Answer:
Johnny is wrong.
[tex]A \cup B = \left\{a,b,c,d,e,f,g,h\right\}[/tex]
Step-by-step explanation:
Johnny is wrong.
A better definition would be: [tex]A \cup B[/tex] is the set of all elements that belong to at least one A or B.
So, the elements that belong to both A and B, like c and d in this exercise, also belong to [tex]A \cup B[/tex].
So:
[tex]A \cup B=\left\{a,b,c,d,e,f,g,h\right\}[/tex]
What of the following basic rules is true about geometry?
A. Opposite angles are equal when two straight lines intersect
B. Supplementary angles total 180°
C. Complementary angles total 90°
D. A, B, and C
E. None of the above
Answer:
D. A, B, and C
Step-by-step explanation:
Option (A) is true because when two straight lines intersect to each other we get two pair of vertically opposite angles and the angles opposite to each other is always equal.
Option (B) is also correct as If the sum of two angles is equal to 180°, then they are supplementary to each other.
Option (C) is also correct as If the sum of the two angles is equal to 90°, then they are Complementary to each other.
Hence, Option (D) is correct.
there are 3 times as many boys as girls. If there are 24 more boys than girls, how many children are there?
How would I put this into a bar diagram? I can do it with algebra but I am having trouble drawing a bar diagram to help represent this problem.
Answer:
There are 48 children.
Step-by-step explanation:
Given :There are 3 times as many boys as girls.
There are 24 more boys than girls,
To Find : how many children are there?
Solution:
Let the number of girls be x
Now we are given that there are 3 times as many boys as girls.
So, no. of boys = 3x
Now we are given that there are 24 more boys than girls.
So, [tex]3x-x=24[/tex]
[tex]2x=24[/tex]
[tex]x=12[/tex]
So, no. of girls = 12
No. of boys = 3x = 3(12) = 36
Now the total no. of children = 12+36 = 48
Hence there are 48 children.
At a grocery store, Daniel wants to buy 3 lb of ham.
What decimal should the digital scale show?
Write 3 as a fraction and then divide.
The scale should read
The digital scale should display 3.0 when weighing 3 lbs of ham. This is because 3 pounds exactly can be displayed as the decimal 3.0 after converting the number into a fraction, 3/1, and dividing the numerator by the denominator.
Explanation:When Daniel wants to buy 3 lb of ham, the digital scale at the grocery store should display the decimal 3.0. This is because 3 pounds exactly translates to 3.0 in decimal terms.
The process of converting a number like 3 into a fraction would begin by writing it as 3/1 (as any number can be written over 1).
To convert that into decimal form, you would divide the top number (numerator) by the bottom number (denominator), so 3 ÷ 1 = 3.0.
Thus, the digital scale should read 3.0 when Daniel weighs out his 3 lbs of ham.
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Tour players Harry, Ron, Harmione and Ginny are playing a card game. A deck of 52 cards are dealt out equally. If Harmione and Ginny have a total of 8 spades among them, what is the probability that Harry has 3 of the remaining 5 spades?
Answer: 0.339
Step-by-step explanation:
Given : Tour players Harry, Ron, Harmione and Ginny are playing a card game.
. A deck of 52 cards are dealt out equally.
Then, the number of card each person has = [tex]\dfrac{52}{4}=13[/tex]
If Harmione and Ginny have a total of 8 spades among them, then the total cards the total spades left = 13-8=5
Now, the number of ways to get 3 of 5 spades : [tex]^5C_3=\dfrac{5!}{3!2!}=10[/tex]
Number of ways to draw remaining 10 cards : [tex]^{21}C_{10}=\dfrac{21!}{10!11!}=352716[/tex]
Also, the total cards Harmione and Ginny have = 13+13=26
Then the total cards left = 26
The number of ways to get 13 cards for Harry :
[tex]^{26}C_{13}=\dfrac{26!}{13!(26-13)!}\\\\=\dfrac{26!}{13!13!}=10400600[/tex]
Now, the probability that Harry has 3 of the remaining 5 spades :_
[tex]\dfrac{^5C_3\times ^{21}C_{10}}{^{26}C_{13}}\\\\=\dfrac{10\times352716}{10400600}\\\\=0.339130434783\approx0.339[/tex]
Hence, the probability that Harry has 3 of the remaining 5 spades= 0.339 (approx)
A hemispherical bowl of radius 12 inches is filled to a depth of h inches, where 0less than or equalshless than or equals12. Find the volume of water in the bowl as a function of h. (Check the special cases hequals0 and hequals12.)
