute
Question 2: A 15-foot statue casts a 20-foot shadow. How tall is a person who casts a 4-foot-long shadow?
A)0.33 feet
B)3.75 feet
C)3 feet
D)5 feet
Question 3: A triangle has sides with lengths 17, 12, and 9. Is the triangle right, acute, or obtuse?
A)Acute
B)Right
C)Can't be determined
D)Obtuse
Question 4: Two friends are standing at opposite corners of a rectangular courtyard. The dimensions of the courtyard are 12 ft. by 25 ft. How far apart are the friends?
A)21.34 ft.
B)21.93 ft.
C)27.73 ft.
D)19.21 ft.
Answer:
Question 1 = D) Acute
Question 2 = C)3 feet
Question 3 = D) Obtuse
Question 4 = C)27.73 ft.
Step-by-step explanation:
Question 1: A triangle has sides with lengths 5, 6, and 7. Is the triangle right, acute, or obtuse?
In order to be able to accurately classify that a triangle with 3 given sides is either a right , acute or obtuse angle, we use the Pythagoras Theorem
Where:
If a² + b² = c² = Right angle triangle
If a² +b² > c² = Acute triangle.
If a² +b² < c² = Obtuse triangle.
It is important to note that the length ‘‘c′′ is always the longest.
Therefore, for the above question, we have lengths
5 = a, 6 = b and c = 7
a² + b² = c²
5² + 6² = 7²
25 + 36 = 49
61 = 49
61 ≠ 49, Hence 61 > 49
Therefore, this is an Acute Triangle
Question 2: A 15-foot statue casts a 20-foot shadow. How tall is a person who casts a 4-foot-long shadow?
This is question that deals with proportion.
The formula to solve for this:
Height of the statue/ Length of the shadow of the person = Height of the person/ Length of the shadow of the person
Height of the statue = 15 feet
Length of the shadow of the person = 20 feet
Height of the person = unknown
Length of the shadow of the person = 4
15/ 20 = Height of the person/4
Cross Multiply
15 × 4 = 20 × Height of the person
Height of the person = 15 × 4/20
= 60/20
Height of the person = 3 feet
Therefore, the person is 3 feet tall.
Question 3: A triangle has sides with lengths 17, 12, and 9. Is the triangle right, acute, or obtuse?
In order to be able to accurately classify that a triangle with 3 given sides is either a right , acute or obtuse angle, we use the Pythagoras Theorem
Where:
If a² + b² = c² = Right angle triangle
If a² +b² > c² = Acute triangle.
If a² +b² < c² = Obtuse triangle.
It is important to note that the length ‘‘c′′ is always the longest.
Therefore, for the above question, we have lengths 17, 12, 9
9 = a, 12 = b and c = 17
a² + b² = c²
9² + 12² = 17²
81 + 144 = 289
225 = 289
225 ≠ 289
225 < 289
Hence, This is an Obtuse Triangle.
Question 4: Two friends are standing at opposite corners of a rectangular courtyard. The dimensions of the courtyard are 12 ft. by 25 ft. How far apart are the friends?
To calculate how far apart the two friends are we use the formula
Distance = √ ( Length² + Breadth²)
We are given dimensions: 12ft by 25ft
Length = 12ft
Breadth = 25ft
Distance = √(12ft)² + (25ft)²
Distance = √144ft²+ 625ft²
Distance = √769ft²
Distance = 27.730849248ft
Approximately ≈27.73ft
Therefore, the friends are 27.73ft apart.
You are ordering two pizzas. A pizza can be small, medium, large, or extra large, with any combination of 8 possible toppings (getting no toppings is allowed, as is getting all 8). How many possibilities are there for your two pizzas
Answer:
1048576
Step-by-step explanation:
Given the following :
Pizza order :
Size = small, medium, large, or extra large = 4 possible sizes
Toppings = any combination of 8 possible toppings (getting no toppings is allowed, as is getting all 8).
