Answer:
[tex]\text{Density}=\frac{0.556\text{ Grams}}{\text{ cm}^3}[/tex]
Step-by-step explanation:
We are asked to find the density of a cylinder whose volume is 1000.0 cubic cm and mass is 556 grams.
[tex]\text{Density}=\frac{\text{Mass}}{\text{Volume}}[/tex]
Substitute the given values:
[tex]\text{Density}=\frac{556\text{ Grams}}{1000.0\text{ cm}^3}[/tex]
[tex]\text{Density}=\frac{0.556\text{ Grams}}{\text{ cm}^3}[/tex]
Therefore, the density of the metallic cylinder is 0.556 grams per cubic centimeter.
• 10 17 (10 complete) Find an equation of the line in the form ax + byc whose x-interceptis 4 and y-intercept is 2 where a, b, and care integers with no factor common to all three and a 20 The equation of the line is IN (Type an equation)
Answer:
y+0.5x-2=0
Step-by-step explanation:
Given,
X-intercept of line = 4
So, the line intersect the x axis at (4,0)
Y-intercept of the line = 2
So, the line intersect the y axis at (0,2)
So, the slope of the line can be given by
[tex]m\ =\ \dfrac{y_2-y_1}{x_2-x_1}[/tex]
[tex]=\ \dfrac{2-0}{0-4}[/tex]
= -0.5
Hence, the equation of line can be given by
y=mx+c
where, m=slope of the line
c= y-intercept of line
So, after putting the value of m and c in above equation, the equation of line will be
y = -0.5x + 2
=> y+0.5x-2=0
So, the equation of line will be y+0.5x-2=0.
You buy g gallons of gasoline at $3.05 per gallon and pay $36.60. Write an equation to find the number of gallons purchased. Then find the number of gallons of gasoline that you purchased.
Answer:
The equation to find the number of gallons purchased is:
[tex]C(n) = 3.05n[/tex]
You purchased 12 gallons of gasoline
Step-by-step explanation:
This problem can be modeled by the following first order function
[tex]C(n) = P_{G}n[/tex]
Where C(n) is the cost in function to the number of gallons, P is the price of the gallon and n is the number of gallons
The problem states that the price of gasoline is $3.05 per gallon, so P = 3.05
The equation to find the number of gallons purchased is:
[tex]C(n) = 3.05n[/tex]
If you pay $36.60, you have C = 36.60, and want to find n, so:
[tex]36.60 = 3.05n[/tex]
[tex]n = \frac{36.60}{3.05}[/tex]
[tex]n = 12[/tex]
You purchased 12 gallons of gasoline
Find the Cartesian Equation of the plane passing through P(8, -2,0) and perpendicular to a- 5i+3j-k What is the distance of this plane to the point 0(2,2, 2)? (a) (b)
Answer:
equation of plane, 5x+3y-z-36=0
Distance of point (2,2,2) from plane = 4.05 units
Step-by-step explanation:
Given,
Plane passing through the point = (8, -2, 0)
Let's say, [tex]x_1\ =\ 8[/tex]
[tex]y_1\ =\ -2[/tex]
[tex]z_1\ =\ 0[/tex]
Plane perpendicular to the vector, a= 5i + 3j- k
Since, the vector is perpendicular to the plane, hence the equation of plane can be given by
[tex](5i + 3j- k).((x-x_1)i+(y-y_1)j+(z- z_1)k)=\ 0[/tex]
[tex]=>(5i + 3j- k).((x-8)i+(y+2)j+(z-0)k)=\ 0[/tex]
[tex]=>\ 5(x-8)+3(y+2)-z=0[/tex]
[tex]=>\ 5x\ -\ 40\ +\ 3y\ +\ 6\ -\ z\ =\ 0[/tex]
[tex]=>\ 5x\ +\ 3y\ -\ z\ -\ 36\ =\ 0[/tex]
Hence, the equation of plane can be given by, 5x+3y-z-36=0
Now, we have to calculate the distance of the point O(2,2,2) from the plane 5x+3y-z-36=0
Let's say,
a= 5, b= 3, c= -1, d=-36
[tex]x_0=2,\ y_0=2,\ z_0=2[/tex]
So, distance of a point from the plane can be given by,
[tex]d=\dfrac{ax_0+by_0+cz_0+d}{\sqrt{a^2+b^2+c^2}}[/tex]
[tex]=\dfrac{\left |5\times 2+3\times 2+(-1)\times 2-36\right |}{\sqrt{5^2+3^2+(-1)^2}}[/tex]
[tex]=\dfrac{24}{\sqrt{35}}[/tex]
= 4.05 units
So, the distance of the point O(2,2,2) from the given plane will be 4.05 units.
