The dad had 90 cents.
Multiply the 90 cents by 5%:
90 x 0.05 = 4.5 cents.
Subtract that from 90:
90 - 4.5 = 85.5 cents.
The lowest guess the son could say was 86 cents to be within 5%
Since the son guessed lower than that he did not get the money.
1. A dad holds five coins in his hand. He tells his son that if he can guess the amount of money he is holding within 5% error, he can have the money. The son guesses that dad is holding 81 cents. The dad opens his hand and displays 90 cents. Did the son guess close enough to get the money?
yes
If 20600 dollars is invested at an interest rate of 10 percent per year, find the value of the investment at the end of 5 years for the following compounding methods, to the nearest cent.
(a) Annual: $
(b) Semiannual: $
(c) Monthly: $
(d) Daily: $
Answer:
a) The value of the investment is $33176.506
b) The value of the investment is $33555.53
c) The value of the investment is $33893.36
d) The value of the investment is $33961.33
Step-by-step explanation:
This is a compound interest problem
Compound interest formula:
The compound interest formula is given by:
[tex]A = P(1 + \frac{r}{n})^{nt}[/tex]
A: Amount of money(Balance)
P: Principal(Initial deposit)
r: interest rate(as a decimal value)
n: number of times that interest is compounded per unit t
t: time the money is invested or borrowed for.
In our problem, we have:
A: the value we want to find
P = 20600(the value invested)
r = 0.1
n: Will change for each letter
t = 5.
a) If the interest is compounded anually, n = 1. So.
[tex]A = 20600(1 + \frac{0.1}{1})^{1*5}[/tex]
[tex]A = 33176.506[/tex]
The value of the investment is $33176.506
b) If the interest is compounded semianually, it happens twice a year, which means n = 2. So:
[tex]A = 20600(1 + \frac{0.1}{2})^{2*5}[/tex]
[tex]A = 33555.23[/tex]
The value of the investment is $33555.53
c) If the interest is compounded monthly, it happens 12 times a year, which means n = 12. So:
[tex]A = 20600(1 + \frac{0.1}{12})^{12*5}[/tex]
[tex]A = 33893.36[/tex]
The value of the investment is $33893.36
d) If the interest is compounded monthly, it happens 365 times a year, which means n = 365. So:
[tex]A = 20600(1 + \frac{0.1}{365})^{365*5}[/tex]
[tex]A = 33961.33[/tex]
The value of the investment is $33961.33
Final answer:
The future value of $20,600 invested at a 10% annual interest rate after 5 years varies based on the compounding frequency: annually it will be $33,186.35, semiannually $33,644.31, monthly $33,949.69, and daily $34,030.18.
Explanation:
Calculating Future Value with Different Compounding Methods
To calculate the future value of an investment with compound interest, we use the formula [tex]FV =PV(1+r/n)^{nt}[/tex],where FV is the future value, P is the principal amount, r is the annual interest rate, n is the number of times the interest is compounded per year, and t is time in years.
Given a principal amount of $20,600 and an annual interest rate of 10%, we will calculate the value of the investment at the end of 5 years for the following compounding methods:
Annual compounding (n=1): FV = 20600(1 + 0.10/1)⁽¹ˣ⁵⁾
Semiannual compounding (n=2): FV = 20600(1 + 0.10/2)⁽²ˣ⁵⁾
Monthly compounding (n=12): FV = 20600(1 + 0.10/12)⁽¹²ˣ⁵⁾
Daily compounding (n=365): FV = 20600(1 + 0.10/365)⁽³⁶⁶ˣ⁵⁾
Using a calculator or spreadsheet software, we can find the values to the nearest cent:
(a) Annual: $33,186.35
(b) Semiannual: $33,644.31
(c) Monthly: $33,949.69
(d) Daily: $34,030.18
The compounding effect shows that the more frequently the interest is compounded, the greater the final value of the investment.
he diameters of red delicious apples of an orchard have a normal distribution with a mean of 3 inches and a standard deviation of 0.5 inch. One apple will be randomly chosen. What is the probability of picking an apple with diameter between 2.5 and 4.25 inches?
Answer: 0.8351
Step-by-step explanation:
Given :Mean : [tex]\mu=\text{ 3 inches}[/tex]
Standard deviation : [tex]\sigma =\text{ 0.5 inch}[/tex]
The formula for z -score :
[tex]z=\dfrac{x-\mu}{\sigma}[/tex]
For x= 2.5 ,
[tex]z=\dfrac{2.5-3}{0.5}=-1[/tex]
For x= 4.25 ,
[tex]z=\dfrac{4.25-3}{0.5}=2.5[/tex]
The p-value = [tex]P(-1<z<2.5)=P(z<2.5)-P(z<-1)[/tex]
[tex]=0.9937903-0.1586553=0.835135\approx0.8351[/tex]
Hence, the probability of picking an apple with diameter between 2.5 and 4.25 inches =0.8351.
To find the probability, we standardize the values, convert them into z-scores, look up the z-scores in a z-table, and subtract the lower cumulative probability from the higher one. The probability of picking an apple with diameter between 2.5 inches and 4.25 inches is 83.51%.
Explanation:The problem involves a normal distribution where the diameter of Red Delicious apples has a mean of 3 inches and a standard deviation of 0.5 inch. The objective is to find the probability of picking an apple with a diameter between 2.5 and 4.25 inches.
We first standardize the given values to convert them into z-scores by subtracting the mean from the given value and dividing by the standard deviation. For 2.5 inches, z = (2.5 - 3) / 0.5 = -1. For 4.25 inches, z = (4.25 - 3) / 0.5 = 2.5.