Answer:
Suppose the bowl is situated such that the rim of the bowl touches the x axis, and the semicircular cross section of the bowl lies below the x-axis (in (iii) and (iv) quadrant ). Then the equation of the cross section of the bowl would be [tex]x^2+y^2=144[/tex], where y≤ 0,
⇒ [tex]y=-\sqrt{144-x^2}[/tex]
Here, h represents the depth of water,
Thus, by using shell method,
The volume of the disk would be,
[tex]V(h) = \pi \int_{-12}^{-12+h} x^2 dx[/tex]
[tex]= \pi \int_{-12}^{-12+h} (144-y^2) dy[/tex]
[tex]= \pi |144y-\frac{y^3}{3}|_{-12}^{-12+h}[/tex]
[tex]=\pi [ (144(-12+h)-\frac{(-12+h)^3}{3}-144(-12)+\frac{(-12)^3}{3}}][/tex]
[tex]=\pi [ -1728 + 144h - \frac{1}{3}(-1728+h^3+432h-36h^2)+1728-\frac{1728}{3}][/tex]
[tex]=\pi [ 144h - \frac{1}{3}(h^3+432h-36h^2}{3}][/tex]
[tex]=\pi [ 144h - \frac{h^3}{3} - 144h + 12h^2][/tex]
[tex]=\pi ( 12h^2 - \frac{h^3}{3})[/tex]
Special cases :
If h = 0,
[tex]V(0) = 0[/tex]
If h = 12,
[tex]V(12) = \pi ( 1728 - 576) = 1152\pi [/tex]
Help! BRANLIEST and 300 points!!! Plz show ALL your steps.
PB is a line segment on a number line. It has endpoints at -2 and 12. What is the coordinate of its midpoint?
Step-by-step explanation:
(12 + -2 )/2
10/2
5 im pretty sure
Sixty-seven percent of adults have looked at their credit score in the past six months. If you select 31 customers, what is the probability that at least 20 of them have looked at their score in the past six months? 0.450 0.550 0.692 0.142
Answer:
0.692.
Step-by-step explanation:
This is a Binomial Probability of Distribution with P(success) = 0.67. Prob success >= 20) , 31 trials.
From Binomial Tables we see that the required probability = 0.692.
You have decided to invest $1000 in a savings bond that pays 4% interest, compounded semi-annually. What will the bond be worth if you cash it in 10 years from now?
N= I/Y= PV= PMT= FV= P/Y=
Answer:
$2191.12
Step-by-step explanation:
We are asked to find the value of a bond after 10 years, if you invest $1000 in a savings bond that pays 4% interest, compounded semi-annually.
[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.
Since interest is compounded semi-annually, so 'n' will be 2 times 10 that is 20.
[tex]4\%=\frac{4}{100}=0.04[/tex]
[tex]FV=\$1,000\times (1+0.04)^{20}[/tex]
[tex]FV=\$1,000\times (1.04)^{20}[/tex]
[tex]FV=\$1,000\times 2.1911231430334194[/tex]
[tex]FV=\$2191.1231430334194[/tex]
[tex]FV\approx \$2191.12[/tex]
Therefore, the bond would be $2191.12 worth in 10 years.
The cost, in dollars, of making x items is given by the function C(x)=25x+300
. Find C(1500), the cost of making 1500 items.
Question 6 options:
a)
$4050
b)
$25,300
c)
$30,300
d)
$37,800
e)
none
Answer:
d) $37800
Step-by-step explanation:
Cost of making x items = [tex]C(x)=25x + 300[/tex]
Cost of making [tex]1500[/tex] items = [tex]C(1500)=25(1500) + 300\\C(1500)= 37500 + 300\\C(1500)= 37800[/tex]
Cost of making [tex]1500[/tex] items = $37800
d) $37800 is the correct answer
Solve the Following Initial Value Problem: 2XYY'+Y^2-4X^3=0. where Y(1)=2
The answer is y= sqrt((x^4+3)/x)
[tex]2xyy'+y^2-4x^3=0[/tex]
Let [tex]z(x)=y(x)^2[/tex], so that [tex]z'(x)=2y(x)y'(x)[/tex] (which appears in the first term on the left side):
[tex]xz'+z=4x^3[/tex]
This ODE is linear in [tex]z[/tex], and we don't have to find any integrating factor because the left side is already the derivative of a product:
[tex](xz)'=4x^3\implies xz=x^4+C\implies z=\dfrac{x^4+C}x[/tex]
[tex]\implies y(x)=\sqrt{\dfrac{x^4+C}x}[/tex]
With [tex]y(1)=2[/tex], we get
[tex]2=\sqrt{1+C}\implies C=3[/tex]
so the solution is as given in your post.
In experiment two, I measured a block to be 4.45 cm by 3.35 em by 6.15 cm and a mass of 155.147g. What is the density of this block in kg/L? 3.