Let x,y \epsilon R. Use mathmatical induction to prove the identity.
x^{n+1}-y^{n+1}=(x-y)(x^{n}+x^{n-1}y+...+xy^{n-1}+y^{n})
Step-by-step explanation:
We will prove by mathematical induction that, for every natural n,
[tex](x-y)(x^{n}+x^{n-1}y+...+xy^{n-1}+y^{n})=x^{n+1}-y^{n+1}[/tex]
We will prove our base case (when n=1) to be true:
Base case:
[tex](x-y)(x^{n}+x^{n-1}y+...+xy^{n-1}+y^{n})=(x-y)(x^{1}+y^{1})=x^2-y^2=x^{1+1}-y^{1+1}[/tex]
Inductive hypothesis:
Given a natural n,
[tex]x^{n+1}-y^{n+1}=(x-y)(x^{n}+x^{n-1}y+...+xy^{n-1}+y^{n})[/tex]
Now, we will assume the inductive hypothesis and then use this assumption, involving n, to prove the statement for n + 1.
Inductive step:
Observe that, for y=0 the conclusion is clear. Then we will assume that [tex]y\neq 0.[/tex]
[tex](x-y)(x^{n+1}+x^{n}y+...+xy^{n}+y^{n+1})=(x-y)y(\frac{x^{n+1}}{y}+x^{n}+...+xy^{n-1}+y^{n})=(x-y)y(\frac{x^{n+1}}{y})+(x-y)y(x^{n}+...+xy^{n-1}+y^{n})=(x-y)y(\frac{x^{n+1}}{y})+y(x^{n+1}-y^{n+1})=(x-y)x^{n+1}+y(x^{n+1}-y^{n+1})=x^{n+2}-yx^{n+1}+yx^{n+1}-y^{n+2}=x^{n+2}-y^{n+2}\\[/tex]
With this we have proved our statement to be true for n+1.
In conlusion, for every natural n,
[tex](x-y)(x^{n}+x^{n-1}y+...+xy^{n-1}+y^{n})=x^{n+1}-y^{n+1}[/tex]
Use the Babylonian method of false position to solve the following
problem,
taken from a clay tablet found in Susa: Let the width of a
rectangle measure a
quarter less than the length. Let 40 be the length of the diagonal.
What are
the length and width? Begin with the assumption that 1 (or 60) is
the length
of the rectangle.
Answer:
length: 32width: 24Step-by-step explanation:
Assume a solution
Assume that 60 is the length. The width is then 1/4 less, or 60 -60/4 = 45.
The diagonal of this rectangle is found using the Pythagorean theorem:
d = √(60² +45²) = √5625 = 75
Make the adjustment
This is a factor of 75/40 larger than the actual diagonal, so the actual dimensions must be 40/75 = 8/15 times those we assumed.
length = (8/15)×60 = 32
width = (8/15)×45 = 24
The length and width of the rectangle are 32 and 24, respectively.
_____
Comment on this solution method
This method is suitable for problems where variables are linearly related. If we were concerned with the area, for example, instead of the diagonal, we would have to adjust the linear dimensions by the square root of the ratio of desired area to "false" area.
Final answer:
This detailed answer explains how to use the Babylonian method of false position to find the length and width of a rectangle.
Explanation:
The Babylonian method of false position involves making an initial assumption about the solution and then iteratively honing in on the correct answer.
Step-by-step:
Let's start with an assumption: Length = 60. Then calculate the width based on the given conditions.
Adjust your assumption based on the calculated width until you reach the correct solution.
By using this method, you can find the length and width of the rectangle described in the problem.
The amount of money spent on red balloon in a certain college town when the football team is in town is a normal random variable with mean $50000 and a standard deviation of $3000. What proportion of home football game days in this town is less than $45000 worth of red balloons sold?
Answer: 0.0475
Step-by-step explanation:
Given : The amount of money spent on red balloon in a certain college town when the football team is in town is a normal random variable with
[tex]\mu=\$50000[/tex] and [tex]\sigma=\$3000[/tex]
Let x be the random variable that represents the amount of money spent on red balloon.
Using formula [tex]z=\dfrac{x-\mu}{\sigma}[/tex], the z-score corresponding to x= 45000 will be :_
[tex]z=\dfrac{45000-50000}{3000}\approx-1.67[/tex]
Now, by using the standard normal distribution table for z, we have
P value : [tex]P(z<-1.67)=1-P(z<1.67)=1-0.9525=0.0475[/tex]
∴The proportion of home football game days in this town is less than $45000 worth of red balloons sold = 0.0475
A ream of paper contains 500 sheets of paper. Norm has 373 sheets of paper left from a ream. Express the portion of a ream Norm has as a fraction and as a decimal.
Answer: In fraction : [tex]\dfrac{373}{500}[/tex]
In Decimal : 0.746
Step-by-step explanation:
Given : A ream of paper contains 500 sheets of paper.