Using a z-table, the z-score of -1 corresponds to a cumulative probability of 0.1587 and the z-score of 2.5 corresponds to a cumulative probability of 0.9938. To find the probability between these two diameters, we subtract the cumulative probability of -1 from the cumulative probability of 2.5.
Therefore, the probability of picking an apple with diameter between 2.5 inches and 4.25 inches is 0.9938 - 0.1587 = 0.8351 or 83.51%.
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Draw fraction rectangles on dot paper to solve the problems.
1) Subtract: 3/5 - 1/4
2) Add: 5/6 + 1/4
3) Manny and Frank ordered pizza. Manny ate 1/4 of the pizza and Frank ate 5/8 of the pizza. How much of the whole pizza was eaten?
Answer:
1) [tex]\frac{7}{20}[/tex]
2) [tex]1\frac{1}{12}[/tex]
3) [tex]\frac{7}{8}[/tex]
Step-by-step explanation:
1) Given problem,
[tex]\frac{3}{5}-\frac{1}{4}[/tex]
LCM(5,4) = 20,
[tex]\frac{3}{5}=\frac{3\times 4}{5\times 4}=\frac{12}{20}[/tex]
[tex]\frac{1}{4}=\frac{1\times 5}{4\times 5}=\frac{5}{20}[/tex]
[tex]\implies \frac{3}{5}-\frac{1}{4}=\frac{12}{20}-\frac{5}{20}=\frac{7}{20}[/tex]
2) Given problem,
[tex]\frac{5}{6}+\frac{1}{4}[/tex]
LCM(6, 4) = 12,
[tex]\frac{5}{6}=\frac{5\times 2}{6\times 2}=\frac{10}{12}[/tex]
[tex]\frac{1}{4}=\frac{1\times 3}{4\times 3}=\frac{3}{12}[/tex]
[tex]\implies \frac{5}{6}+\frac{1}{4}=\frac{10}{12}+\frac{3}{12}=\frac{13}{12}=1\frac{1}{12}[/tex]
3) ∵ Pizza ate by Manny = [tex]\frac{1}{4}[/tex]
Pizza ate by Frank = [tex]\farc{5}{8}[/tex]
∴ Total pizza eaten = [tex]\frac{1}{4}+\frac{5}{8}[/tex]
LCM(4, 8) = 8,
[tex]\frac{1}{4}=\frac{2}{8}[/tex]
Thus, total pizza eaten = [tex]\frac{2}{8}+\frac{5}{8}=\frac{7}{8}[/tex]
If (x,y) is a solution to the system of equations shown below, what is the product of the y-coordinates of the solutions? x^2+4y^2=40 x+2y=8
Answer:
The product of the y-coordinates of the solutions is equal to 3
Step-by-step explanation:
we have
[tex]x^{2}+4y^{2}=40[/tex] -----> equation A
[tex]x+2y=8[/tex] ------> equation B
Solve by graphing
Remember that the solutions of the system of equations are the intersection point both graphs
using a graphing tool
The solutions are the points (2,3) and (6,1)
see the attached figure
The y-coordinates of the solutions are 3 and 1
therefore
The product of the y-coordinates of the solutions is equal to
(3)(1)=3
For the following functions from R -> R, determine if function is one to one, onto, or both. Explain.
a) f(x)=3x-4
b)g(x)=(x^2)-2
c) h(x)=2/x
d) k(x)=ln(x)
e) l(x) = e^x
Answer with explanation:
a. f(x)=3x-4
Let [tex]f(x_1)=f(x_2)[/tex]
[tex]3x_1-4=3x_2-4[/tex]
[tex]3x_1=3x_2-4+4[/tex]
[tex]3x_1=3x_2[/tex]
[tex]x_1=x_2[/tex]
Hence, the function one-one.
Let f(x)=y
[tex]y=3x-4[/tex]
[tex]3x=y+4[/tex]
[tex]x=\frac{y+4}{3}[/tex]
We can find pre image in domain R for every y in range R.
Hence, the function onto.
b.g(x)=[tex]x^2-2[/tex]
Substiute x=1
Then [tex]g(x)=1-2=-1[/tex]
Substitute x=-1
Then g(x)=1-2=-1
Hence, the image of 1 and -1 are same . Therefore, the given function g(x) is not one-one.
The given function g(x) is not onto because there is no pre image of -2, -3,-4...... R.
Hence, the function neither one-one nor onto on given R.
c.[tex]h(x)=\frac{2}{x}[/tex]
The function is not defined for x=0 .Therefore , it is not a function on domain R.
Let [tex]h(x_1)=h(x_2)[/tex]
[tex] \frac{2}{x_1}=\frac{2}{x_2}[/tex]
By cross mulitiply
[tex]x_1= \frac{2\times x_2}{2}[/tex]
[tex]x_1=x_2[/tex]
Hence, h(x) is a one-one function on R-{0}.
We can find pre image for every value of y except zero .Hence, the function
h(x) is onto on R-{0}.
Therefore, the given function h(x) is both one- one and onto on R-{0} but not on R.
d.k(x)= ln(x)
We know that logarithmic function not defined for negative values of x. Therefore, logarithmic is not a function R.Hence, the given function K(x) is not a function on R.But it is define for positive R.
Let[tex]k(x_1)=k(x_2)[/tex]
[tex] ln(x_1)=ln(x_2)[/tex]
Cancel both side log then
[tex]x_1=x_2[/tex]
Hence, the given function one- one on positive R.
We can find pre image in positive R for every value of [tex]y\in R^+[/tex].