Answer:
1.69224581396 Kg/L
Step-by-step explanation:
We are given the measure of the block as 4.45 cm × 3.35 cm × 6.15 cm.
Volume of block = 4.45 cm × 3.35 cm × 6.15 cm = 91.681125 cm cube = 91.681125 × 0.001 L = 0.091681125 L
We did the above step to convert the volume of block into Liter.
Mass of block is given as 155.147 gram = 155.147 × 0.001 kg = 0.155147 kg
We converted the mass of block into kilograms because we need density in Kg/L.
Density is defined as mass per unit volume
Density = [tex]\frac{Mass}{Volume}[/tex]
= [tex]\frac{0.155147 }{0.091681125}[/tex]]
= 1.69224581396 Kg/L
The density is found to be approximately 1.688 kg/L.
To find the density of the block, we need to use the density formula:
Density = Mass / Volume
The given dimensions of the block are:
Length = 4.45 cmWidth = 3.35 cmHeight = 6.15 cmFirst, calculate the volume:
Volume = Length × Width × Height
Volume = 4.45 cm × 3.35 cm × 6.15 cm
Volume ≈ 91.88925 cubic centimeters (cm)
Next, convert mass to kilograms and volume to liters:
Mass = 155.147 g = 0.155147 kg (since 1 kg = 1000 g)Volume = 91.88925 cm³ × 0.001 L/cm³ ≈ 0.09188925 L (since 1 cm³ = 0.001 L)Finally, calculate the density in kg/L:
Density = Mass / Volume
Density ≈ 0.155147 kg / 0.09188925 L
Density ≈ 1.688 kg/L
Thus, the density of the block is approximately 1.688 kg/L.
The charge to rent a trailer is $15 for up to 2 hours plus $8 per additional hour or portion of an hour. Find the cost to rent a trailer for 2.8 hours, 3 hours, and 8.5 hours. Then graph al ordered pairs, (hours, cost), for the function a. What is the cost to rent a trailer for 2.8 hours?
Answer:
The cost to rent a trailer for 2.8 hours is $21.4.
The cost to rent a trailer for 3 hours is $23.
The cost to rent a trailer for 8.5 hours is $67.
Step-by-step explanation:
Let x be the number of hours.
It is given that the charge to rent a trailer is $15 for up to 2 hours plus $8 per additional hour or portion of an hour.
The cost to rent a trailer for x hours is defined as
[tex]C(x)=\begin{cases}15 & \text{ if } x\leq 2 \\ 15+8(x-2) & \text{ if } x>2 \end{cases}[/tex]
For x>2, the cost function is
[tex]C(x)=15+8(x-2)[/tex]
We need to find the cost to rent a trailer for 2.8 hours, 3 hours, and 8.5 hours.
Substitute x=2.8 in the above function.
[tex]C(2.8)=15+8(2.8-2)=15+8(0.8)=21.4[/tex]
The cost to rent a trailer for 2.8 hours is $21.4.
Substitute x=3 in the above function.
[tex]C(3)=15+8(3-2)=15+8(1)=23[/tex]
The cost to rent a trailer for 3 hours is $23.
Substitute x=8.5 in the above function.
[tex]C(8.5)=15+8(8.5-2)=15+8(6.5)=67[/tex]
The cost to rent a trailer for 8.5 hours is $67.
Written all the ordered pairs in the form of (hours, cost).
(2.8,21.4), (3,23) and (8.5,67)
Plot these points on coordinate plane.
Final answer:
To find the cost to rent a trailer for 2.8 hours, we consider the flat fee of $15 for the first 2 hours and add the additional cost of $8 for the partial hour beyond 2 hours, resulting in a total cost of $23.
Explanation:
The cost to rent a trailer for a given number of hours is determined by a flat fee of $15 for the first 2 hours and an additional cost of $8 for each extra hour or partial hour. For 2.8 hours, since this exceeds the initial 2-hour period, we calculate the cost as follows:
Flat fee for the first 2 hours: $15Extra time: 0.8 hours, which counts as an additional full hour for charging purposesAdditional cost for 0.8 hours (rounded up to 1 hour): $8Total cost = Flat fee + cost for additional hoursTotal cost = $15 + $8 = $23Therefore, the cost to rent a trailer for 2.8 hours is $23.
An airplane left airport A flying on a course of 72 degrees.
After flying 495 km, the plane was just able to pick up
communication signals from airport B which is due east of airport
A. If airport B's communication signals can be picked up within a
radius of 300 km from airport B, for how many kilometers can the
airplane fly and still be in contact with airport B? Give you
answer to one decimal place.
Distance for which Aeroplane can be in contact with Airport B is = 396.34 km
Step-by-step explanation:In the question,
We have an Airport at point A and another at point B.
Now,
Airplane flying at the angle of 72° with vertical catches signals from point D.
