Answer:
6/50
Step-by-step explanation:
There are 50 tiles
6 purple
18 pink
26 orange
P( purple) = purple/ total
= 6/50
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]
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 = 1mLThe dot plot represents a sampling of ACT scores: dot plot titled ACT Scores with Score on the x axis and Number of Students on the y axis with 1 dot over 24, 3 dots over 26, 3 dots over 27, 5 dots over 28, 3 dots over 30, 3 dots over 32, 1 dot over 35 Which box plot represents the dot plot data? box plot titled ACT Score with a minimum of 24, quartile 1 of 25, median of 26, quartile 3 of 29, and maximum of 35 box plot titled ACT Score with a minimum of 23, quartile 1 of 25, median of 26, quartile 3 of 29, and maximum of 36 box plot titled ACT Score with a minimum of 23, quartile 1 of 27, median of 30, quartile 3 of 34, and maximum of 36 box plot titled ACT Score with a minimum of 24, quartile 1 of 27, median of 28, quartile 3 of 30, and maximum of 35
Answer:
box plot titled ACT Score with a minimum of 24, quartile 1 of 27, median of 28, quartile 3 of 30, and maximum of 35
Step-by-step explanation:
The scores of the students represented on the dot plot are:
1 dot => 24
3 dots => 26, 26, 26
3 dots => 27, 27, 27
5 dots => 28, 28, 28, 28, 28
3 dots => 30, 30, 30
3 dots => 32, 32, 32
1 dot => 35
Quickly, we can ascertain 3 values from these data points of which we can use to find out which box plot represents the dot plot data.
The minimum score = 24
The maximum score = 35
The median score is the 10th value, which is the middle value of the data point = 28
Therefore, we can conclude that: "box plot titled ACT Score with a minimum of 24, quartile 1 of 27, median of 28, quartile 3 of 30, and maximum of 35".
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]
Salaries 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).
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.
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.
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.
If the normality requirement is not satisfied (that is, np(1p) is not at least 10), then a 95% confidence interval about the population proportion will include the population proportion in ________ 95% of the intervals. (This is a reading assessment question. Be certain of your answer because you only get one attempt on this question.)
Answer:
less than
Step-by-step explanation:
If the normality requirement is not satisfied (that is, np(1 - p) is not at least 10), then a 95% confidence interval about the population proportion will include the population proportion in _less than__ 95% of the intervals.
The confidence interval consist of all reasonable values of a population mean. These are value for which the null hypothesis will not be rejected.
So, let assume that If the 95% confidence interval contains the value for the hypothesized mean, then the sample mean is reasonably close to the hypothesized mean. The effect of this is that the p- value is going to be greater than 0.05, so we fail to reject the null hypothesis.
On the other hand,
If the 95% confidence interval do not contains the value for the hypothesized mean, then the sample mean is far away from the hypothesized mean. The effect of this is that the p- value is going to be lesser than 0.05, so we reject the null hypothesis.
A research center claims that % of adults in a certain country would travel into space on a commercial flight if they could afford it. In a random sample of adults in that country, % say that they would travel into space on a commercial flight if they could afford it. At , is there enough evidence to reject the research
Complete Question
A research center claims that 30% of adults in a certain country would travel into space on a commercial flight if they could afford it. In a random sample of 700 adults in that country, 34% say that they would travel into space on a commercial flight if they could afford it. At , is there enough evidence to reject the research center's claim
Answer:
Yes there is sufficient evidence to reject the research center's claim.
Step-by-step explanation:
From the question we are told that
The population proportion is p = 0.30
The sample proportion is [tex]\r p = 0.34[/tex]
The sample size is n = 700
The null hypothesis is [tex]H_o : p = 0.30[/tex]
The alternative hypothesis is [tex]H_a : p \ne 0.30[/tex]
Here we are going to be making use of level of significance = 0.05 to carry out this test
Now we will obtain the critical value of [tex]Z_{\alpha }[/tex] from the normal distribution table , the value is [tex]Z_{\alpha } = 1.645[/tex]
Generally the test statistics is mathematically represented as
[tex]t = \frac{ \r p - p }{ \sqrt{ \frac{ p (1-p)}{n} } }[/tex]
substituting values
[tex]t = \frac{ 0.34 - 0.30 }{ \sqrt{ \frac{ 0.30 (1-0.30 )}{ 700} } }[/tex]
[tex]t = 2.31[/tex]
Looking at the values of t and [tex]Z_{\alpha }[/tex] we see that [tex]t > Z_{\alpha }[/tex] hence the null hypothesis is rejected
Thus we can conclude that there is sufficient evidence to reject the research center's claim.
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!
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.
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 :)
logx-log(x-l)^2=2log(x-1)
Answer:
x = 1.00995066776
x = 2.52925492433
Step-by-step explanation:
This sort of equation is best solved using a graphing calculator. For that purpose, I like to rewrite the equation as a function whose zeros we're seeking. Here, that becomes ...