Therefore, the function k(x) is one-one and onto on [tex]R^+[/tex] but not on R.
e.l(x)=[tex]e^x[/tex]
Using horizontal line test if we draw a line y=-1 then it does not cut the graph at any point .If the horizontal line cut the graph atmost one point the function is one-one.Hence, the horizontal line does not cut the graph at any point .Therefore, the function is one-one on R.
If a horizontal line cut the graph atleast one point then the function is onto on a given domain and codomain.
If we draw a horizontal line y=-1 then it does not cut the graph at any point .Therefore, the given function is not onto on R.
What is the optimal solution for the following problem?
Minimize
P = 3x + 15y
subject to
2x + 4y ? 12
5x + 2y ? 10
and
x ? 0, y ? 0.
(x, y) = (2, 0)
(x, y) = (0, 3)
(x, y) = (0, 0)
(x, y) = (1, 2.5)
(x, y) = (6, 0)
Answer:Find the slope of the line that passes through the points shown in the table.
The slope of the line that passes through the points in the table is
.
Step-by-step explanation:
By substitifying the given points into the objective function, we can evaluate the minimum P. The point (x, y) = (0, 0) gives the minimum value of P = 0, which is the optimal solution for this problem.
Explanation:This problem is a classic example of a linear programming problem, a method to achieve the best outcome in a mathematical model whose requirements are represented by linear relationships. In this case, we are asked to minimize P = 3x + 15y subject to the constraints [tex]2x + 4y \leq 12, 5x + 2y \leq 10, and ,x \geq 0, y \geq 0.[/tex] In other words, we are looking for values of x and y that satisfy the constraints and result in the smallest possible value of P.
By substituting our given points into the equation for P we can compare the results. The smallest value for P corresponds to the point (x, y) = (0, 0) with P = 0. This is the optimal solution for this problem because it results in the lowest value for P while still satisfying all the constraints.
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Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. Find the probability that a given score is between negative 2.13 and 3.88 and draw a sketch of the region.
Answer: 0.9834
Step-by-step explanation:
Given : The test scores are normally distributed with
Mean : [tex]\mu=\ 0[/tex]
Standard deviation :[tex]\sigma= 1[/tex]
The formula to calculate the z-score :-
[tex]z=\dfrac{x-\mu}{\sigma}[/tex]
For x = -2.13
[tex]z=\dfrac{-2.13-0}{1}=-2.13[/tex]
For x = 3.88
[tex]z=\dfrac{3.88-0}{1}=3.88[/tex]
The p-value = [tex]P(-2.13<z<3.88)=P(z<3.88)-P(z<-2.13)[/tex]
[tex]0.9999477-0.0165858=0.9833619\approx0.9834[/tex]
Hence, the probability that a given score is between negative 2.13 and 3.88 = 0.9834
solve the system of equation by guess sidle method
8x1 + x2 + x3 = 8
2x1 + 4x2 + x3 = 4
x1 + 3x2 + 5x3 = 5
Answer: The solution is,
[tex]x_1\approx 0.876[/tex]
[tex]x_2\approx 0.419[/tex]
[tex]x_3\approx 0.574[/tex]
Step-by-step explanation:
Given equations are,
[tex]8x_1 + x_2 + x_3 = 8[/tex]
[tex]2x_1 + 4x_2 + x_3 = 4[/tex]
[tex]x_1 + 3x_2 + 5x_3 = 5[/tex],
From the above equations,
[tex]x_1=\frac{1}{8}(8-x_2-x_3)[/tex]
[tex]x_2=\frac{1}{4}(4-2x_1-x_3)[/tex]
[tex]x_3=\frac{1}{5}(5-x_1-3x_2)[/tex]
First approximation,
[tex]x_1(1)=\frac{1}{8}(8-(0)-(0))=1[/tex]
[tex]x_2(1)=\frac{1}{4}(4-2(1)-(0))=0.5[/tex]
[tex]x_3(1)=\frac{1}{5}(5-1-3(0.5))=0.5[/tex]
Second approximation,
[tex]x_1(2)=\frac{1}{8}(8-(0.5)-(0.5))=0.875[/tex]
[tex]x_2(2)=\frac{1}{4}(4-2(0.875)-(0.5))=0.4375[/tex]
[tex]x_3(2)=\frac{1}{5}((0.875)-3(0.4375))=0.5625[/tex]
Third approximation,
[tex]x_1(3)=\frac{1}{8}(8-(0.4375)-(0.5625))=0.875[/tex]
[tex]x_2(3)=\frac{1}{4}(4-2(0.875)-(0.5625))=0.421875[/tex]
[tex]x_3(3)=\frac{1}{5}(5-(0.875)-3(0.421875))=0.571875[/tex]
Fourth approximation,
[tex]x_1(4)=\frac{1}{8}(8-(0.421875)-(0.571875))=0.875781[/tex]
[tex]x_2(4)=\frac{1}{4}(4-2(0.875781)-(0.571875))=0.419141[/tex]
[tex]x_3(4)=\frac{1}{5}(5-(0.875781)-3(0.419141))=0.573359[/tex]
Fifth approximation,
[tex]x_1(5)=\frac{1}{8}(8-(0.419141)-(0.573359))=0.875938[/tex]
[tex]x_2(5)=\frac{1}{4}(4-2(0.875938)-(0.573359))=0.418691[/tex]
[tex]x_3(5)=\frac{1}{5}(5-(0.875938)-3(0.418691))=0.573598[/tex]
Hence, by the Gauss Seidel method the solution of the given system is,
[tex]x_1\approx 0.876[/tex]
[tex]x_2\approx 0.419[/tex]
[tex]x_3\approx 0.574[/tex]
What is the value of x?
Enter your answer in the box.
