Answer:
(a) The probability that X is at most 30 is 0.9726.
(b) The probability that X is less than 30 is 0.9554.
(c) The probability that X is between 15 and 25 (inclusive) is 0.7406.
Step-by-step explanation:
We are given that 11% of all steel shafts produced by a certain process are nonconforming but can be reworked. A random sample of 200 shafts is taken.
Let X = the number among these that are nonconforming and can be reworked
The above situation can be represented through binomial distribution such that X ~ Binom(n = 200, p = 0.11).
Here the probability of success is 11% that this much % of all steel shafts produced by a certain process are nonconforming but can be reworked.
Now, here to calculate the probability we will use normal approximation because the sample size if very large(i.e. greater than 30).
So, the new mean of X, [tex]\mu[/tex] = [tex]n \times p[/tex] = [tex]200 \times 0.11[/tex] = 22
and the new standard deviation of X, [tex]\sigma[/tex] = [tex]\sqrt{n \times p \times (1-p)}[/tex]
= [tex]\sqrt{200 \times 0.11 \times (1-0.11)}[/tex]
= 4.42
So, X ~ Normal([tex]\mu =22, \sigma^{2} = 4.42^{2}[/tex])
(a) The probability that X is at most 30 is given by = P(X < 30.5) {using continuity correction}
P(X < 30.5) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{30.5-22}{4.42}[/tex] ) = P(Z < 1.92) = 0.9726
The above probability is calculated by looking at the value of x = 1.92 in the z table which has an area of 0.9726.
(b) The probability that X is less than 30 is given by = P(X [tex]\leq[/tex] 29.5) {using continuity correction}
P(X [tex]\leq[/tex] 29.5) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{29.5-22}{4.42}[/tex] ) = P(Z [tex]\leq[/tex] 1.70) = 0.9554
The above probability is calculated by looking at the value of x = 1.70 in the z table which has an area of 0.9554.
(c) The probability that X is between 15 and 25 (inclusive) is given by = P(15 [tex]\leq[/tex] X [tex]\leq[/tex] 25) = P(X < 25.5) - P(X [tex]\leq[/tex] 14.5) {using continuity correction}
P(X < 25.5) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{25.5-22}{4.42}[/tex] ) = P(Z < 0.79) = 0.7852
P(X [tex]\leq[/tex] 14.5) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{14.5-22}{4.42}[/tex] ) = P(Z [tex]\leq[/tex] -1.70) = 1 - P(Z < 1.70)
= 1 - 0.9554 = 0.0446
The above probability is calculated by looking at the value of x = 0.79 and x = 1.70 in the z table which has an area of 0.7852 and 0.9554.
Therefore, P(15 [tex]\leq[/tex] X [tex]\leq[/tex] 25) = 0.7852 - 0.0446 = 0.7406.
Match the base to the corresponding height.
Base (b)
Height (h)
b
h
h
b
The base 1 is matched with height 2, base 2 is matched with height 3 and base 3 is matched with height 1. The base to the corresponding height is matched in the attached figure.
What is a triangle?Triangle is the closed shaped polygon which has 3 sides and 3 interior angles. The height of the triangle is the dimension of the elevation from the opposite peak to the length of the base.
Thus, the base 1 is matched with height 2, base 2 is matched with height 3 and base 3 is matched with height 1. The base to the corresponding height is matched in the attached figure.
In the given figure, three triangles is shown with base and height. Here,
The base 1 is matched with height 2, as the height shown in figure 2 is the dimension of the elevation from the opposite peak to the length of the base 1.Similarly, base 2 is matched with height 3.Base 3 is matched with height 1.
Thus, the base 1 is matched with height 2, base 2 is matched with height 3 and base 3 is matched with height 1. The base to the corresponding height is matched in the attached figure.
Learn more about the base and height of the triangle here;
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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.
I need help please help meee I don’t understand
Answer:
204
Step-by-step explanation:
To simplify the shape, you can do multiple things. I've opted to shave down both prongs to take it from a 'T' shape to a rectangular prism.