Combination of Toppings = 2^8
Four different sizes of pizza = 4
Number of possibilities in ordering for a single pizza :
(4 * 2^8) = 4 * 256 = 1024
Number of possibilities in ordering two pizzas :
(4 * 2^8)^2
(2^2 * 2^8)^2
From indices :
[2^(2+8)]^2
[2^(10)]^2
2^(10*2)
2^20
= 1048576
Let X denote the day she gets enrolled in her first class and let Y denote the day she gets enrolled in both the classes. What is the distribution of X
Answer:
X is uniformly distributed.
Step-by-step explanation:
Uniform Distribution:
This is the type of distribution where all outcome of a certain event have equal likeliness of occurrence.
Example of Uniform Distribution is - tossing a coin. The probability of getting a head is the same as the probability of getting a tail. The have equal likeliness of occurrence.
(21x-3)+21=23x+6 solve
Answer:
False
Step-by-step explanation:
You Cnat solve it
Answer:
you cannot solve it
Step-by-step explanation:
false
Please help me solve for the median !!!
Answer:
50.93
Step-by-step explanation:
Add up the frequencies:
2 + 5 + 14 + 15 + 21 + 18 + 15 + 9 + 2 = 101
Divide by 2: 101/2 = 50.5
So the median is the 51st number, with 50 below and 50 above.
Add up the frequencies until you find the interval that contains the 51st number.
2 + 5 + 14 + 15 = 36
2 + 5 + 14 + 15 + 21 = 57
So the median is in the group 49.5 − 51.5. To estimate the median, we use interpolation. Find the slope of the line from (36, 49.5) to (57, 51.5).
m = (51.5 − 49.5) / (57 − 36)
m = 2/21
So at x = 51:
2/21 = (y − 49.5) / (51 − 36)
y = 50.93
A machine used to fill gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of ounces and a standard deviation of ounce. You randomly select cans and carefully measure the contents. The sample mean of the cans is ounces. Does the machine need to be reset? Explain your reasoning. ▼ Yes No , it is ▼ very unlikely likely that you would have randomly sampled cans with a mean equal to ounces, because it ▼ lies does not lie within the range of a usual event, namely within ▼ 1 standard deviation 2 standard deviations 3 standard deviations of the mean of the sample means.
Complete question is;
A machine used to fill gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of 128 ounces and a standard deviation of 0.20 ounce. You randomly select 35 cans and carefully measure the contents. The sample mean of the cans is 127.9 ounces. Does the machine need to be? reset? Explain your reasoning.
(yes/no)?, it is (very unlikely/ likely) that you would have randomly sampled 35 cans with a mean equal to 127.9 ?ounces, because it (lies/ does not lie) within the range of a usual? event, namely within (1 standard deviation, 2 standard deviations 3 standard deviations) of the mean of the sample means.
Answer:
Yes, we should reset the machine because it is unusual to have a mean equal to 127.9 from a random sample of 35 as the mean of 127.9 doesn't fall within range of a usual event with 2 standard deviations of the mean of the sample means.
Step-by-step explanation:
We are given;
Mean: μ = 128
Standard deviation; σ = 0.2
n = 35
Now, formula for standard error of mean is given as;
se = σ/√n
se = 0.2/√35
se = 0.0338
Normally, the range of values should be within 2 standard deviations of mean. In this case, normal range of values will be;
μ ± 2se = 128 ± 0.0338
This gives; 127.9662, 128.0338
So, Yes, we should reset the machine because it is unusual to have a mean equal to 127.9 from a random sample of 35 as the mean of 127.9 doesn't fall within range of a usual event with 2 standard deviations of the mean of the sample means.