Answer:
x = 25
Step-by-step explanation:
Step 1: Identify the similar triangles
Triangle DQC and triangle DBR are similar
Step 2: Identify the parallel lines
QC is parallel to BR
Step 3: Find x
DQ/QB = DC/CR
40/24 = x/15
x = 25
!!
Answer: [tex]x=25[/tex]
Step-by-step explanation:
In order to calculate the value of "x", you can set up de following proportion:
[tex]\frac{BQ+QD}{QD}=\frac{RC+CD}{CD}\\\\\frac{24+40}{40}=\frac{15+x}{x}[/tex]
Now, the final step is to solve for "x" to find its value.
Therefore, its value is the following:
[tex]1.6=\frac{15+x}{x}\\\\1.6x=x+15\\\\1.6x-x=15\\\\0.6x=15\\\\x=\frac{15}{0.6}\\\\x=25[/tex]
A farmer builds a fence to enclose a rectangular pasture. He uses 160 feet of fence. Find the total area of the pasture if it is 50 feet long
Answer:
1500 ft²
Step-by-step explanation:
The sum of two adjacent sides of the pasture is half the perimeter (160 ft/2 = 80 ft), so the side adjacent to the 50 ft side will be 80 ft - 50 ft = 30 ft.
The product of adjacent sides of a rectangle gives the area of the rectangle. That area will be ...
area = (50 ft)(30 ft) = 1500 ft²
Find all the zeros of the polynomial function f(x) = x + 2x² - 9x - 18 a) (-3) b) (-3. -2,3) c) (-2) d) (-3.2.3) e) none
Answer:x=-3,-2,3
Step-by-step explanation:
Given equation of polynomial is
[tex]x^{3}+2x^2-9x-18=0[/tex]
taking [tex]x^3[/tex] and -9x together and remaining together we get
[tex]x^3-9x+2x^2-18=0[/tex]
[tex]x\left ( x^2-9\right )+2\left ( x^2-9\right )[/tex]
[tex]x\left ( \left ( x+3\right )\left ( x-3\right )\right )+2\left ( \left ( x+3\right )\left ( x-3\right )\right )[/tex]
[tex]taking \left ( x+3\right )\left ( x-3\right ) as common[/tex]
[tex]\left ( x+2\right )\left ( x+3\right )\left ( x-3\right )=0[/tex]
therefore
x=-3,-2,3
Please help me with this
Answer: first option.
Step-by-step explanation:
By definition, the measure of any interior angle of an equilateral triangle is 60 degrees.
Then, we can find the value of "y". This is:
[tex]2y+6=60\\\\y=\frac{54}{2}\\\\y=27[/tex]
Since the three sides of an equilateral triangle have the same length, we can find the value of "x". This is:
[tex]x+4=2x-3\\\\4+3=2x-x\\\\x=7[/tex]
According to a recent study, 9.3% of high school dropouts are 16- to 17-year-olds. In addition, 6.3% of high school dropouts are white 16- to 17-year-olds. What is the probability that a randomly selected dropout is white, given that he or she is 16 to 17 years old?
Answer: [tex]\dfrac{21}{31}[/tex]
Step-by-step explanation:
Let A represents the event that high school dropouts are 16- to 17-year-olds and B be the event that high school dropouts are white.
Given : The probability of high school dropouts are 16- to 17-year-olds :[tex]P(A)=0.093[/tex]
The probability of of high school dropouts are white 16- to 17-year-olds :
[tex]P(A\cap B)=0.063[/tex]
Then , the conditional probability that a randomly selected dropout is white, given that he or she is 16 to 17 years old is given by :-
[tex]P(B|A)=\dfrac{P(A\cap B)}{P(A)}\\\\\Rightarrow\ P(B|A)=\dfrac{0.063}{0.093}=\dfrac{21}{31}[/tex]
The linear correlation between violent crime rate and percentage of the population that has a cell phone is minus 0.918 for years since 1995. Do you believe that increasing the percentage of the population that has a cell phone will decrease the violent crime rate? What might be a lurking variable between percentage of the population with a cell phone and violent crime rate? Will increasing the percentage of the population that has a cell phone decrease the violent crime rate? Choose the best option below. No Yes
Answer:
The correct option is No and lurking variable is "Economy".
Step-by-step explanation:
Consider the provided information.
It is given that the linear correlation between violent crime rate and percentage of the population that has a cell phone is minus 0.918 for years since 1995.
There is a lurking variable between the proportion of the cell phone population and the rate of violent crime, the correlation between the two factors does not indicate causation.
Once the economy become stronger, crime rate tends to decrease.
Thus the correct option is NO.
Now, we need to identify the lurking variable between percentage of the population with a cell phone and violent crime rate.
Lurking variable is the variable which is unknown and not controlled.
Here, if we observe we can identify that economy plays a vital role. If economy become stronger, crime rate tends to decrease and population are better to buy phones. Also it is difficult to controlled over it.
Thus, the lurking variable is "Economy".
The number of measles cases increased 13.6 % to 57 cases this year, what was the number of cases prior to the increase? (Express your answer rounded correctly to the nearest whole number)
Answer:
The number of cases prior to the increase is 50.
Step-by-step explanation:
It is given that the number of measles cases increased by 13.6% and the number of cases after increase is 57.
We need to find the number of cases prior to the increase.
Let x be the number of cases prior to the increase.
x + 13.6% of x = 57
[tex]x+x\times \frac{13.6}{100}=57[/tex]
[tex]x+0.136x=57[/tex]
[tex]1.136x=57[/tex]
Divide both the sides by 1.136.