For height of the prongs, take 4 from 6.
6 - 4 = 2
Divide by 2 as there are 2 prongs.
2 / 2 = 1
Remember L * W * H
6 * 3 * 1 = 18
Remember that there are two prongs!
3 + 4 = 7
6 * 7 * 4 = 168
168 + 2(18) = 204
A line passes through (-5, -3) and is parallel to -3x - 7y = 10. The equation of the line in slope-intercept form is _____
Answer:
-3x - 7y = 36
Step-by-step explanation:
The given line -3x - 7y = 10 has an infinite number of parallel lines, all of the form -3x - 7y = C.
If we want the equation of a line parallel to -3x - 7y = 10 that passes through (-5, -3), we substitute -5 for x in -3x - 7y = 10 and substitute -3 for y in -3x - 7y = 10:
-3(-5) - 7(-3) = C, or
15 + 21 = C, or C = 36
Then the desired equation is -3x - 7y = 36.
Find (fºg)(2) and (f+g)(2) when f(x)= 1/x and g(x) = 4x +9
[tex](f\circ g)(2)=\dfrac{1}{4\cdot2+9}=\dfrac{1}{17}\\\\(f+g)(2)=\dfrac{1}{2}+4\cdot2+9=\dfrac{1}{2}+17=\dfrac{1}{2}+\dfrac{34}{2}=\dfrac{35}{2}[/tex]
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
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.
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]
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
(16 points) Find the radius of convergence and the interval of convergence of the power series. g
Answer:
The equation to be solved is missing in the question.
I will explain power series and ways to find the radius and interval of convergence of a powers series in the attached image.
Step-by-step explanation:
Understand the power seriesFind radius of convergenceDetermine interval of convergenceA box contains 40 identical discs which are either red or white if probably picking a red disc is 1/4. Calculate the number of;
1. White disc.
2. red disc that should be added such that the probability of picking a red disc will be 1/4
please help !! Solve –2.5x ≤ 25
Answer:
x ≥-10
Step-by-step explanation:
–2.5x ≤ 25
Divide each side by -2.5, remembering to flip the inequality
–2.5x/-2.5 ≥ 25 /-2.5
x ≥-10
Answer:
[tex]x\leq -10[/tex]
Step-by-step explanation:
[tex]-2.5x\leq 25[/tex]-----> Multiply by -1:
[tex]2.5x\geq -25[/tex]-----> Divide by 2.5:
[tex]x\geq -10[/tex]
Hope this helps!
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
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]
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).
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!
one third multiplied by the sum of a and b
Answer:
1/3(a+b)
hope it helps :>
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.
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
Twice the difference of a number and 9 is 3. Use the variable b for the unknown number.
Answer:
b = 10.5
Step-by-step explanation:
2(b-9) = 3
then:
2*b + 2*-9 = 3
2b - 18 = 3
2b = 3 + 18
2b = 21
b = 21/2
b = 10.5
check:
2(10.5 - 9) = 3
2*1.5 = 3
Find usubscript10 in the sequence -23, -18, -13, -8, -3, ...
Step-by-step explanation:
utilise the formula a+(n-1)d
a is the first number while d is common difference
Answer:
22
Step-by-step explanation:
Using the formular, Un = a + (n - 1)d
Where n = 10; a = -23; d = 5
U10 = -23 + (9)* 5
U10 = -23 + 45 = 22
Compute (3/4)*(8/9)*(15/16)*(24/25)*(35/36)*(48/49)*(63/64)*(80/81)*(99/100) Express your answer in the simplest way possible. (Suggestion: First, try computing 3/4*8/9 then 3/4*8/9*15/16 and so on. Look for patterns.