Norm has 373 sheets of paper left from a ream.
Then, the fraction of ream Norm has will be :-
[tex]\dfrac{\text{Number of sheets left from ream }}{\text{Total sheets in a ream }}\\\\=\dfrac{373}{500}[/tex]
To convert in decimal we divide 373 by 500, we get 0.746.
Hence, The portion of a ream Norm has as = [tex]\dfrac{373}{500}[/tex] or 0.746
Estimate the product or quotient.
4/7 x 1/6
Answer: I'm sure its 2/21
Step-by-step explanation: you just need to multiply cross sides then divide by any number that works on both of them.
I hope that I answer your question.
What sine function represents an amplitude of 4, a period of pi over 2, no horizontal shift, and a vertical shift of −3?
f(x) = −3 sin 4x + 4
f(x) = 4 sin 4x − 3
f(x) = 4 sin pi over 2x − 3
f(x) = −3 sin pi over 2x + 4
Answer:
[tex]f(x) = 4sin(\frac{\pi}{2}x) - 3[/tex], the third one
Step-by-step explanation:
Explaining the sine function:
The sine function is defined by:
[tex]S = Asin(p(x - x_{0})) + V[/tex]
In which A is the amplitude, [tex]p = \frac{2\pi}{N}[/tex] is the period, [tex]x_{0}[/tex] is the horizontal shift and V is the vertical shift.
So, in your problem:
The amplitude is 4, so A = 4.
The period is [tex]\frac{\pi}{2}[/tex], so [tex]p = \frac{\pi}{2}[/tex].
There is no horizontal shift, so [tex]x_{0} = 0[/tex].
The vertical shift is −3, so V = -3.
The sine function that represents these following conditions is
[tex]f(x) = 4sin(\frac{\pi}{2}x) - 3[/tex], the third one
A researcher wants to compare student loan debt for students who attend four-year public universities with those who attend four –year private universities. She plans to take a random sample of 100 recent graduates of public universities and 100 recent graduates of private universities. Which type of random sampling is utilized in her study design?
Answer:
A simple random sample.
Step-by-step explanation:
A simple random sample is an statistical sample in which each member of a group has the same probability of being chosen. Since the researcher doesn't really have specific characteristics added to the sample other than being from public or private universities, this would be a simple random sample.
A researcher wants to provide a rabbit exactly 162 units ofprotein, 72 units of carbohydrates, and 30 units of vitamin A. The rabbit is fed three types of food. Each gram of Food A has 5 units of protein, 2 units of carbohydrates, and 1unit of vitamin A. Each gram of Food B contains 11 units of protein, 5 units ofcarbohydrates, and 2 units of vitamin A. Each gram of Food C contains 23 units of protein, 11 units of carbohydrates, and 4 units of vitamin A. How many grams of each food should the rabbit be fed?
Answer:
The rabbit should be fed:
[tex]6 + 2z[/tex] grams of food A
[tex]12 - 3z[/tex] grams of food B
[tex]z[/tex] grams of food C
For [tex]z \leq 4[/tex].
Step-by-step explanation:
This can be solved by a system of equations.
I am going to say that x is the number of grams of food A, y is the number of grams of food B and z is the number of grams of Food C.
The problem states that:
A researcher wants to provide a rabbit exactly 162 units of protein:
There are 5 units of protein in each gram of food A, 11 units of protein in each gram of food B and 23 units of protenin in each gram of food C. So
[tex]5x + 11y + 23z = 162[/tex]
A researcher wants to provide a rabbit exactly 72 units of carbohydrates:
There are 2 units of carbohydrates in each gram of food A, 5 units of carbohydrates in each gram of food B and 11 units of carbohydrates in each gram of food C. So:
[tex]2x + 5y + 11z = 72[/tex]
A researcher wants to provide a rabbit exactly 30 units of Vitamin A:
There is 1 unit of Vitamin A in each gram of food A, 2 units of Vitamin A in each gram of food B and 4 units of Vitamin A in each gram of food C. So:
[tex]x + 2y + 4z = 30[/tex].
We have to solve the following system of equations:
[tex]5x + 11y + 23z = 162[/tex]
[tex]2x + 5y + 11z = 72[/tex]
[tex]x + 2y + 4z = 30[/tex].
I think that the easier way to solve this is reducing the augmented matrix of this system.