The age of some lecturers are 42,54,50,54,50,42,46,46,48 and 48 calculate the mean age and standard deviation
Answer:
Mean age: 48
Standard deviation: 4
Step-by-step explanation:
a) Mean
The formula for Mean = Sum of terms/ Number of terms
Number of terms
= 42 + 54 + 50 + 54 + 50 + 42 + 46 + 46 + 48+ 48/ 10
= 480/10
= 48
The mean age is 48
b) Standard deviation
The formula for Standard deviation =
√(x - Mean)²/n
Where n = number of terms
Standard deviation =
√[(42 - 48)² + (54 - 48)² + (50 - 48)² +(54 - 48)² + (50 - 48)² +(42 - 48)² + (46 - 48)² + (46 - 48)² + (48 - 48)² + (48 - 48)² / 10]
= √-6² + 6² + 2² + 6² + 2² + -6² + -2² + -2² + 0² + 0²/10
=√36 + 36 + 4 + 36 + 4 + 36 + 4 + 4 + 0 + 0/ 10
=√160/10
= √16
= 4
The standard deviation of the ages is 4
Please help. I’ll mark you as brainliest if correct!
Answer:
9 3 -7 -13
4 -4 11 8
0 9 2 -4
Step-by-step explanation:
9 3 -7 -13
4 -4 11 8
0 9 2 -4
Answer: 9 3 -7 -13
4 -4 11 8
0 9 2 -4
Step-by-step explanation:
5x+4(-x-2)=-5x+2(x-1)+12
Answer:
x=9/2
Step-by-step explanation:
Let's solve your equation step-by-step.
5x+4(−x−2)=−5x+2(x−1)+12
Step 1: Simplify both sides of the equation.
5x+4(−x−2)=−5x+2(x−1)+12
5x+(4)(−x)+(4)(−2)=−5x+(2)(x)+(2)(−1)+12 (Distribute)
5x+−4x+−8=−5x+2x+−2+12
(5x+−4x)+(−8)=(−5x+2x)+(−2+12) (Combine Like Terms)
x+−8=−3x+10
x−8=−3x+10
Step 2: Add 3x to both sides.
x−8+3x=−3x+10+3x
4x−8=10
Step 3: Add 8 to both sides.
4x−8+8=10+8
4x=18
Step 4: Divide both sides by 4.
4x/4=18/4
x=9/2
Hey market sales six cans of food for every seven boxes of food the market sold a total of 26 cans and boxes today how many of each kind did the market sale
Answer:
It sold 14 cans boxes of food and 12 cans of food.
Step-by-step explanation:
The factor for the food cans depend upon every seven food boxes .So, the same no. of sets of food cans will be sold.
Let the no. of sets of food boxes be x.
According to the question,
6x+7x=26
13x=26
x=26/13
x=2
No. of food cans =6x=6×2=12 cans
No. of food boxes=7x=7×2=14 boxes
Please mark brainliest ,if it is truly the best ! Thank you!
You are studying for your final exam of the semester up to this point you received 3 exam scores of 61% 62% and 86% to receive a grade of c and the class you must have an average exam score between 70% and 79% for all four exams including the final find the widest range of scores that you can get on the final exam in order to receive a grade of C for the class 63 to 100% 71 to 100% 68 to 97
There will be a total of 4 test scores including the final exam. To get a 70, the 4 tests need to equal 4 x 70 = 280 points , to be 79, they have to equal 4 x 79 = 316 points.
The 3 already done = 61 + 62 + 86 = 209 points.
The final exam needs to be between :
280 -209 = 71
316 -209 = 107. The answer would be between 71 and 100%
A regular polygon inscribed in a circle can be used to derive the formula for the area of a circle. The polygon area can be expressed in terms of the area of a triangle. Let s be the side length of the polygon, let r be the hypotenuse of the right triangle, let h be the height of the triangle, and let n be the number of sides of the regular polygon. polygon area = n(12sh) Which statement is true? As h increases, s approaches r so that rh approaches r². As r increases, h approaches r so that rh approaches r². As s increases, h approaches r so that rh approaches r². As n increases, h approaches r so that rh approaches r².
Answer:
Option (D)
Step-by-step explanation:
Formula to get the area of a regular polygon in a circle will be,
Area = [tex]n[\frac{1}{2}\times (\text{Base})\times (\text{Height})][/tex]
= [tex]n[\frac{1}{2}\times (\text{s})\times (\text{h})][/tex]
Here 'n' is the number of sides.