[tex]\frac{1.136x}{1.136}=\frac{57}{1.136}[/tex]
[tex]x=50.176[/tex]
[tex]x\approx 50[/tex]
Therefore the number of cases prior to the increase is 50.
*Asymptotes*
g(x) =2x+1/x-3
Give the domain and x and y intercepts
Answer: Assuming the function is [tex]g(x)=\frac{2x+1}{x-3}[/tex]:
The x-intercept is [tex](\frac{-1}{2},0)[/tex].
The y-intercept is [tex](0,\frac{-1}{3})[/tex].
The horizontal asymptote is [tex]y=2[/tex].
The vertical asymptote is [tex]x=3[/tex].
Step-by-step explanation:
I'm going to assume the function is: [tex]g(x)=\frac{2x+1}{x-3}[/tex] and not [tex]g(x)=2x+\frac{1}{x}-3[/tex].
So we are looking at [tex]g(x)=\frac{2x+1}{x-3}[/tex].
The x-intercept is when y is 0 (when g(x) is 0).
Replace g(x) with 0.
[tex]0=\frac{2x+1}{x-3}[/tex]
A fraction is only 0 when it's numerator is 0. You are really just solving:
[tex]0=2x+1[/tex]
Subtract 1 on both sides:
[tex]-1=2x[/tex]
Divide both sides by 2:
[tex]\frac{-1}{2}=x[/tex]
The x-intercept is [tex](\frac{-1}{2},0)[/tex].
The y-intercept is when x is 0.
Replace x with 0.
[tex]g(0)=\frac{2(0)+1}{0-3}[/tex]
[tex]y=\frac{2(0)+1}{0-3}[/tex]
[tex]y=\frac{0+1}{-3}[/tex]
[tex]y=\frac{1}{-3}[/tex]
[tex]y=-\frac{1}{3}[/tex].
The y-intercept is [tex](0,\frac{-1}{3})[/tex].
The vertical asymptote is when the denominator is 0 without making the top 0 also.
So the deliminator is 0 when x-3=0.
Solve x-3=0.
Add 3 on both sides:
x=3
Plugging 3 into the top gives 2(3)+1=6+1=7.
So we have a vertical asymptote at x=3.
Now let's look at the horizontal asymptote.
I could tell you if the degrees match that the horizontal asymptote is just the leading coefficient of the top over the leading coefficient of the bottom which means are horizontal asymptote is [tex]y=\frac{2}{1}[/tex]. After simplifying you could just say the horizontal asymptote is [tex]y=2[/tex].
Or!
I could do some division to make it more clear. The way I'm going to do this certain division is rewriting the top in terms of (x-3).
[tex]y=\frac{2x+1}{x-3}=\frac{2(x-3)+7}{x-3}=\frac{2(x-3)}{x-3}+\frac{7}{x-3}[/tex]
[tex]y=2+\frac{7}{x-3}[/tex]
So you can think it like this what value will y never be here.
7/(x-3) will never be 0 because 7 will never be 0.
So y will never be 2+0=2.
The horizontal asymptote is y=2.
(Disclaimer: There are some functions that will cross over their horizontal asymptote early on.)
7% of items in a shipment are known to be defective. If a sample of 5 items is randomly selected from this shipment, what is the probability that at least one defective item will be observed in this sample? Round your result to 2 significant places after the decimal (For example, 0.86732 should be entered as 0.87).
Answer: 0.30
Step-by-step explanation:
Binomial distribution formula :-
[tex]P(x)=^nC_xp^x(1-p)^{n-x}[/tex], where P(x) is the probability of getting success in x trials , n is the total number of trials and p is the probability of getting success in each trial.
Given : The probability that a shipment are known to be defective= 0.07
If a sample of 5 items is randomly selected from this shipment,then the probability that at least one defective item will be observed in this sample will be :-
[tex]P(X\geq1)=1-P(0)\\\\=1-(^5C_0(0.07)^0(1-0.07)^{5-0})\\\\=1-(0.93)^5=0.3043116307\approx0.30[/tex]
Hence, the probability that at least one defective item will be observed in this sample =0.30
Suppose you just received a shipment of nine televisions. Three of the televisions are defective. If two televisions are randomly selected, compute the probability that both televisions work. What is the probability at least one of the two televisions does not work?
a)
The probability that both televisions work is: 0.42
b)
The probability at least one of the two televisions does not work is:
0.5833
Step-by-step explanation:There are a total of 9 televisions.
It is given that:
Three of the televisions are defective.
This means that the number of televisions which are non-defective are:
9-3=6
a)
The probability that both televisions work is calculated by:
[tex]=\dfrac{6_C_2}{9_C_2}[/tex]
( Since 6 televisions are in working conditions and out of these 6 2 are to be selected.
and the total outcome is the selection of 2 televisions from a total of 9 televisions)
Hence, we get:
[tex]=\dfrac{\dfrac{6!}{2!\times (6-2)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{\dfrac{6!}{2!\times 4!}}{\dfrac{9!}{2!\times 7!}}\\\\\\=\dfrac{5}{12}\\\\\\=0.42[/tex]
b)
The probability at least one of the two televisions does not work:
Is equal to the probability that one does not work+probability both do not work.
Probability one does not work is calculated by:
[tex]=\dfrac{3_C_1\times 6_C_1}{9_C_2}\\\\\\=\dfrac{\dfrac{3!}{1!\times (3-1)!}\times \dfrac{6!}{1!\times (6-1)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{3\times 6}{36}\\\\\\=\dfrac{1}{2}\\\\\\=0.5[/tex]
and the probability both do not work is:
[tex]=\dfrac{3_C_2}{9_C_2}\\\\\\=\dfrac{1}{12}\\\\\\=0.0833[/tex]
Hence, Probability that atleast does not work is:
0.5+0.0833=0.5833
To find the probability that both televisions work, use the combination formula to determine the number of ways to select 2 working televisions out of the total number of televisions. Divide this number by the total number of ways to select 2 televisions.