This system has the following augmented matrix:
[tex]\left[\begin{array}{cccc}5&11&23&162\\2&5&11&72\\1&2&4&30\end{array}\right][/tex]
To help reduce this matrix, i am going to swap the first line with the third
[tex]L_{1} <-> L_{3}[/tex]
Now we have the following matrix:
[tex]\left[\begin{array}{cccc}1&2&4&30\\2&5&11&72\\5&11&23&162\end{array}\right][/tex]
Now i am going to do these following operations, to reduce the first row:
[tex]L_{2} = L_{2} - 2L_{1}[/tex]
[tex]L_{3} = L_{3} - 5L_{1}[/tex]
Now we have
[tex]\left[\begin{array}{cccc}1&2&4&30\\0&1&3&12\\0&1&3&12\end{array}\right][/tex]
Now, to reduce the second row, i do:
[tex]L_{3} = L_{3} - L_{2}[/tex]
The matrix is:
[tex]\left[\begin{array}{cccc}1&2&4&30\\0&1&3&12\\0&0&0&0\end{array}\right][/tex]
This means that z is a free variable, so we are going to write y and x as functions of z.
From the second line, we have
[tex]y + 3z = 12[/tex]
[tex]y = 12 - 3z[/tex]
From the first line, we have
[tex]x + 2y + 4z = 30[/tex]
[tex]x + 2(12 - 3z) + 4z = 30[/tex]
[tex]x + 24 - 6z + 4z = 30[/tex]
[tex]x = 6 + 2z[/tex]
Our solution is: [tex]x = 6 + 2z, y = 12 - 3z, z = z[/tex].
However, we can not give a negative number of grams of a food. So
[tex]y \geq 0[/tex]
[tex]12 - 3z \geq 0[/tex]
[tex]-3z \geq -12 *(-1)[/tex]
[tex]3z \leq 12[/tex]
[tex]z \leq 4[/tex]
The rabbit should be fed:
[tex]6 + 2z[/tex] grams of food A
[tex]12 - 3z[/tex] grams of food B
[tex]z[/tex] grams of food C
For [tex]z \leq 4[/tex].
U fill containers with an average of 12 ounces of oil with
astanderd deviation of .25 ounces. take a random sample of 40
cans,what is the probability that the sample mean,X is grater
then12.05.
Answer: 0.1038
Step-by-step explanation:
We assume that oil in each container is filled will normal distribution.
Given : Population mean : [tex]\mu=12[/tex]
Standard deviation: [tex]\sigma=0.25[/tex]
Sample size : [tex]n=40[/tex]
Let x be the random variable that denotes the amount of oil filled in container.
z-score : [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
For x= 12.05
[tex]z=\dfrac{12.05-12}{\dfrac{0.25}{\sqrt{40}}}=1.26491106407\approx1.26[/tex]
Now by using the standard normal table for z, we have the probability that the sample mean,X is greater then 12.05:-
[tex]P(z>1.26)=1-0.8961653=0.1038347\approx0.1038[/tex]
Hence, the probability that the sample mean,X is greater then 12.05 = 0.1038
4 1/6 divided by 1 1/3
Answer:
[tex]\frac{25}{8}[/tex]
Step-by-step explanation:
A fraction is a part of a whole .
A proper fraction is a fraction whose numerator is less than denominator .
For example [tex]\frac{2}{3}\,,\,\frac{3}{4}[/tex]
An improper fraction is a fraction whose numerator is greater than denominator .
For example [tex]\frac{5}{4}\,,\,\frac{8}{7}[/tex]
A mixed fraction is made up of whole number and a proper fraction .
Given : [tex]4\frac{1}{6}\,,1\frac{1}{3}[/tex]
Solution :
[tex]4\frac{1}{6}=\frac{4\times 6+1}{6}=\frac{25}{6}\\1\frac{1}{3}=\frac{3\times 1+1}{3}=\frac{4}{3}[/tex]
We need to divide [tex]4\frac{1}{6}[/tex] by [tex]1\frac{1}{3}[/tex] .
[tex]4\frac{1}{6}\div 1\frac{1}{3}=\frac{25}{6}\div\frac{4}{3} \\\Rightarrow 4\frac{1}{6}\div 1\frac{1}{3}= \frac{25}{6}\times \frac{3}{4}=\frac{25}{8}[/tex]
Consider a bag containing five red marbles, two green ones, one transparent one, four yellow ones, and two orange ones How many possible sets of five marbles are there in which none of them are red or green? sets Need Help? Tente Tutor
Answer: 21
Step-by-step explanation:
Given : A bag containing five red marbles, two green ones, one transparent one, four yellow ones, and two orange ones .
Total marbles other than red or green = 1+4+2=7
Now, the number of combinations to select five marbles from the set of 7 will be :-
[tex]7C_5=\dfrac{7!}{5!(7-5)!}=\dfrac{7\times6\times5!}{5!\times2!}=21[/tex]
Hence, the number of possible sets of five marbles are there in which none of them are red or green =21
Consider strings of length 10 which contain only letters from the set {A, E, I, O, U} and digits from {1, 3, 5, 7, 9}. Suppose repitition of letters is not allowed.
How many different strings are there?
