Divide the total cards by 18:
236 / 18 = 13.111
Use the whole number and multiply by 18:
13 x 18 = 234
This means 234 cards would be in storage boxes.
236 - 234 = 2 cards would be left over.
A group of college students are going to a lake house for the weekend and plan on renting small cars and large cars to make the trip. Each small car can hold 5 people and each large car can hold 7 people. The students rented 3 times as many small cars as large cars, which altogether can hold 44 people. Write a system of equations that could be used to determine the number of small cars rented and the number of large cars rented. Define the variables that you use to write the system.
Answer:
The equation 7 m + 15 m = 44 is the equation that can be used to determine the number of small cars rented and the number of large cars rented.
where, m : Number of large cars rented
Step-by-step explanation:
The number of people small car can hold = 5
The number of people large car can hold = 7
Let us assume the number of large cars rented = m
So, the number of smaller cars rented = 3 x ( Number of large cars rented)
= 3 m
Now, the number of people in m large cars = m x ( Capacity of 1 large car)
= m x ( 7) = 7 m
And, the number of people in 3 m small cars = 3 m x ( Capacity of 1 small car) = 3 m x ( 5) = 15 m
Total people altogether going for the plan = 44
⇒ The number of people in ( Small +Large) car = 44
or, 7 m + 15 m = 44
Hence, the equation 7 m + 15 m = 44 is the equation that can be used to determine the number of small cars rented and the number of large cars rented.
We define variable as
Let x be the number of small cars and y be the number of large cars.
Since ,
Each small car can hold 5 people and each large car can hold 7 people.
i.e. Number of people in x cars = 5x
Number of people in y cars = 7y
The students rented 3 times as many small cars as large cars, implies
y=3(x)
They altogether can hold 44 people.
i.e. 5x+7y=44
Thus , the system of equations that could be used to determine the number of small cars rented and the number of large cars rented :
[tex]y=3(x)[/tex]
[tex]5x+7y=44[/tex]
I need help with my math homework. The questions is: Find all solutions of the equation in the interval [0,2π).
√3 tan(x- π/8)-1=0
Answer:
[tex]\frac{7\pi}{24}[/tex] and [tex]\frac{31\pi}{24}[/tex]
Step-by-step explanation:
[tex]\sqrt{3} \tan(x-\frac{\pi}{8})-1=0[/tex]
Let's first isolate the trig function.
Add 1 one on both sides:
[tex]\sqrt{3} \tan(x-\frac{\pi}{8})=1[/tex]
Divide both sides by [tex]\sqrt{3}[/tex]:
[tex]\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}[/tex]
Now recall [tex]\tan(u)=\frac{\sin(u)}{\cos(u)}[/tex].
[tex]\frac{1}{\sqrt{3}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}[/tex]
or
[tex]\frac{1}{\sqrt{3}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}[/tex]
The first ratio I have can be found using [tex]\frac{\pi}{6}[/tex] in the first rotation of the unit circle.
The second ratio I have can be found using [tex]\frac{7\pi}{6}[/tex] you can see this is on the same line as the [tex]\frac{\pi}{6}[/tex] so you could write [tex]\frac{7\pi}{6}[/tex] as [tex]\frac{\pi}{6}+\pi[/tex].
So this means the following:
[tex]\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}[/tex]
is true when [tex]x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi[/tex]
where [tex]n[/tex] is integer.
Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.
So now we have a linear equation to solve:
[tex]x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi[/tex]
Add [tex]\frac{\pi}{8}[/tex] on both sides:
[tex]x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi[/tex]
Find common denominator between the first two terms on the right.
That is 24.
[tex]x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi[/tex]
[tex]x=\frac{7\pi}{24}+n \pi[/tex] (So this is for all the solutions.)
Now I just notice that it said find all the solutions in the interval [tex][0,2\pi)[/tex].
So if [tex]\sqrt{3} \tan(x-\frac{\pi}{8})-1=0[/tex] and we let [tex]u=x-\frac{\pi}{8}[/tex], then solving for [tex]x[/tex] gives us:
[tex]u+\frac{\pi}{8}=x[/tex] ( I just added [tex]\frac{\pi}{8}[/tex] on both sides.)
So recall [tex]0\le x<2\pi[/tex].
Then [tex]0 \le u+\frac{\pi}{8}<2 \pi[/tex].