Explanation:To find the probability that both televisions work, we need to first determine the number of ways we can select 2 televisions out of the 9 available. This can be done using the combination formula, which is C(n, r) = n!/(r!(n-r)!), where n is the total number of items and r is the number of items being selected. In this case, n = 9 and r = 2.
Now we need to determine the number of ways we can select 2 working televisions out of the 6 working televisions. Again, we can use the combination formula with n = 6 and r = 2.
The final step is to divide the number of ways we can select 2 working televisions by the total number of ways we can select 2 televisions to get the probability that both televisions work.
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what is the LCM of 8 and 10 .
Answer:
40
Step-by-step explanation:
8: 8, 16, 24, 32, 40
10: 10 20 30 40
Answer:
The least common multiple of 8 and 10 is 40.
Step-by-step explanation:
8: 8*1=8, 8*2=16, 8*3=24, 8*4=32, 8*5=40
10: 10*1=10, 10*2=20, 10*3=30, 10*4=40
The least common factor of 8 and 10 is 40.
8. 8 + (-2) – 9 – (-7)
A.24
B.-8
C.4
D.-10
Answer:
4
Step-by-step explanation:
8+(-2) is 66-9 is -3-3 -(-7) is -3+7 which is 4Find the decimal form of 2/4
Answer:
Step-by-step explanation:
.5
Answer is provided in image attached.
To fill out a function's ___ ___, you will need to use test numbers before and after each of the function's ___ and asymtopes
A). Sign chart; Values
B). rational equation; values
C). sign chart; zeroes
D). rational equation; zeroes
Answer:
C). sign chart; zeroes
Step-by-step explanation:
A function potentially changes sign at each of its zeros and vertical asymptotes. So, to fill out a sign chart, you need to determine what the sign is on either side of each of these points. You can do that using test numbers, or you can do it by understanding the nature of the zero or asymptote.
Examples:
f1(x) = (x -3) . . . . changes sign at the zero x=3. Is positive for x > 3, negative for x < 3.
f2(x) = (x -4)^2 . . . . does not change sign at the zero x=4. It is positive for any x ≠ 4. This will be true for any even-degree binomial factor.
f3(x) = 1/(x+2) . . . . has a vertical asymptote at x=-2. It changes sign there because the denominator changes sign there.
f4(x) = 1/(x+3)^2 . . . . has a vertical asymptote at x=-3. It does not change sign there because the denominator is of even degree and does not change sign there.
Suppose a random sample of 90 companies taken in 2006 showed that 14 offered high-deductible health insurance plans to their workers. A separate random sample of 120 firms taken in 2007 showed that 30 offered high-deductible health insurance plans to their workers. Based on the sample results, can you conclude that there is a higher proportion of companies offering high-deductible health insurance plans to their workers in 2007 than in 2006? Conduct your hypothesis test at a level of significance alphaequals0.01.
Answer:
Step-by-step explanation:
Given that a random sample of 90 companies taken in 2006 showed that 14 offered high-deductible health insurance plans to their workers. A separate random sample of 120 firms taken in 2007 showed that 30 offered high-deductible health insurance plans to their workers.
H0: p1=p2
Ha: p1 <p2
(Two tailed test at99%)
Difference 14.44 %
Chi-squared 5.883
DF 1
Significance level P = 0.0153
Since p >0.01, our alpha reject null hypothesis.
NO. Based on the sample results, you can not conclude that there is a higher proportion of companies offering high-deductible health insurance plans to their workers in 2007 than in 2006
Use the alternative curvature formula K=|a x v|/|v|^3 to find the curvature of the following parameterized curves. to find the curvature of the following parameterized curve. r(t)=<6+5t^2, t, 0>
Answer:
[tex]k=\frac{10}{(100t^2+1)^{\frac{3}{2}}}[/tex]
Or
[tex]k=\frac{10\sqrt{100t^2+1}}{(100t^2+1)^{2}}[/tex]
Step-by-step explanation:
We want to compute the curvature of the parameterized curve, [tex]r(t)=\:<\:6+5t^2,t,0)\:>\:[/tex] using the alternative formula:
[tex]k=\frac{|a\times v|}{|v|^3}[/tex].
We first compute the required ingredients.
The velocity vector is [tex]v=r'(t)=<\:10t,1,0\:>[/tex]
The acceleration vector is given by [tex]a=r''(t)=<\:10,0,0\:>[/tex]
The magnitude of the velocity vector is [tex]|v|=\sqrt{(10t)^2+1^2+0^2}=\sqrt{100t^2+1}[/tex]
The cross product of the velocity vector and the acceleration vector is
[tex]a\times v=\left|\begin{array}{ccc}i&j&k\\10&0&0\\10t&1&0\end{array}\right|=10k[/tex]
We now substitute ingredients into the formula to get:
[tex]k=\frac{|10k|}{(\sqrt{100t^2+1})^3}[/tex].
[tex]k=\frac{10}{(100t^2+1)^{\frac{3}{2}}}[/tex]
Or
[tex]k=\frac{10\sqrt{100t^2+1}}{(100t^2+1)^{2}}[/tex]
The correct conversion from metric system to household system is
A. 5 ml equals 1 tablespoon
B. 15 ml equals 1 teaspoon
C. 30 ml equals 1 fluid ounce
D. 500 ml equals 1 measuring cup
Answer:
The closest conversion would be C. 30 ml equals 1 fluid ounce , it is only off by 0.43 ml
Step-by-step explanation:
Great question, it is always good to ask away in order to get rid of any doubts you may be having.