How many different strings are there if the letters, i.e. A, E, I, O, U and the digits, i.e. 1,3, 5, 7, 9 must alternate?
How many different strings are there if all five letters must be adjacent in each string?
Answer:
a) There are 3,628,800 different strings.
b) There are 28,800 different strings if the letters ad digits must alternate.
c)There are 86,400 different string if all five letters must be adjacent in each string.
Step-by-step explanation:
There are 10 digits.
Our strings have the following format:
C1 - C2 - C3 - C4 - C5 - C6 - C7 - C8 - C9 - C10
repitition of letters is not allowed.
a) How many different strings are there?
C1 can be any of the 10, C2 can be 9, C3 can be 8, ...
So in total there are:
[tex]T = 10*9*8*7*6*5*4*3*2*1 = 3,628,800[/tex]
There are 3,628,800 different strings.
b) How many different strings are there if the letters, i.e. A, E, I, O, U and the digits, i.e. 1,3, 5, 7, 9 must alternate?
There are the following possiblities:
5(l) - 5(d) - 4(l) - (4d) - ...
Or
5(d) - 5(l) - 4(d) - 4(l) - ...
So:
[tex]T = 2*(5*5*4*4*3*3*2*2*1*1) = 28,800[/tex]
There are 28,800 different strings if the letters ad digits must alternate.
c) How many different strings are there if all five letters must be adjacent in each string?
L - L - L - L - L - D - D - D - D - D
D - L - L - L - L - L - D - D - D - D
D - D - L - L - L - L - L - D - D - D
D - D - D - L - L - L - L - L - D - D
D - D - D - D - L - L - L - L - L - D
D - D - D - D - D - L - L - L - L - L
There are [tex]T = 6*(5*5*4*4*3*3*2*2*1*1) = 86,400[/tex]
There are 86,400 different string if all five letters must be adjacent in each string.
Is 57.3 a whole number
Answer:
No, whole numbers are numbers that are whole, which would mean that they are not a decimal or a fraction. This is because a fraction or a decimal is a portion of a number, which would mean that the decimal or fraction is not complete/ not whole!
Eric wants to estimate the percentage of elementary school children who have a social media account. He surveys 450 elementary school children and finds that 280 have a social media account. Identify the values needed to calculate a confidence interval at the 99% confidence level. Then find the confidence interval.
The confidence interval for population mean is given by :-
[tex]\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
, where [tex]\hat{p}[/tex] is the sample proportion, n is the sample size , [tex]z_{\alpha/2}[/tex] is the critical z-value.
The values needed to calculate a confidence interval at the 99% confidence level are :
Given : Significance level : [tex]\alpha:1-0.99=0.01[/tex]
Sample size : n=450
Critical value : [tex]z_{\alpha/2}=2.576[/tex]
Sample proportion: [tex]\hat{p}=\dfrac{280}{450}\approx0.62[/tex]
Now, the 99% confidence level will be :
[tex]\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\\\\=0.62\pm(2.576)\sqrt{\dfrac{0.62(1-0.62)}{450}}\\\\\approx0.62\pm0.023\\\\=(0.62-0.023,\ 0.62+0.023)=(0.597,\ 0.643)[/tex]
Hence, the 99% confidence interval is [tex](0.597,\ 0.643)[/tex]
To calculate a 99% confidence interval for the true proportion, find the values needed and apply the formula. In this case, the confidence interval is between 0.5601 and 0.6811.
1. Sample proportion [tex](\( \hat{p} \))[/tex]:
[tex]\[ \hat{p} = \frac{x}{n} = \frac{280}{450} \approx 0.6222 \][/tex]
2. Margin of error E :
For a 99% confidence level, z = 2.576.
[tex]\[ E = z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]\[ E = 2.576 \times \sqrt{\frac{0.6222 \times (1-0.6222)}{450}} \]\[ E = 2.576 \times \sqrt{\frac{0.6222 \times 0.3778}{450}} \]\[ E = 2.576 \times \sqrt{\frac{0.235052}{450}} \]\[ E \approx 2.576 \times \sqrt{0.00052233} \]\[ E \approx 2.576 \times 0.022854 \]\[ E \approx 0.058893 \][/tex]
3. Confidence interval:
[tex]\[ \text{Confidence interval} = \hat{p} \pm E \]\[ \text{Confidence interval} = 0.6222 \pm 0.0589 \][/tex]
Therefore, the confidence interval for the percentage of elementary school children who have a social media account at the 99% confidence level is approximately (0.5633, 0.6811) .
Two sides of a rectangle are 4cm in length. The other two sides are 6cm in length. What is the perimeter of the rectangle? Include the abbreviation for millimeter as the units.
Answer: 200 mm
Step-by-step explanation:
The perimeter of rectangle is given by :-
[tex]P=2(l+w)[/tex], where l is length and w is width of the rectangle.