[tex]f(x)=\log{(x)}-\log{(x-1)}^2-2\log{(x-1)}[/tex]
The attached graph shows zeros at
x = 1.00995066776 and 2.52925492433
_____
Comment on the equation
Note that we have taken the middle term to be the square of the log, rather than the log of a square. For the latter interpretation, see mberisso's answer at https://brainly.com/question/17210068
Comment on the answer refinement
We have used Newton's method iteration to refine the solutions to this equation. The solution near 1.00995 requires the initial guess be very close for that method to work properly. Fortunately, the 1.01 value shown on the graph is sufficient for the purpose.
find the area of square whose side is 2.5 cm
Answer:
6.25
Step-by-step explanation:
2.5 *2.5=6.25
Answer:
6.25cm^2.
Step-by-step explanation:
To find the area of a square, you multiply the two sides, 2.5✖️2.5.
This gives the area of 6.25cm^2.
Hope this helped!
Have a nice day:)
If f(x)=x/2-3and g(x)=4x^2+x-4, find (f+g)(x)
Step-by-step explanation:
(f+g)(x) = f(x) + g(x)
= x/2-3 + 4x²+x+4
= ..........
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
one third multiplied by the sum of a and b
Answer:
1/3(a+b)
hope it helps :>
I NEED this answered within the next 30 minutes! Please it is simple. There is an error in this. What is it?
Answer:
(a). x = 80°
(b). x = 7.2 units
Step-by-step explanation:
Angle formed between the tangents from a point outside the circle measure the half of the difference of intercepted arcs.
(a). Here the intercepted arcs are,
Measure of major arc = 360° - 100°
= 260°
Measure of minor arc = 100°
x° = [tex]\frac{1}{2}[m(\text{Major arc})-m(\text{Minor arc})][/tex]
= [tex]\frac{1}{2}(260-100)[/tex]
x = 80°
(b). If a secant and tangent are drawn form a point outside the circle, then square of the measure of tangent is equal to the product of the measures of the secant segment and and its external segment.
x² = 4(4 + 9)
x² = 4 × 13
x² = 52
x = √52
x = 7.211 ≈ 7.2 units
Max believes that the sales of coffee at his coffee shop depend upon the weather. He has taken a sample of 5 days. Below you are given the results of the sample.
Cups of Coffee Sold Temperature
350 50
200 60
210 70
100 80
60 90
40 100
A. Which variable is the dependent variable?
B. Compute the least squares estimated line.
C. Compute the correlation coefficient between temperature and the sales of coffee.
D. Predict sales of a 90 degree day.
Answer:
1. cups of coffee sold
2.Y = 605.7 - 5.943x
3. -0.952
4. 70.84
Step-by-step explanation:
1. the dependent variable in this question is the cups of coffee sold
2. least square estimation line
Y = a+bx
we have y as the cups of coffee sold
x as temperature.
first we will have to solve for a and then b
∑X = 450
∑Y = 960
∑XY = 61600
∑X² = 35500
∑Y² = 221800
a = ∑y∑x²-∑x∑xy/n∑x²-(∑x)²
a = 960 * 35500-450*61600/6*35500-450²
a = 6360000/10500
= 605.7
b = n∑xy - ∑x∑y/n∑x²-(∑x)²
= 6*61600 - 450*960/6*35500 - 450²
= -5.943
the regression line
Y = a + bx
Y = 605.7 - 5.943x
3. we are to find correlation coefficient
r = n∑xy - ∑x∑y multiplied by√(n∑x²-(∑x)² * (n∑y² - (∑y)²)
= 6*61600 -960*450/√(6*35500 - 450²)*(6*221800 - 960²)
=-62400/√4296600000
= -62400/65548.5
= -0.952
4. we have to predict sales of a 90 degree day fro the regression line
Y = 605.7 - 5.943x
y = 605.7 - 5.943(90)
y = 605.7 - 534.87
= 70.84
88 feet/second = 60 miles/hour. How many feet per second is 1 mile/hour? (Hint: divide both sides of the equation
by the same amount.)
Round to the nearest thousandth.
One mile per hour is equivalent to
ao feet/second
Two sides of a triangle are equal length. The length of the third side exceeds the length of one of the other sides by 3 centimeters. The perimeter of the triangle is 93 centimeters. Find the length of each of the shorter sides of the triangle
Answer:
30 cm
Step-by-step explanation:
let x be the lenght of the two sides of equal lenghts, so the other is x+3
and the perimeter is x+x +x +3
P=3x+3
P=3(x+1)
93=3(x+1)
31=x+1
x=30
so the shorter sides are of 30 centimeters and the longest is 33
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
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]
The quotient of 8 and the difference of three and a number.
Answer: 8÷(3-x)
Answer:
Below
Step-by-step explanation:
● 8 ÷ (3-x)
Dividing by 3-x is like multiplying by 1/(3-x)
● 8 × (1/3-x)
● 8 /(3-x)
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.
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 !
:)
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]
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.