Subtract [tex]\frac{\pi}{8}[/tex] on both sides:
[tex]-\frac{\pi}{8}\le u <2 \pi-\frac{\pi}{8}[/tex]
Simplify:
[tex]-\frac{\pi}{8}\le u <\pi (2-\frac{1}{8})[/tex]
[tex]-\frac{\pi}{8}\le u<\frac{15\pi}{8}[/tex]
So we want to find solutions to:
[tex]\tan(u)=\frac{1}{\sqrt{3}}[/tex] with the condition:
[tex]-\frac{\pi}{8}\le u<\frac{15\pi}{8}[/tex]
That's just at [tex]\frac{\pi}{6}[/tex] and [tex]\frac{7\pi}{6}[/tex]
So now adding [tex]\frac{\pi}{8}[/tex] to both gives us the solutions to:
[tex]\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}[/tex] in the interval:
[tex]0\le x<2\pi[/tex].
The solutions we are looking for are:
[tex]\frac{\pi}{6}+\frac{\pi}{8}[/tex] and [tex]\frac{7\pi}{6}+\frac{\pi}{8}[/tex]
Let's simplifying:
[tex](\frac{1}{6}+\frac{1}{8})\pi[/tex] and [tex](\frac{7}{6}+\frac{1}{8})\pi[/tex]
[tex]\frac{7}{24}\pi[/tex] and [tex]\frac{31}{24}\pi[/tex]
[tex]\frac{7\pi}{24}[/tex] and [tex]\frac{31\pi}{24}[/tex]
Giving a test to a group of students, the grades and gender are summarized below A B C Total Male 6 18 3 27 Female 13 20 12 45 Total 19 38 15 72 If one student is chosen at random,
If one student is chosen at random,
Find the probability that the student got a B:
Find the probability that the student was female AND got a "C":
Find the probability that the student was female OR got an "B":
If one student is chosen at random, find the probability that the student got a 'B' GIVEN they are male:
Answer:
P(Got a B) = [tex]\frac{19}{36}[/tex]
P(Female AND got a C) = [tex]\frac{1}{6}[/tex]
P(Female or Got an B = [tex]\frac{7}{8}[/tex]
P(got a 'B' GIVEN they are male) = [tex]\frac{9}{19}[/tex]
Step-by-step explanation:
Given:
A B C Total
Male 6 18 3 27
Female 13 20 12 45
Total 19 38 15 72
We know that the probability = The number of favorable outcomes ÷ The total number of possible outcomes.
Total number = 72
1) If one student is chosen at random, Find the probability that the student got a B:
Got B = 38
P(Got a B) = [tex]\frac{38}{72}[/tex]
Simplifying the above probability, we get
P(Got a B) = [tex]\frac{19}{36}[/tex]
2) Find the probability that the student was female AND got a "C":
Female AND got a C = 12
P(Female AND got a C ) = [tex]\frac{12}{72}[/tex]
Simplifying the above probability, we get
P(Female AND got a C) = [tex]\frac{1}{6}[/tex]
3) Find the probability that the student was female OR got an "B":
Female OR got an B = Total number of female + students got B - Female got 20
= 45 + 38 - 20
= 73 - 20
= 63
P(Female OR got an B ) = [tex]\frac{63}{72}[/tex]
P(Female or Got an B = [tex]\frac{7}{8}[/tex]
4) If one student is chosen at random, find the probability that the student got a 'B' GIVEN they are male:
Total number of students who got B = 38
Student got a B given they are male = 18
P(got a 'B' GIVEN they are male) = [tex]\frac{18}{38}[/tex]
P(got a 'B' GIVEN they are male) = [tex]\frac{9}{19}[/tex]
The probability study includes: Probability of a student getting 'B' is 0.528; a student being female and getting 'C' is 0.167; being female or getting 'B' is 0.875 and the probability of a male getting 'B' is 0.667.
Explanation:To answer these questions, we need to use the concept of probability in mathematics. Probability is calculated by dividing the number of desired outcomes by the total number of outcomes.
For the probability that the student got a 'B', we divide the total number of 'B' grades (38) by the total number of students (72). This results in a probability of 38/72 or 0.528. The probability that the student was female AND got a 'C': we have 12 students fitting this description, and a total of 72 students. So, the probability would be 12/72 or 0.167. The probability that the student was female OR got a 'B', we add the number of females (45) to the number of B's (38), subtract the overlap (girls who got B's, 20). That leaves us with 45 + 38 - 20 = 63. We divide this by total students, 72, for a probability of 63/72 or 0.875. The probability that the student got a 'B' GIVEN they are male: Here we focus only on the boys, of whom there are 27. 18 of these received a 'B', giving us a probability of 18/27 or 0.667.Learn more about Probability here:
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One container of Tums® costs 4.00 dollars. Each container has eighty 1.00 g tablets. Assume each Tums® is 40.0 percent CaCO₃ by mass. Using only Tums®, you are required to neutralize 0.500 L of 0.400 M HCl. How much does this cost? Assume you are able to purchase individual tablets. Express your answer in dollars.
Answer:
The total cost is 1.25 dollars.