Given : Two sides of a rectangle are 4 cm in length. The other two sides are 6 cm in length.
The perimeter of the rectangle will be :_
[tex]P=2(4+6)=2(10)=20\ cm[/tex]
We know that 1 cm = 10 mm
Therefore, perimeter of the rectangle = [tex]20\times10=200\ mm[/tex]
Two simple statements are connected with "AND." You're constructing the truth table of this compound statement. How many rows does the truth table x will have?
Answer:
4 rows
Step-by-step explanation:
It is given that two statements two be connected with AND, the statements may be either true or false.
The output of the AND will be true if both the statements will be true otherwise false.
We can construct the table as follows:
statement 1 statement 2 output
false false false
false true false
true false false
true true true
Hence, the number of rows the truth will have = 4
You deposit $3000 into a money-market savings account which pays 4.8% compounded quarterly, and you make no withdrawals from or further deposits into this account for 3 years. How much money is in your account at the end of those 3 years?
Give answer in dollars rounded to the nearest cent. Do NOT enter "$" sign in answer.
Answer:
$5265.71
Step-by-step explanation:
We have been given that you deposit $3000 into a money-market savings account which pays 4.8% compounded quarterly.
We will use future value formula to solve our given problem.
[tex]FV=C_0\times (1+r)^n[/tex], where,
[tex]C_0=\text{Initial amount}[/tex],
r = Rate of return in decimal form,
n = Number of periods.
[tex]4.8\%=\frac{4.8}{100}=0.048[/tex]
[tex]n=3\times 4=12[/tex]
[tex]FV=\$3,000\times (1+0.048)^{12}[/tex]
[tex]FV=\$3,000\times (1.048)^{12}[/tex]
[tex]FV=\$3,000\times 1.7552354909370114[/tex]
[tex]FV=\$5265.7064\approx \$5265.71[/tex]
Therefore, there will be $5265.71 in your account at the end of those 3 years.
Let A be the matrix: [130 024 154 11-4] Find a basis for the nullspace of A.
Answer:
The basis for the null space of A is [tex]{\left[\begin{array}{c}-1&-1&1&0\end{array}\right],\left[\begin{array}{c}-1&1&0&1\end{array}\right]}[/tex]
Step-by-step explanation:
The first step is to find the reduced row echelon form of the matrix:[tex]\left[\begin{array}{cccc}1&0&1&1\\3&2&5&1\\0&4&4&-4\end{array}\right][/tex]
Make zeros in column 1 except the entry at row 1, column 1. Subtract row 1 multiplied by 3 from row 2 [tex]\left(R_2=R_2-\left(3\right)R_1\right)[/tex][tex]\left[\begin{array}{cccc}1&0&1&1\\0&2&2&-2\\0&4&4&-4\end{array}\right][/tex]
Make zeros in column 2 except the entry at row 2, column 2. Subtract row 2 multiplied by 2 from row 3 [tex]\left(R_3=R_3-\left(2\right)R_2\right)[/tex][tex]\left[\begin{array}{cccc}1&0&1&1\\0&2&2&-2\\0&0&0&0\end{array}\right][/tex]
Multiply the second row by 1/2 [tex]\left(R_2=\left(1/2\right)R_2\right)[/tex][tex]\left[\begin{array}{cccc}1&0&1&1\\0&1&1&-1\\0&0&0&0\end{array}\right][/tex]
2. Convert the matrix equation back to an equivalent system and solve the matrix equation
[tex]1x_{1} +x_{3} +1x_{4}=0\\ 1x_{2} +x_{3} -1x_{4}=0\\0=0[/tex]
[tex]\left[\begin{array}{cccc}1&0&1&1\\0&1&1&-1\\0&0&0&0\end{array}\right] \left[\begin{array}{c}x_{1} &x_{2} &x_{3}&x_{4} \end{array}\right]=\left[\begin{array}{c}0&0&0\end{array}\right][/tex]
If we take [tex]x_{3}=t, x_{4}=s[/tex] then [tex]x_{1}=-s-t,x_{2}=s-t,x_{3}=t,x_{4}=s[/tex]
Therefore,
[tex]\boldsymbol{x}=\left[\begin{array}{c}-s-t&s-t&t&s\end{array}\right]=\left[\begin{array}{c}-1&-1&1&0\end{array}\right]t+\left[\begin{array}{c}-1&1&0&1\end{array}\right]s\\\boldsymbol{x}=\left[\begin{array}{c}-1&-1&1&0\end{array}\right]x_{3} +\left[\begin{array}{c}-1&1&0&1\end{array}\right]x_{4}[/tex]
The null space has a basis formed by the set {[tex]{\left[\begin{array}{c}-1&-1&1&0\end{array}\right],\left[\begin{array}{c}-1&1&0&1\end{array}\right]}[/tex]}
If one 20-mL ampul contains 0.5 g of aminophylline, how many milliliters should be administered to provide a 25-mg dose of aminophylline?