If n increases, h approaches r so that 'rh' approaches r².
In other words, if the number of sides of the polygon gets increased, area of the polygon approaches the area of the circle.
Therefore, Option (4) will be the answer.
In this exercise it is necessary to have knowledge about polygons, so we have to:
Letter D
Then using the formula for the area of a regular polygon we find that:
[tex]A=n(1/2*B*H)\\=n(1/2*S*H)[/tex]
So from this way we were not able to identify the option that best corresponds to this alternative.
See more about polygons at brainly.com/question/17756657
Three out of every ten dentists recommend a certain brand of fluoride toothpaste. Which assignment of random digits would be used to simulate the random sampling of dentists who prefer this fluoride toothpaste?
Answer:
eddfdgdccggģdffcdrrfxddxcvgfx
10) How many possible outfit combinations come from six shirts, three
slacks, and five ties? *
A 15
B 18
C 30
D 90
Answer:
The answer is D)90
Hope I helped
The cost of a daily rental car is as follows: The initial fee is $39.99 for the car, and it costs $0.20 per mile. If Julie's final bill was $100.00 before taxes, how many miles did she drive?
Answer:
300.05 miles
Step-by-step explanation:
initial fee= $39.99
final bill = $ 100
cost =$ 0.20 per mile
remaining amount = $ 60.01
solution,
she drive = remaining amount / cost
=60.01/0.20
=300.05 miles
Answer:
500 miles
Step-by-step explanation:
Let us use cross multiplication to find the unknown amount.
Given:
1) Cost for 1 mile=$0.20
2)Cost for x miles=$100
Solution:
No of miles Cost
1) 1 $0.20
2)x $100
By cross multiplying,
100 x 1= 0.20x
x=100/0.20
x=500 miles
Thank you!
The cost, C, in United States Dollars ($), of cleaning up x percent of an oil spill along the Gulf Coast of the United States increases tremendously as x approaches 100. One equation for determining the cost (in millions $) is:
Complete Question
On the uploaded image is a similar question that will explain the given question
Answer:
The value of k is [tex]k = 214285.7[/tex]
The percentage of the oil that will be cleaned is [tex]x = 80.77\%[/tex]
Step-by-step explanation:
From the question we are told that
The cost of cleaning up the spillage is [tex]C = \frac{ k x }{100 - x }[/tex] [tex]x \le x \le 100[/tex]
The cost of cleaning x = 70% of the oil is [tex]C = \$500,000[/tex]
Now at [tex]C = \$500,000[/tex] we have
[tex]\$ 500000 = \frac{ k * 70 }{100 - 70 }[/tex]
[tex]\$ 500000 = \frac{ k * 70 }{30 }[/tex]
[tex]\$ 500000 = \frac{ k * 70 }{30 }[/tex]
[tex]k = 214285.7[/tex]
Now When [tex]C = \$900,000[/tex]
[tex]x = 80.77\%[/tex]
Foram prescritos 500mg de dipirona para uma criança com febre.Na unidade tem disponivel ampola de 1g/2ml.Quantos g vão ser administrados no paciente
De acordo com a disponibilidade da unidade, há apenas a seguinte dosagem: 1g/2mL - ou seja, uma grama de dipirona a cada 2mL
O enunciado está meio mal formulado, pois é dito que foram prescritos 500mg de dipirona e é essa quantidade de farmaco que a criança tem que tomar. Deseja-se saber quantos mL deverao ser administrados.
Fazendo a classica regra de 3, podemos chegar no volume desejado:
(atentar que 500mg = 0,5g)
g mL
1 --------- 2
0,5 --------- X
1 . X = 0,5 . 2
X = 1mLSalaries of 42 college graduates who took a statistics course in college have a mean, , of . Assuming a standard deviation, , of $, construct a % confidence interval for estimating the population mean .
Answer:
The 99% confidence interval for estimating the population mean μ is ($60,112.60, $68087.40).