The metric system is a decimal system of measurement while the household system is a system of measurement usually found with kitchen utensils. The correct conversions are the following.
5 ml equals 0.33814 tablespoon
15 ml equals 3.04326 teaspoon
29.5735 ml equals 1 fluid ounce
236.588 ml equals 1 measuring cup
So the closest conversion would be C. 30 ml equals 1 fluid ounce , it is only off by 0.43 ml
I hope this answered your question. If you have any more questions feel free to ask away at Brainly.
The correct conversion between the metric and household system provided in the choices is 30ml equals 1 fluid ounce. However, 5ml is equivalent to 1 teaspoon, 15 ml to 1 tablespoon, and 250 ml to 1 measuring cup.
Explanation:The correct conversion from the metric system to the household system among the options given is C. 30 ml equals 1 fluid ounce. The rationale behind this is that 30 ml is universally accepted as being equal to 1 fluid ounce in the household system.
Option A, B and, D are incorrect conversions. More accurate conversions would be: A. 5 ml equals 1 teaspoon; B. 15 ml equals 1 tablespoon; D. 250 ml equals 1 measuring cup.
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Given P(A and B) 0.20, P(A) 0.49, and P(B) = 0.41 are events A and B independent or dependent? 1) Dependent 2) Independent
Answer: The correct option is (1) Dependent.
Step-by-step explanation: For two events, we are given the following values of the probabilities :
P(A ∩ B) = 0.20, P(A) = 0.49 and P(B) = 0.41.
We are to check whether the events A and B are independent or dependent.
We know that
the two events C and D are said to be independent if the probabilities of their intersection is equal to the product of their probabilities.
That is, P(C ∩ D) = P(C) × P(D).
For the given two events A and B, we have
[tex]P(A)\times P(B)=0.49\times0.41=0.2009\neq P(A\cap B)=0.20\\\\\Rightarrow P(A\cap B)\neq P(A)\times P(B).[/tex]
Therefore, the probabilities of the intersection of two events A and B is NOT equal to the product of the probabilities of the two events.
Thus, the events A and B are NOT independent. They are dependent events.
Option (1) is CORRECT.
14. Let R^2 have inner product defined by ((x1,x2), (y,, y2)) 4x1y1 +9x2y2 A. Determine the norm of (-1,2) in this space B. Determine the norm of (3,2) in this space.
The norm of a vector [tex]\vec x[/tex] is equal to the square root of the inner product of [tex]\vec x[/tex] with itself.
a. [tex]\|(-1,2)\|=\sqrt{\langle(-1,2),(-1,2)\rangle}=\sqrt{4(-1)^2+9(2)^2}=\sqrt{40}=2\sqrt{10}[/tex]
b. [tex]\|(3,2)\|=\sqrt{\langle(3,2),(3,2)\rangle}=\sqrt{4(3)^2+9(2)^2}=\sqrt{72}=6\sqrt2[/tex]
Let Z be a standard normal random variable and calculate the following probabilities, drawing pictures wherever appropriate. (Round your answers to four decimal places.) (a) P(0 ≤ Z ≤ 2.38) .4913 (b) P(0 ≤ Z ≤ 1) .3413 (c) P(−2.70 ≤ Z ≤ 0) .4965 (d) P(−2.70 ≤ Z ≤ 2.70) .9931 (e) P(Z ≤ 1.62) .9474 (f) P(−1.55 ≤ Z) .9394 (g) P(−1.70 ≤ Z ≤ 2.00) .9327 (h) P(1.62 ≤ Z ≤ 2.50) .0464 (i) P(1.70 ≤ Z) .0445 (j) P(|Z| ≤ 2.50) .9876
The correct answers are:
(a) [tex]P(0\leq Z\leq 2.38)=0.4913[/tex]
(b) [tex]P(0\leq Z\leq 1)=0.3413[/tex]
(c) [tex]P(-2.70\leq Z\leq 0)=0.4965[/tex]
(d) [tex]P(-2.70\leq Z\leq 2.70)=0.9926[/tex]
(e) [tex]P(Z\leq 1.62)=0.9474[/tex]
(f) [tex]P(-1.55\leq Z)=0.9394[/tex]
(g) [tex]P(-1.70\leq Z\leq 2.00)=0.9326[/tex]
(h) [tex]P(1.62\leq Z\leq 2.50)=0.0457[/tex]
(i) [tex]P(1.70\leq Z)=0.0446[/tex]
(j) [tex]P(|Z|\leq 2.50)=1.9862[/tex]
Let's calculate these probabilities step by step using the standard normal distribution table (also known as the z-table).
For reference, the standard normal distribution table provides the probabilities associated with the standard normal random variable (Z), which has a mean of [tex]0[/tex] and a standard deviation of [tex]1[/tex].
We'll use the standard normal distribution table to find the probabilities corresponding to the given Z-values.