Answer:
1mL should be administered to provide a 25-mg dose of aminophylline.
Step-by-step explanation:
The problem states that one 20-mL ampul contains 0.5 g of aminophylline, and asks how many milliliters should be administered to provide a 25-mg dose of aminophylline.
The first step is converting 0.5g to mg.
Each g has 1000mg, so:
1g - 1000mg
0.5g - xmg
x = 1000*0.5
x = 500mg.
Now we have that one 20-mL ampul contains 500mg of aminophylline. How many milliliters should be administered to provide a 25-mg dose of aminophylline?
As the dose increases, so does the quantity of aminophylline. It means that we have a direct rule of a three, there is a cross multiplication. So:
20mL - 500mg
x mL - 25mg
500x = 500
[tex]x = \frac{500}{500}[/tex]
x = 1 mL
1mL should be administered to provide a 25-mg dose of aminophylline.
Final answer:
1 mL of the aminophylline solution should be administered to provide a 25-mg dose. This calculation was made by first converting the total amount of aminophylline to milligrams and then establishing the amount per milliliter to solve for the necessary volume.
Explanation:
To determine how many milliliters of aminophylline should be administered to provide a 25-mg dose, we need to use the following information:
1 ampul = 20 mL contains 0.5 g of aminophyllineDesired dose = 25 mg of aminophyllineFirst, we need to convert 0.5 g of aminophylline to milligrams:
0.5 g × 1000 mg/g = 500 mg
Now that we have the total amount of aminophylline in milligrams, we can determine the amount per milliliter:
500 mg / 20 mL = 25 mg/mL
To find the volume that contains 25 mg, we set up a proportion:
(25 mg of aminophylline) / (X mL) = (25 mg/mL)
By solving for X, we find:
X = 1 mL
So, 1 mL of the aminophylline solution should be administered to provide a 25-mg dose.
A commercial development project requires annual outlays of $65,000 for 10 years. Net cash inflows beginning in year 11 are expected to be $170,000 per year for 20 years. If the developer requires a rate of return of 16% , compute the net present value of the project.
Answer:
Net Present Value = - $99,360
Step-by-step explanation:
As provided,
Cash outlay = $65,000 each year for 10 years
Since the first outlay will be immediately, the cumulative discounting factor for cash outlay will be @ 16% = 1 for year 0 + 4.606 for 9 years = 5.606
Therefore, cumulative present value of total cash outlay = $65,000 [tex]\times[/tex] 5.606 = $364,390
Cash inflows beginning in year 11 = $170,000 for another continuous 20 years.
these cash flow will occur in the beginning of year 11 and end of year 10
Discounting factor will be [tex]\frac{1}{(1+0.16)^1^0}[/tex] = 0.2267
For, consecutive 20 years = 1.559
Therefore, value of inflows = $170,000 [tex]\times[/tex] 1.559 = $265,030
Net Present Value = Present Value of Cash Inflows - Present Value of Cash Outflows = $265,030 - $364,390 = - $99,360
Rewrite the subtraction number sentence as an addition number sentence.
5- (-2)
Answer:
5 + 2
Step-by-step explanation:
We have to rewrite the given statement in addition form.
The integers have property of:
Negative(-) Negative(-) = Positive(+)
Positive(+) Positive(+) = Positive(+)
Positive(+) Negative(-) = Negative(-)
Negative(-) Positive(+) = Negative(-)
The given statement is:
5-(-2)
Since we have two negative together, it is converted into a positive.
Thus, the given statement can be written in positive form as
5 + 2
Suppose a company wants to determine the current percentage of customers who are subjected to their advertisements online. How many customers should the company survey in order to be 98% confident that the estimated (sample) proportion is within 3 percentage points of the true population proportion of customers who are subjected to their advertisements online? z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576
Answer: 1503
Step-by-step explanation:
Given : Significance level : [tex]\alpha:1-0.98=0.02[/tex]
Critical value : [tex]z_{\alpha/2}=2.326[/tex]
Margin of error : [tex]E=\pm0.03[/tex]
We know that the formula to find the sample size (the prior true proportion is not available) is given by :-
[tex]n=0.25(\dfrac{z_{\alpha/2}}{E})^2[/tex]
i.e. [tex]n=0.25(\dfrac{2.326}{0.03})^2=1502.85444444\approx1503[/tex]
Hence, the minimum sample size should be 1503.
Workers at paper company count the number of boxes of paper in a warehouse each month. In january, there were 160341 boxes of paper. In February, there were 32698 boxes of paper. How does the digit 6 in February compare to the digit 6 in january?