Step-by-step explanation:
The complete question is:
Salaries of 42 college graduates who took a statistics course in college have a mean, [tex]\bar x[/tex] of, $64, 100. Assuming a standard deviation, σ of $10,016 construct a 99% confidence interval for estimating the population mean μ.
Solution:
The (1 - α)% confidence interval for estimating the population mean μ is:
[tex]CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]
The critical value of z for 99% confidence interval is:
[tex]z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.57[/tex]
Compute the 99% confidence interval for estimating the population mean μ as follows:
[tex]CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]
[tex]=64100\pm 2.58\times\frac{10016}{\sqrt{42}}\\\\=64100+3987.3961\\\\=(60112.6039, 68087.3961)\\\\\approx (60112.60, 68087.40)[/tex]
Thus, the 99% confidence interval for estimating the population mean μ is ($60,112.60, $68087.40).
Gail paid a total of $12,000 for stock that was $6 per share. If she sold all her shares for $18,000, how much profit on each share did she make?
A
$9
B
$3
С.
S2000
D
$6.000
Answer:
$3
Step-by-step explanation:
Given
Total Cost Price: $12,000
Unit Cost Price= $6
Total Selling Price = $18,000
Required
Determine the profit on each share
First, we need to determine the units of share bought;
Units = Total cost price / Unit Cost Price
[tex]Units = \frac{\$12000}{\$6}[/tex]
[tex]Units = 2000[/tex]
Next is to determine the selling price of each share; This is calculated as follows;
Unit Selling Price = Total Selling Price / Units Sold
[tex]Unit\ Selling\ Price = \frac{\$18000}{\$2000}[/tex]
[tex]Unit\ Selling\ Price = \$9[/tex]
The profit is the difference between the unit cost price and unit selling price
[tex]Profit = Unit\ Selling\ Price - Unit\ Cost\ Price[/tex]
[tex]Profit = \$9 - \$6[/tex]
[tex]Profit = \$3[/tex]
Use Lagrange multipliers to minimize the function subject to the following two constraints. Assume that x, y, and z are nonnegative. Question 18 options: a) 192 b) 384 c) 576 d) 128 e) 64
Complete Question
The complete question is shown on the first uploaded image
Answer:
Option C is the correct option
Step-by-step explanation:
From the question we are told that
The equation is [tex]f (x, y , z ) = x^2 +y^2 + z^2[/tex]
The constraint is [tex]P(x, y , z) = x + y + z - 24 = 0[/tex]
Now using Lagrange multipliers we have that
[tex]\lambda = \frac{ \delta f }{ \delta x } = 2 x[/tex]
[tex]\lambda = \frac{ \delta f }{ \delta y } = y[/tex]
[tex]\lambda = \frac{ \delta f }{ \delta z } = 2 z[/tex]
=> [tex]x = \frac{ \lambda }{2}[/tex]
[tex]y = \frac{ \lambda }{2}[/tex]
[tex]z = \frac{ \lambda }{2}[/tex]
From the constraint we have
[tex]\frac{\lambda }{2} + \frac{\lambda }{2} + \frac{\lambda }{2} = 24[/tex]
=> [tex]\frac{3 \lambda }{2} = 24[/tex]
=> [tex]\lambda = 16[/tex]
substituting for x, y, z
=> x = 8
=> y = 8
=> z = 8
Hence
[tex]f (8, 8 , 8 ) = 8^2 +8^2 + 8^2[/tex]
[tex]f (8, 8 , 8 ) = 192[/tex]
The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100, and the standard deviation is 2. You wish to test H0: μ = 100 versus H1: μ ≠ 100 with a sample of n = 9 specimens.
A. If the acceptance region is defined as 98.5 le x- 101.5, find the type I error probability alpha.
B. Find beta for the case where the true mean heat evolved is 103.
C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?