Distance travelled by Airplane, AD = 495 km
Now, Let us say,
AB = x
So,
In triangle ABD, Using Cosine Rule, we get,
[tex]cos(90-72) =cos18= \frac{AB^{2}+AD^{2}-BD^{2}}{2.AD.AB}[/tex]
So,
On putting the values, we get,
[tex]cos18 = \frac{x^{2}+495^{2}-300^{2}}{2(495)(x)}\\0.951(990x)=x^{2}+245025-90000\\x^{2}-941.54x+155025=0\\[/tex]
Therefore, x is given by,
x = 212.696, 728.844
So,
The value of x can not be 212.696 as the length of LB (radius) itself is 300 km.
So,
x = 728.844 km
So,
AL = AB - BL
AL = x - 300
AL = 728.844 - 300
AL = 428.844 km
Now, in the circle from a property of secants we can say that,
AL x AM = AD x AC
So,
428.844 x (728.844 + 300) = 495 x AC
441213.576 = 495 x AC
AC = 891.34 km
So,
The value of CD is given by,
CD = AC - AD
CD = 891.34 - 495
CD = 396.34 km
Therefore, the distance for which the Aeroplane can still be in the contact with Airport B is 396.34 km.
Dr. Fitzgerald has graded 15 of 26 exams for Epi 501. (a) What proportion of all exams has Dr. Fitzgerald graded? (b) What was the ratio of graded to ungraded tests?
Answer: a) 15:26, and b) 15:11.
Step-by-step explanation:
Since we have given that
Number of graded tests = 15
Number of total tests = 26
Number of ungraded tests is given by
[tex]26-15\\\\=11[/tex]
a) Proportion of all exams has Dr. Fitxgerald graded is given by
15:26.
b) Ratio of graded to ungraded tests is given by 15:11
Hence, a) 15:26, and b) 15:11.
(a) The proportion of all exams graded by Dr. Fitzgerald is [tex]\(\frac{15}{26}\)[/tex].
(b) The ratio of graded to ungraded tests is [tex]\(\frac{15}{26 - 15}\) or \(\frac{15}{11}\)[/tex].
(a) To find the proportion of exams graded by Dr. Fitzgerald, we divide the number of exams graded by the total number of exams. This gives us the fraction:
[tex]\[ \text{Proportion graded} = \frac{\text{Number of exams graded}}{\text{Total number of exams}} = \frac{15}{26} \][/tex]
This fraction represents the part of the whole set of exams that has been graded.
(b) To find the ratio of graded to ungraded tests, we take the number of exams that have been graded and divide it by the number of exams that have not been graded. The number of ungraded exams is the total number of exams minus the number of graded exams:
[tex]\[ \text{Number of ungraded exams} = \text{Total number of exams} - \text{Number of exams graded} = 26 - 15 = 11 \][/tex]
Now, we can find the ratio:
[tex]\[ \text{Ratio of graded to ungraded tests} = \frac{\text{Number of exams graded}}{\text{Number of exams ungraded}} = \frac{15}{11} \][/tex]
This ratio tells us how many times greater the number of graded exams is compared to the number of ungraded exams.
Determine whether the following possible responses should be classified as ratio, interval, nominal or ordinal data.
? Ratio Ordinal Nominal Interval 1. The college (Arts and Science, Business, etc.) you are enrolled in
? Ratio Ordinal Nominal Interval 2. The number of students in a statistics course
? Ratio Ordinal Nominal Interval 3. The age of each of your classmates
? Ratio Ordinal Nominal Interval 4. Your hometown
Answer:
1. The college (Arts and Science, Business, etc.) you are enrolled in
Nominal
2. The number of students in a statistics course Ratio
3. The age of each of your classmates Ratio
4. Your hometown Nominal
Step-by-step explanation:
Nominal, ordinal, interval, or ratio data are the four fundamental levels of measurement scales that are used to capture data.
Nominal, are used for labeling variables, without any quantitative value.
Ordinal, the order of the values is what is significant, but the differences between each one is not really known.
Interval, we know both, the order and the exact differences between the values
Ratio, they have the order, the exact value between units, and have an absolute zero
in the following ordinary annuity, the interest is compounded with each payment, and the payment made at the end of the compounding period. find the accumulated amount of the annuity. (round your answer to the nearest cent)
$2000 monthy at 6.3% for 20 years
Answer: $957646.07
Step-by-step explanation:
The formula we use to find the accumulated amount of the annuity is given by :-
[tex]FV=m(\frac{(1+\frac{r}{n})^{nt})-1}{\frac{r}{n}})[/tex]
, where m is the annuity payment deposit, r is annual interest rate , t is time in years and n is number of periods.