Answer:
The digit 6 in February is worth 1/100 of its value in January.
Step-by-step explanation:
The value of a given digit in a number can be found by setting the other digits to zero.
In February, the digit 6 has a value of 00600 = 600.
In January, the digit 6 has a value of 060000 = 60,000.
The ratio of the value in February to the value in January is ...
600/60000 = 1/100
The digit 6 in February is one hundredth of the digit 6 in January.
Assume the random variable X is normally distributed with meanmu equals 50μ=50and standard deviationsigma equals 7σ=7.Compute the probability. Be sure to draw a normal curve with the area corresponding to the probability shaded.Upper P left parenthesis 34 less than Upper X less than 63 right parenthesisP(34
Answer: 0.9575465
Step-by-step explanation:
Let the random variable X is normally distributed with mean [tex]\mu=50[/tex] and standard deviation[tex]\sigma=7[/tex] .
Using the formula , [tex]z=\dfrac{x-\mu}{\sigma}[/tex] , we have the z-value for x= 34
[tex]z=\dfrac{34-50}{7}\approx-2.29[/tex]
For x= 63
[tex]z=\dfrac{63-50}{7}\approx1.86[/tex]
P-value : P(34<x<63)=P(-2.29<z<1.86)
[tex]=P(z<1.86)-P(z<-2.29)\\\\=0.9685572-(1-P(z<2.29))\\\\1=0.9685572-(1-0.9889893)\\\\=0.9575465[/tex]
Hence, the required probability = 0.9575465
Drug A has a concentration of 475 mg/10 mL. How many grams are in 100 mL of Drug A? (Round to the nearest tenth if applicable).
Answer: There are 4.8 grams in 100 mL of Drug A
Step-by-step explanation:
In order to determinate how many grams are in 100 mL of Drug A you can multiply and divide the expression of the concentration by 10 (To obtain 100 mL in the denominator)
Notice that the value remains unaltered.
(475 mg/10 mL )(10/10) = 4750 mg/ 100 mL
But the question is how many grams are in 100 mL, so you have to convert the value from mg to g.
The prefix m (mili) is equivalent to 0,001 so you can use 0.001 instead of the prefix
4750(0.001) g/ 100 mL
4.75 g/ 100 mL
The rounded result is 4.8 g/ 100 mL
Suppose H,K C G are subgroups of orders 5 and 8, respectively. Prove that H K = {e}.
Step-by-step explanation:
Consider the provided information.
We have given that H and k are the subgroups of orders 5 and 8, respectively.
We need to prove that H∩K = {e}.
As we know "Order of element divides order of group"
Here, the order of each element of H must divide 5 and every group has 1 identity element of order 1.
1 and 5 are the possible order of 5 order subgroup.
For subgroup order 8: The possible orders are 1, 2, 4 and 8.
Now we want to find the intersection of these two subgroups.
Clearly both subgroup H and k has only identity element in common.
Thus, H∩K = {e}.
Construct a truth table for the logical operator NAND.
Answer:
The output for NAND gate is false if all the inputs are true. it is true if either A or B is False.
Step-by-step explanation:
Step1
NAND operator:
NAND is a logic or Boolean expression and is a combination of NOT and AND Gate.
NAND gate is built using various junction diodes and transistors.
Step2
The output of NAND can be 0 or 1. Here, LOW (0) and HIGH (1). 0 or 1 is commonly named as FALSE or TRUE respectively.
It is basically the complement of AND gate. If all its inputs are True or 1, then the output is False or 0 else if either A or B is 0 or 1 then it will give the output as TRUE.
Step3
Here, A and B are 2 inputs used to make the truth table for NAND.
In 1st row, when A and B are both 0, the output for NAND is 1.
Likewise, in 2st row and 3rd row, when either A or B is 0 or 1, the output for NAND is 1 that is TRUE.
But in last row, when both inputs A and B are true, then the output for NAND gate is FALSE or 0.
The diagram and truth table for NAND is shown below.
Final answer:
A truth table for the NAND logical operator lists all possible binary input combinations of variables A and B, applies the AND operation, and then negates the output to present the NAND result. The resulting truth table will show that a NAND gate outputs true for all cases except when both inputs are true.
Explanation:
Constructing a truth table for the logical operator NAND is a way to visualize how this operator works with different input values.
The NAND operator is essentially the negation of the AND operator, meaning that it outputs the opposite of what an AND gate would. Here is how you construct a truth table for NAND:
List all possible input combinations of the variables A and B in binary (00, 01, 10, 11).Under the AND column, apply the AND operation to each pair of inputs (true only if both A and B are true).Finally, under the NAND column, negate the output of the AND operation (true if the AND output is false).The resulting truth table for the NAND operation is as follows:
A B A AND B A NAND B
false false false true
false true false true
true false false true
true true true false