Answer:
A.the type 1 error probability is [tex]\mathbf{\alpha = 0.0244 }[/tex]
B. β = 0.0122
C. β = 0.0000
Step-by-step explanation:
Given that:
Mean = 100
standard deviation = 2
sample size = 9
The null and the alternative hypothesis can be computed as follows:
[tex]\mathtt{H_o: \mu = 100}[/tex]
[tex]\mathtt{H_1: \mu \neq 100}[/tex]
A. If the acceptance region is defined as [tex]98.5 < \overline x > 101.5[/tex] , find the type I error probability [tex]\alpha[/tex] .
Assuming the critical region lies within [tex]\overline x < 98.5[/tex] or [tex]\overline x > 101.5[/tex], for a type 1 error to take place, then the sample average x will be within the critical region when the true mean heat evolved is [tex]\mu = 100[/tex]
∴
[tex]\mathtt{\alpha = P( type \ 1 \ error ) = P( reject \ H_o)}[/tex]
[tex]\mathtt{\alpha = P( \overline x < 98.5 ) + P( \overline x > 101.5 )}[/tex]
when [tex]\mu = 100[/tex]
[tex]\mathtt{\alpha = P \begin {pmatrix} \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{\overline 98.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} + \begin {pmatrix}P(\dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} > \dfrac{101.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} }[/tex]
[tex]\mathtt{\alpha = P ( Z < \dfrac{-1.5}{\dfrac{2}{3}} ) + P(Z > \dfrac{1.5}{\dfrac{2}{3}}) }[/tex]
[tex]\mathtt{\alpha = P ( Z <-2.25 ) + P(Z > 2.25) }[/tex]
[tex]\mathtt{\alpha = P ( Z <-2.25 ) +( 1- P(Z < 2.25) })[/tex]
From the standard normal distribution tables
[tex]\mathtt{\alpha = 0.0122+( 1- 0.9878) })[/tex]
[tex]\mathtt{\alpha = 0.0122+( 0.0122) })[/tex]
[tex]\mathbf{\alpha = 0.0244 }[/tex]
Thus, the type 1 error probability is [tex]\mathbf{\alpha = 0.0244 }[/tex]
B. Find beta for the case where the true mean heat evolved is 103.
The probability of type II error is represented by β. Type II error implies that we fail to reject null hypothesis [tex]\mathtt{H_o}[/tex]
Thus;
β = P( type II error) - P( fail to reject [tex]\mathtt{H_o}[/tex] )
∴
[tex]\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }[/tex]
Given that [tex]\mu = 103[/tex]
[tex]\mathtt{\beta = P( \dfrac{98.5 -103}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-103}{\dfrac{2}{\sqrt{9}}}) }[/tex]
[tex]\mathtt{\beta = P( \dfrac{-4.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-1.5}{\dfrac{2}{3}}) }[/tex]
[tex]\mathtt{\beta = P(-6.75 \leq Z \leq -2.25) }[/tex]
[tex]\mathtt{\beta = P(z< -2.25) - P(z < -6.75 )}[/tex]
From standard normal distribution table
β = 0.0122 - 0.0000
β = 0.0122
C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?
[tex]\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }[/tex]
Given that [tex]\mu = 105[/tex]
[tex]\mathtt{\beta = P( \dfrac{98.5 -105}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-105}{\dfrac{2}{\sqrt{9}}}) }[/tex]
[tex]\mathtt{\beta = P( \dfrac{-6.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-3.5}{\dfrac{2}{3}}) }[/tex]
[tex]\mathtt{\beta = P(-9.75 \leq Z \leq -5.25) }[/tex]
[tex]\mathtt{\beta = P(z< -5.25) - P(z < -9.75 )}[/tex]
From standard normal distribution table
β = 0.0000 - 0.0000
β = 0.0000
The reason why the value of beta is smaller here is that since the difference between the value for the true mean and the hypothesized value increases, the probability of type II error decreases.
Please help ! I’ll mark you as brainliest if correct.
Answer:
D = -87Dx = 174Dy = -435Dz = 0(x, y, z) = (-2, 5, 0)Step-by-step explanation:
The determinant of the coefficient matrix is ...