(a [tex]P(0\leq Z\leq 2.38)[/tex] From the z-table, [tex]P(Z\leq 2.38)=0.9913 P(0\leq Z\leq 2.38)=0.9913-0.5=0.4913[/tex] (subtracting the cumulative probability up to [tex]0[/tex] from the cumulative probability up to [tex]2.38[/tex])
(b) [tex]P(0\leq Z\leq 1)[/tex] From the z-table, [tex]P(Z\leq 1)=0.8413 P(0\leq Z\leq 1)=0.8413-0.5=0.3413[/tex]
(c) [tex]P(-2.70\leq Z\leq 0)[/tex] From the z-table, [tex]P(Z\leq 0)=0.5[/tex] and [tex]P(Z\leq -2.70)=0.0035 P(-2.70\leq Z\leq 0)=0.5-0.0035=0.4965[/tex]
(d) [tex]P(-2.70\leq Z\leq 2.70)[/tex] From the z-table, [tex]P(Z\leq 2.70)=0.9961[/tex] and [tex]P(Z\leq -2.70)=0.0035 P(-2.70\leq Z\leq 2.70)=0.9961-0.0035=0.9926[/tex]
(e) [tex]P(Z\leq 1.62)[/tex] From the z-table, [tex]P(Z\leq 1.62)=0.9474[/tex]
(f) [tex]P(-1.55\leq Z)[/tex] From the z-table, [tex]P(Z\leq -1.55)=0.0606 P(-1.55\leq Z)=1-0.0606=0.9394[/tex] (subtracting the cumulative probability up to [tex]-1.55[/tex] from [tex]1[/tex])
(g [tex]P(-1.70\leq Z\leq 2.00)[/tex] From the z-table, [tex]P(Z\leq 2.00)=0.9772[/tex] and [tex]P(Z\leq -1.70)=0.0446 P(-1.70\leq Z\leq 2.00)=0.9772-0.0446=0.9326[/tex]
(h) [tex]P(1.62\leq Z\leq 2.50)[/tex] From the z-table, [tex]P(Z2.50)=0.9931[/tex] and [tex]P(Z\leq 1.62)=0.9474 P(1.62\leq Z\leq 2.50)=0.9931-0.9474=0.0457[/tex]
(i) [tex]P(1.70\leq Z)[/tex][tex]From the z-table, P(Z\leq 1.70)=0.9554 , P(1.70\leq Z)=1-0.9554=0.0446[/tex] (subtracting the cumulative probability up to [tex]1.70[/tex] from [tex]1[/tex])
(j) [tex]P(|Z|\leq 2.50)[/tex] Since the standard normal distribution is symmetric, [tex]P(|Z|\leq 2.50)=2*P(0\leq Z\leq 2.50)=2*0.9931=1.9862[/tex] (multiply by [tex]2[/tex] because the probability of Z being between [tex]-2.50[/tex] and [tex]2.50[/tex] is twice the probability of Z being between [tex]0[/tex] and [tex]2.50[/tex] )
Let Z be a standard normal random variable and calculate the following probabilities, drawing pictures wherever appropriate. (Round your answers to four decimal places.)
(a) [tex]P(0 \leq Z \leq 2.38) .4913[/tex]
(b) [tex]P(0 \leq Z\leq 1) .3413[/tex]
(c) [tex]P(-2.70 \leq Z \leq 0) .4965[/tex]
(d) [tex]P(-2.70 \leq Z \leq 2.70) .9931[/tex]
(e) [tex]P(Z \leq 1.62) .9474[/tex]
(f) [tex]P(-1.55 \leq Z) .9394[/tex]
(g) [tex]P(-1.70 \leq Z \leq 2.00) .9327[/tex]
(h) [tex]P(1.62 \leq Z \leq 2.50) .0464[/tex]
(i) [tex]P(1.70 \leq Z) .0445[/tex]
(j) [tex]P(|Z| \leq 2.50) .9876[/tex]
A basketball team sells tickets that cost $10, $20, or, for VIP seats, $30. The team has sold 3142 tickets overall. It has sold 207 more $20 tickets than $10 tickets. The total sales are $59,670. How many tickets of each kind have been sold?
Answer:
1,084 tickets were sold that cost $10
1,291 tickets were sold that cost $20
767 tickets were sold that cost $30
Step-by-step explanation:
Let
x ----> the number of tickets that cost $10 sold
y ----> the number of tickets that cost $20 sold
z ----> the number of tickets that cost $30 sold
we know that
x+y+z=3,142 -----> equation A
10x+20y+30z=59,670 ----> equation B
y=x+207 ----> equation C
substitute equation C in equation A and equation B
x+(x+207)+z=3,142 ----> 2x+z=2,935 ----> equation D
10x+20(x+207)+30z=59,670 ---> 30x+30z=55,530 ----> equation E
Solve the system of equations D and E by graphing
The solution is the intersection point both graphs
The solution is the point (1,084,767)
so
x=1,084, z=767
see the attached figure
Find the value of y
y=x+207 ----> y=1,084+207=1,291
therefore
1,084 tickets were sold that cost $10
1,291 tickets were sold that cost $20
767 tickets were sold that cost $30
Thirty-five more than the limit of weight (w) in an elevator is greater then 1,050 pounds.
A. Model with an inequality.
B. Solve the inequality.
C. Where would the solution for the inequality be graphed on the number line?
Answer:
Part A) [tex]w+35 > 1,050[/tex]
Part B) [tex]w > 1,015\ pounds[/tex]
Part C) The graph in the attached figure
Step-by-step explanation:
Part A) Model with an inequality.
Let
w -----> the limit of weight in an elevator
we know that
The inequality that represent this situation is
[tex]w+35 > 1,050[/tex]
Part B) Solve the inequality
[tex]w+35 > 1,050[/tex]
solve for w
Subtract 35 sides
[tex]w+35-35 > 1,050-35[/tex]
[tex]w > 1,015\ pounds[/tex]
All real numbers greater than 1,015 pounds
Part C) Where would the solution for the inequality be graphed on the number line?
[tex]w > 1,015\ pounds[/tex]
The solution is the interval -------> (1,015,∞)
In a number line is the shaded area at right of the number x=1.015 (open circle)
see the attached figure