Given : m= $2000 ; n= 12 [∵12 in a year] ; t= 20 years ; r= 0.063
Now substitute all these value in the formula , we get
[tex]FV=(2000)(\frac{(1+\frac{0.063}{12})^{12\times20})-1}{\frac{0.063}{12}})[/tex]
i.e. [tex]FV=(2000)(\frac{(1+0.00525)^{240})-1}{0.00525})[/tex]
i.e. [tex]FV=(2000)(\frac{(3.51382093497)-1}{0.00525})[/tex]
i.e. [tex]FV=(2000)(\frac{2.51382093497}{0.00525})[/tex]
i.e. [tex]FV=(2000)(478.823035232)[/tex]
i.e. [tex]FV=957646.070464\approx957646.07\ \ \ \text{ [Rounded to the nearest cent]}[/tex]
Hence, the accumulated amount of the annuity= $957646.07
The future value or accumulated amount of an ordinary annuity is calculated using the formula where P is the periodic payment, r is the interest rate per period, n is the number of compounding periods per year, and t is the time in years. Given P = $2000, r = 6.3%, n = 12 and t = 20 years, substituting these values into the formula gives the accumulated amount
Explanation:To find the future value or accumulated amount of an ordinary annuity, we use the formula: FV = P * (((1 + r)^nt - 1) / r), where P is the periodic payment, r is the interest rate per period, n is the number of compounding periods per year, and t is the time in years.
In the given problem, P = $2000, r = 6.3% or 0.063 (in decimal), n = 12 (since the payments are monthly), and t = 20 years.
Substituting these into the formula, FV = $2000 * (((1 + 0.063 /12)^(12*20) - 1) / (0.063/12)).
Calculating the equation, we'll get the accumulated amount to the nearest cent.
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Employment data at a large company reveal that 74% of the workers are married, 42% are college graduates, and that 56% are married, given that they are college graduates. Which of the following statements are true about the events married and college graduate?
(A) These events are pairwise disjoint.
(B) These events are independent events.
(C) These events are both independent and pairwise disjoint.
(D) A worker is either married or a college graduate always.
(E) None of these above are true.
Answer:
(E) None of these above are true.
Step-by-step explanation:
Married = 74% or 0.74
College graduates = 42% or 0.42
pr(married | college graduates) = 0.56
(A) These events are pairwise disjoint. This is false. Pairwise disjoint are also known as mutually exclusive events. Here we can see that both events are occurring at same time.
(B) These events are independent events. This is also false.
(C) These events are both independent and pairwise disjoint. False
(D) A worker is either married or a college graduate always. False
Here Probability(A or B) shall be 1
= Pr(A) + Pr(B) - Pr( A and B) = 0.74 + 0.42 - 0.56 * 0.42 = 0.9248
This is not equal to 1.
(E) None of these above are true. This is true.
Jay said that by the quotient of powers property zero to the fifth power divided by zero to the second power equals zero to the fifth power -2 equals zero to the third power which equals zero is this correct explain
Answer:
No, Jay is not correct.
Step-by-step explanation:
Quotient of powers property:
For any non-zero number a and any integer x and y:
[tex]\frac{a^x}{a^y}=a^{x-y}[/tex]
According to by the quotient of powers property
[tex]\frac{0^5}{0^2}=0^{5-2}\Rightarrow 0^3=0[/tex]
We need to check whether Jay is correct or not.
No, Jay is not correct because quotient of powers property is used for non-zero numbers.
[tex]\frac{0^m}{0^n}=\frac{0}{0}=unde fined[/tex]
Therefore, Jay is not correct.
Prove that x-1 is a factor of x^n-1 for any positive integer n.
Answer:
[tex]x-1[/tex] is a factor of [tex]x^n - 1[/tex]
Step-by-step explanation:
[tex]x-1[/tex] is a factor of [tex]x^n - 1[/tex]
We will prove this with the help of principal of mathematical induction.
For n = 1, [tex]x-1[/tex] is a factor [tex]x-1[/tex], which is true.
Let the given statement be true for n = k that is [tex]x-1[/tex] is a factor of [tex]x^k - 1[/tex].
Thus, [tex]x^k - 1[/tex] can be written equal to [tex]y(x-1)[/tex], where y is an integer.
Now, we will prove that the given statement is true for n = k+1
[tex]x^{k+1} - 1\\=(x-1)x^k + x^k - 1\\=(x-1)x^k + y(x-1)\\(x-1)(x^k + y)[/tex]
Thus, [tex]x^k - 1[/tex] is divisible by [tex]x-1[/tex].
Hence, by principle of mathematical induction, the given statement is true for all natural numbers,n.
As the owner of a small restaurant, you purchase 5 boxes of napkins for $75.00 every 3 months. Each box contains 525 napkins. To the nearest hundredth, what is the cost for each individual napkin?