[tex]D=\left|\begin{array}{ccc}2&5&3\\4&-1&-4\\-5&-2&6\end{array}\right|\\\\=2(-1)(6)+5(-4)(-5)+3(4)(-2)-2(-4)(-2)-5(4)(6)-3(-1)(-5)\\\\=-12+100-24-16-120-15=\boxed{-87}[/tex]
The other determinants are found in similar fashion after substituting the constants on the right for each of the above matrix columns, in turn.
Those determinants are ...
[tex]D_x=\left|\begin{array}{ccc}21&5&3\\-13&-1&-4\\0&-2&6\end{array}\right|=174[/tex]
[tex]D_y=\left|\begin{array}{ccc}2&21&3\\4&-13&-4\\-5&0&6\end{array}\right|=-435[/tex]
[tex]D_z=\left|\begin{array}{ccc}2&5&21\\4&-1&-13\\-5&-2&0\end{array}\right|=0[/tex]
The solutions are ...
x = 174/-87 = -2
y = -435/-87 = 5
z = 0
That is, (x, y, z) = (-2, 5, 0).
Time
(minutes)
Water
(gallons)
1
16.50
1.5
24.75
2
33
find the constant of proportionality for the second and third row
Answer:
16.50
Step-by-step explanation:
Constant of proportionality = no of gallons of water per 1 minute.
In the first row, we have 16.50 gallons of water per 1 minute.
In the 2nd row, we have 24.75 gallons of water in 1.5 minutes. In 1 minute, we will have 24.75 ÷ 1.5 = 16.50 gallons
In the 3rd row, we have 33 gallons in 2 minutes. In 1 minute, we will have 33 ÷ 2 = 16.50 gallons.
We can see that there seems to be the same constant of proportionality for the 2nd and 3rd row, which is 16.50.
Thus, a relationship between gallons of water (w) and time (t), considering the constant, 16.50, can be written as: [tex] w = 16.50t [/tex]
This means the constant of proportionality, 16.50, is same for all rows.
Find a cubic polynomial with integer coefficients that has $\sqrt[3]{2} + \sqrt[3]{4}$ as a root.
Find the powers [tex]a=\sqrt{2}+\sqrt{3}[/tex]
$a^{2}=5+2 \sqrt{6}$
$a^{3}=11 \sqrt{2}+9 \sqrt{3}$
The cubic term gives us a clue, we can use a linear combination to eliminate the root 3 term $a^{3}-9 a=2 \sqrt{2}$ Square $\left(a^{3}-9 a\right)^{2}=8$ which gives one solution. Expand we have $a^{6}-18 a^{4}-81 a^{2}=8$ Hence the polynomial $x^{6}-18 x^{4}-81 x^{2}-8$ will have a as a solution.
Note this is not the simplest solution as $x^{6}-18 x^{4}-81 x^{2}-8=\left(x^{2}-8\right)\left(x^{4}-10 x^{2}+1\right)$
so fits with the other answers.
Answer:
[tex]y^3 -6y-6[/tex]
Transform the given parametric equations into rectangular form. Then identify the conic.
Answer:
Solution : Option B
Step-by-Step Explanation:
We have the following system of equations at hand here.
{ x = 5 cot(t), y = - 3csc(t) + 4 }
Now instead of isolating the t from either equation, let's isolate cot(t) and csc(t) --- Step #1,
x = 5 cot(t) ⇒ x - 5 = cot(t),
y = - 3csc(t) + 4 ⇒ y - 4 = - 3csc(t) ⇒ y - 4 / - 3 = csc(t)
Now let's square these two equations. We know that csc²θ - cot²θ = 1, so let's subtract the equations as well. --- Step #2
( y - 4 / - 3 )² = (csc(t))²
- ( x - 5 / 1 )² = (cot(t))²
___________________
(y - 4)² / 9 - x² / 25 = 1
And as we are subtracting the two expressions, this is an example of a hyperbola. Therefore your solution is option b.