Answer:
[tex](\frac{3}{4})*(\frac{8}{9})*(\frac{15}{16})*(\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49})*(\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}) = \frac{11}{20}[/tex]
Step-by-step explanation:
Given
[tex](\frac{3}{4})*(\frac{8}{9})*(\frac{15}{16})*(\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49})*(\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100})[/tex]
Required
Simplify
For clarity, group the expression in threes
[tex]((\frac{3}{4})*(\frac{8}{9})*(\frac{15}{16}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]
Evaluate the first group [Divide 8 by 4]
[tex]((\frac{3}{1})*(\frac{2}{9})*(\frac{15}{16}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]
[Divide 9 by 3]
[tex]((\frac{1}{1})*(\frac{2}{3})*(\frac{15}{16}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]
[tex]((\frac{2}{3})*(\frac{15}{16}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]
[Divide 15 by 3]
[tex]((\frac{2}{1})*(\frac{5}{16}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]
[Divide 16 by 2]
[tex]((\frac{1}{1})*(\frac{5}{8}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]
[tex](\frac{5}{8})*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]
Evaluate the second group [Divide 35 and 25 by 5]
[tex](\frac{5}{8})*((\frac{24}{5})*(\frac{7}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]
[Divide 49 by 7]
[tex](\frac{5}{8})*((\frac{24}{5})*(\frac{1}{3})*(\frac{4}{7}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]
[Divide 24 by 3]
[tex](\frac{5}{8})*((\frac{8}{5})*(\frac{1}{1})*(\frac{4}{7}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]
[tex](\frac{5}{8})*((\frac{8}{5})*(\frac{4}{7}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]
Merge the first and second group
[tex]((\frac{5}{8})*(\frac{8}{5})*(\frac{4}{7}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]
[tex](1*(\frac{4}{7}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]
[tex](\frac{4}{7})*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]
Evaluate the last group [Divide 99 by 9]
[tex](\frac{4}{7})*((\frac{63}{64})*(\frac{80}{9})*(\frac{11}{100}))[/tex]
[Divide 63 by 9]
[tex](\frac{4}{7})*((\frac{7}{64})*(\frac{80}{1})*(\frac{11}{100}))[/tex]
[Divide 64 and 80 by 8]
[tex](\frac{4}{7})*((\frac{7}{8})*(\frac{10}{1})*(\frac{11}{100}))[/tex]
[Divide 10 and 4 by 2]
[tex](\frac{4}{7})*((\frac{7}{4})*(\frac{5}{1})*(\frac{11}{100}))[/tex]
[Divide 100 by 5]
[tex](\frac{4}{7})*((\frac{7}{4})*(\frac{1}{1})*(\frac{11}{20}))[/tex]
[tex](\frac{4}{7})*((\frac{7}{4})*(\frac{11}{20}))[/tex]
[tex](\frac{4}{7})*(\frac{7}{4})*(\frac{11}{20})[/tex]
[tex]1*(\frac{11}{20})[/tex]
[tex]\frac{11}{20}[/tex]
Hence;
[tex](\frac{3}{4})*(\frac{8}{9})*(\frac{15}{16})*(\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49})*(\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}) = \frac{11}{20}[/tex]
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]
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:)
An investigator claims, with 95 percent confidence, that the interval between 10 and 16 miles includes the mean commute distance for all California commuters. To have 95 percent confidence signifies that
Answer:
Hello the options to your question is missing below are the options
A) if sample means were obtained for a long series of samples, approximately 95 percent of all sample means would be between 10 and 16 miles
B.the unknown population mean is definitely between 10 and 16 miles
C.if these intervals were constructed for a long series of samples, approximately 95 percent would include the unknown mean commute distance for all Californians
D.the unknown population mean is between 10 and 16 miles with probability .95
Answer : if these intervals were constructed for a long series of samples, approximately 95 percent would include the unknown mean commute distance for all Californians ( c )
Step-by-step explanation:
95% confidence
interval = 10 to 16 miles
To have 95% confidence signifies that if these intervals were constructed for a long series of samples, approximately 95 percent would include the unknown mean commute distance for all Californians
confidence interval covers a range of samples/values in the interval and the higher the % of the confidence interval the more precise the interval is,
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.
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 = 1mLA 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.
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
= ..........