As the owner of a small restaurant, you purchase 5 boxes of napkins for $75.00 every 3 months. Each box contains 525 napkins. To the nearest hundredth, what is the cost for each individual napkin?
A) 0.01
B) 0.02
C) 0.03
D) 0.05
Answer: 0.03
Step-by-step explanation:
Total number of napkins: 5 x 525 = 2,625
75/2626 = 0.02857, which rounds to 0.03
The cost per individual napkin, when rounded to the nearest hundredth, is $0.03.
Explanation:To begin finding the cost per napkin, we first need to find out how many napkins are purchased every 3 months. Since each box contains 525 napkins and you purchase 5 boxes every 3 months, that would be 525 * 5 = 2625 napkins. The cost of these napkins is $75.00.
So, to find the cost per individual napkin, you would divide the total cost by the total number of napkins. That would be 75 / 2625 = $0.028571... When rounded to the nearest hundredth, this becomes $0.03. So, each individual napkin costs $0.03. Therefore, the correct answer is (C) 0.03.
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Cantwell Associates, a real estate developer, is planning to build a new apartment complex consisting of one-bedroom units, two-bedroom townhouses, and three-bedroom townhouses. A total of 216 units is planned. The total number of two- and three-bedroom townhouses will equal the number of one-bedroom units. If the number of one-bedroom units will be 3 times the number of three-bedroom townhouses, find how many units of each type will be in the complex. one-bedroom units units two-bedroom townhouses units three-bedroom townhouses units
Answer:
108 one-bedroom units
72 two-bedroom units
36 three-bedroom units
Step-by-step explanation:
Let x, y, z the number of one-bedroom, two-bedroom and three-bedroom units respectively. Then
1) x+y+z = 216
2) y+z = x
3) x = 3z
Multiplying equation 1) by -1 and adding it to 2), we get
-x = x-216 so, x = 216/2 = 108
x = 108
Replacing this value in 3) we get
z = 108/3 = 36
z = 36
Replacing now in 2)
y+36 = 108, y = 108-36 and
y = 72
In the planned apartment complex, there will be 0 one-bedroom units, 216 two-bedroom townhouses, and 0 three-bedroom townhouses.
Explanation:Let x be the number of one-bedroom units. Since the number of two- and three-bedroom townhouses equals the number of one-bedroom units, let y be the number of two-bedroom townhouses and z be the number of three-bedroom townhouses. We know that x + y + z = 216. Additionally, x = 3z because the number of one-bedroom units will be 3 times the number of three-bedroom townhouses. Substituting x = 3z into the first equation gives 3z + y + z = 216. Simplifying this equation, we get 4z + y = 216.
Now, we can solve this system of equations to find the values of x, y, and z. Subtracting y from both sides of the equation 4z + y = 216 gives 4z = 216 - y. Let's call this equation (1). Substituting x = 3z and y = 216 - 4z into the equation x + y + z = 216 gives 3z + (216 - 4z) + z = 216. Simplifying this equation, we get 4z + 216 = 216. Subtracting 216 from both sides of the equation gives 4z = 0. Let's call this equation (2).
Since equation (1) and equation (2) both have 4z on the left side, we can equate the right sides of the equations. This gives 216 - y = 0. Solving for y, we find y = 216. Plugging this value of y into equation (1), we get 4z = 216 - 216, which simplifies to 4z = 0. Solving for z, we find z = 0. Finally, plugging the value of z into the equation x = 3z, we get x = 3(0), which simplifies to x = 0.
Therefore, there are 0 one-bedroom units, 216 two-bedroom townhouses, and 0 three-bedroom townhouses in the complex.
The CEO of a company that sells car stereos has determined the profit of selling x number of stereos to be: P(x) = –.04x2 + 1000x – 16,500 How much profit should the company expect from selling 12,500 stereos?
Answer:
6233500
Step-by-step explanation:
We are given that CEO of a company that sells car stereos has determined the profit x number of stereos.
The profit of selling x number of stereos is given by
[tex]P(x)=-0.04x^2=100x-16500[/tex]
We have to find the value of profit when the company selling 12500 stereos.
Substitute the value of x=12500
Then, we get
[tex]P(12500)=-.04(12500)^2+1000(12500)-16500[/tex]
[tex]P(12500)=-6250000+12500000-16500=-6266500+12500000[/tex]
[tex]P(12500)=6233500[/tex]
Hence, the company should expect profit 6233500 from selling 12500 stereos.
what is the value of x if the average of 10, 11, 12 and x is 9?
Answer: The value of x is 3.
Step-by-step explanation:
Since we have given that
10,11, 12 and x
Average of above numbers = 9
As we know that
Average is given by
[tex]\dfrac{\text{Sum of observation}}{\text{Number of observation}}\\\\\\\dfrac{10+11+12+x}{4}=9\\\\10+11+12+x=9\times 4\\\\33+x=36\\\\x=36-33\\\\x=3[/tex]
Hence, the value of x is 3.
the forecast calls for a 30% chance of snow today and a 40% chance of snow tomorrow. what are the chances it will rain two days in a row?
Answer:
The chances it will rain two days in a row is 12%
Step-by-step explanation:
The forecast calls for a 30% chance of snow today
So, chance of snowfall today = 30% = 0.3
A 40% chance of snow tomorrow.
So, chance of snowfall tomorrow= 40% = 0.4
The chances it will rain two days in a row = [tex]0.4 \times 0.3[/tex]
= [tex]0.12[/tex]
So, percent it will rain two days in a row = [tex]0.12 \times 100 = 12\%[/tex]
Hence the chances it will rain two days in a row is 12%
Total departmental sales in the Housewares Department were $513000.00. A salesperson made 14% of the total departmental sales of that month and earns 6.5% commission on his sales. Find the dollar amount of commission.
a.
$33345.00
b.
$4691.95
c.
$4668.30
d.
$71820.00
e.
$4683.19
f.
None of the above.
Answer: c. $4668.30
Explanation:
Given:
Sales = $513000
Sales made by an individual = 14% of $513000
Sales made by an individual = [tex]\frac{14}{100}\times 513000[/tex]
Sales made by an individual = $71280
Commission made on this sales = 6.5% of $71280
Commission made on this sales = [tex]\frac{6.5}{100}\times 71280[/tex]
Commission made on this sales = $4668.30
In the xy plane, a quadrilateral has vertices at (-1, 4), (7,4), (7,5), and (-1. 5). What is the perimeter of the quadrilateral? (A) 17 (B) 18 (C) 19 (1) 32 (E) 34
Answer:
(B) 18.
Step-by-step explanation:
We are asked to find the perimeter of a quadrilateral with vertices at (-1, 4), (7,4), (7,5), and (-1. 5).
First of all, we will draw vertices of quadrilateral on coordinate plane and connect the vertices as shown in the attached photo.
We can see that our quadrilateral is a parallelogram, whose parallel sides are equal.
[tex]\text{Perimeter of quadrilateral}=8+1+8+1[/tex]
[tex]\text{Perimeter of quadrilateral}=16+2[/tex]
[tex]\text{Perimeter of quadrilateral}=18[/tex]
Therefore, the perimeter of the given quadrilateral is 18 units.
d) Find the 5th degree Taylor polynomial centered at x = 0 for the function y = \tiny \frac{x}{1+x}
Answer:
[tex]P(x) = x - x^2 + x^3 - x^4+x^5[/tex]
Step-by-step explanation:
Let us first remember how a Taylor polynomial looks like:
Given a differentiable function [tex]f[/tex] then we can find its Taylor series to the [tex]nth[/tex] degree as follows:
[tex]P(x) = f(x_{0}) + f'(x_{0}).(x-x_{0}) + \frac{f''(x_{0})}{2!}.(x-x_{0})^2+.....+\frac{f^n(x_{0})}{n!}.(x-x_{0})^n + R_{n}(x).(x-x_{0})^n[/tex]
Where [tex]R_{n}(x)[/tex] represents the Remainder and [tex]f^n(x)[/tex] is the [tex]nth[/tex] derivative of [tex]f[/tex].
So let us find those derivatives.
[tex]f(x) = \frac{x}{1+x}\\f'(x) = \frac{1}{(1+x)^2}\\f''(x) = \frac{-2}{(1+x)^3}\\f'''(x) = \frac{6}{(1+x)^4}\\f''''(x) = \frac{-24}{(1+x)^5}\\f'''''(x) = \frac{120}{(1+x)^6}[/tex]
The only trick for this derivatives is for the very first one:
[tex]f'(x) = \frac{1}{1+x} - \frac{x}{(1+x)^2}\\f'(x) = \frac{(1+x) - x}{(1+x)^2} = \frac{1}{(1+x)^2}\\[/tex]
Then it's only matter of replacing on the Taylor Series and replacing [tex]x_{0}=0[/tex]
What is the lenght of a diagonal of a square with sides 16feet
long ? Round to the nearest tenth.
Answer:
The length of the diagonal is 22.6 ft.
Step-by-step explanation:
To find the length of the diagonal of a square, multiply the length of one side by the square root of 2:
If the length of one side is x, [tex]length = x\sqrt{2}[/tex] as you can see in the image attached.
This fact is a consequence of applying the Pythagoras' Theorem to find the length of the diagonal if we know the side length of the square.
[tex]length^{2} = x^{2}+x^{2} \\ length=\sqrt{x^{2}+x^{2}} \\ length=\sqrt{2x^{2} } \\ length=x\sqrt{2}[/tex]
We know that the length of one side is 16 ft so [tex]length = 16\sqrt{2}=22.627[/tex] and round to the nearest tenth is 22.6 ft
Find the values of x1 and x2 where the following two constraints intersect.
1) 9x1 + 7x2 ≥ 57
(2) 4x1 + 6x2 ≥ 13
Constraints are simply the subjects of an objective function.
The point of intersection is: [tex]\mathbf{(x_1,y_1) = (9.54,-4.19)}[/tex]
The constraints are given as:
[tex]\mathbf{9x_1 + 7x_2 \ge 57}[/tex]
[tex]\mathbf{4x_1 + 6x_2 \ge 13}[/tex]
Express [tex]\mathbf{4x_1 + 6x_2 \ge 13}[/tex] as an equation
[tex]\mathbf{4x_1 + 6x_2= 13}[/tex]
Subtract 6x2 from both sides
[tex]\mathbf{4x_1 = 13 - 6x_2}[/tex]
Divide through by 4
[tex]\mathbf{x_1 = \frac{1}{4}(13 - 6x_2)}[/tex]
Substitute [tex]\mathbf{x_1 = \frac{1}{4}(13 - 6x_2)} \\[/tex] in [tex]\mathbf{9x_1 + 7x_2 \ge 57}[/tex]
[tex]\mathbf{9 \times \frac{1}{4}(13 - 6x_2) + 7x_2 \ge 57}[/tex]
Open brackets
[tex]\mathbf{29.25 - 13.5x_2 + 7x_2 \ge 57}[/tex]
[tex]\mathbf{29.25-6.5x_2 \ge 57}[/tex]
Collect like terms
[tex]\mathbf{-6.5x_2 \ge 57 - 29.25}[/tex]
[tex]\mathbf{-6.5x_2 \ge 27.25}[/tex]
Divide both sides by -6.5
[tex]\mathbf{x_2 \ge -4.19}[/tex]
Substitute -4.19 for x2 in [tex]\mathbf{4x_1 + 6x_2 \ge 13}[/tex]
[tex]\mathbf{4x_1 + 6 \times -4.19 \ge 13}[/tex]
[tex]\mathbf{4x_1 - 25.14 \ge 13}[/tex]
Add 25.14 to both sides
[tex]\mathbf{4x_1 \ge 38.14}[/tex]
Divide both sides by 4
[tex]\mathbf{x_1 \ge 9.54}[/tex]
Hence, the values are:
[tex]\mathbf{(x_1,y_1) = (9.54,-4.19)}[/tex]
Read more about inequalities at:
https://brainly.com/question/20383699
The solution of the problem involves finding the values of x1 and x2 which satisfy both inequalities when plotted on a graph. This can be done by simplifying the equations and comparing them.
Explanation:To solve this problem, we need to find where the two inequalities intersect. This means that we need to find the values of x1 and x2 which satisfy both inequalities.
Let's start with the first inequality '9x1 + 7x2 ≥ 57'. This means that the sum of 9 times x1 and 7 times x2 should be greater than or equal to 57. You can simplify this inequality by dividing the entire expression by the smallest coefficient which is 9, getting 'x1 + (7/9)x2 ≥ 57/9'.
Similarly, simplifying the second inequality '4x1 + 6x2 ≥ 13' by dividing by the smallest coefficient which is 4, we get 'x1 + (3/2)x2 ≥ 13/4'.
By comparing these two simplified inequalities, you should be able to identify the values of x1 and x2 where both inequalities are satisfied.
Learn more about Inequalities here:https://brainly.com/question/30231190
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The local pet store surveyed 50 people about pets. Eleven of these people owned dogs, 13 owned cats, and 6 owned fish. One person owned all three types of pets, 2 people owned only fish and dogs, 3 people only fish and cats, and 5 people owned only cats and dogs. How many people owned none of these pets?
Answer: 29
Step-by-step explanation:
Let S denotes the total number of people surveyed, A denotes the event of having dog , B denotes the event of having cats and C denotes the event of having fish.
Given : n(S)=50 ;n(A)=11 ; n(B) =13 and n(C)=6
Also, n(A∩B)=5 ; n(A∩C) = 2 and n(B∩C)=3 and n(A∩B∩C)=1
We know that,
[tex]n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A \cap B)-n(A \cap C)-n(B \cap C)-n(A \cap B\cap C)\\\\=11+13+6-5-2-3+1=21[/tex]
Now, the number of people owned none of these pets :-
[tex]n(S)-n(A\cup B\cup C)\\\\=50-21=29[/tex]
Hence, the number of people owned none of these pets =29