Step-by-step explanation:
The reaction between HCl and CaCO₃ is giving by:
2HCl(aq) + CaCO₃(s) → CaCl₂(aq) + CO₂(g) + H₂O(l) (1)
0.500L M: 100.01g/mol
0.400M
According to equation (1), 2 moles of HCl react with 1 mol of CaCO₃, so to neutralize HCl, we need the next amount of CaCO₃:
[tex] m CaCO_{3} = (\frac{1 \cdot mol HCl}{2}) \cdot M_{CaCO_{3}} = (\frac{0.500L \cdot 0.400 \frac {mol}{L}}{2}) \cdot 100.01 \frac{g}{mol} = 10.001 g [/tex]
The CaCO₃ mass of each tablet is:
[tex] m CaCO_{3} = 1 g_{tablet} \cdot \frac{40g CaCO_{3}}{100g_{tablet}} = 0.4g [/tex]
Hence, the number of tablets that we need to neutralize the HCl is:
[tex] number_{tablets} = ( \frac{1 tablet}{0.4 g CaCO_{3}}) \cdot 10.001g CaCO_{3} = 25 [/tex]
Finally, if every 80 tablets costs 4.00 dollars, 25 tablets will cost:
[tex] cost = (\frac {4 dollars}{80 tablets}) \cdot 25 tablets = 1.25 dollars [/tex]
So, the total cost to neutralize the HCl is 1.25 dollars.
I hope it helps you!
A paint manufacturer uses a machine to fill gallon cans with paint (1 galequals128 ounces). The manufacturer wants to estimate the mean volume of paint the machine is putting in the cans within 0.5 ounce. Assume the population of volumes is normally distributed.(a) Determine the minimum sample size required to construct a 90% confidence interval for the population mean. Assume the population standard deviation is 0.80 ounce.(b) The sample mean is 127 ounces. With a sample size of 8, a 90% level of confidence, and a population standard deviation of 0.80 ounce, does it seem possible that the population mean could be exactly 128 ounces? Explain.
Answer:
a) To determine the minimum sample size we need to use the formula shown in the picture 1.
E is the margin of error, which is the distance from the limits to the middle (the mean) of the confidence interval. This means that we have to divide the range of the interval by 2 to find this distance.
E = 0.5/2 = 0.25
Now we apply the formula
n = (1.645*0.80/0.25)^2 = 27.7 = 28
The minimum sample size would be 28.
b) To answer the question we are going to make a 90% confidence interval. The formula is:
(μ - E, μ + E)
μ is the mean which is 127. The formula for E is shown in the picture.
E = 0.80*1.645/√8 = 0.47
(126.5, 127.5)
This means that the true mean is going to be contained in this interval 90% of the time. This is why it doesn't seem possible that the population mean is exactly 128.
(a) Minimum sample size needed for a 90% confidence interval is 7.
(b) With a sample mean of 127 ounces, 128 ounces seems unlikely for the population mean.
To solve this problem, we can use the formula for the confidence interval of the population mean:
[tex]\[ \text{Confidence Interval} = \text{Sample Mean} \pm Z \left( \frac{\text{Population Standard Deviation}}{\sqrt{\text{Sample Size}}} \right) \][/tex]
Where:
- Sample Mean = 127 ounces
- Population Standard Deviation = 0.80 ounce
- Z = Z-score corresponding to the desired confidence level
- Sample Size = n
(a) To determine the minimum sample size required for a 90% confidence interval:
We first need to find the Z-score corresponding to a 90% confidence level. We'll use a Z-table or a calculator. For a 90% confidence level, the Z-score is approximately 1.645.
[tex]\[ \text{Margin of Error} = Z \left( \frac{\text{Population Standard Deviation}}{\sqrt{\text{Sample Size}}} \right) \][/tex]
Given that the margin of error is 0.5 ounce, we can rearrange the formula to solve for the sample size:
[tex]\[ 0.5 = 1.645 \left( \frac{0.80}{\sqrt{n}} \right) \][/tex]
Solving for ( n ):
[tex]\[ \sqrt{n} = \frac{1.645 \times 0.80}{0.5} \][/tex]
[tex]\[ \sqrt{n} = 2.632 \][/tex]
[tex]\[ n = (2.632)^2 \][/tex]
[tex]\[ n \approx 6.92 \][/tex]
Since the sample size must be a whole number, we round up to the nearest whole number. Therefore, the minimum sample size required is 7.
(b) To determine if it's possible that the population mean could be exactly 128 ounces with a sample mean of 127 ounces, a sample size of 8, and a 90% confidence level:
[tex]\[ \text{Confidence Interval} = 127 \pm 1.645 \left( \frac{0.80}{\sqrt{8}} \right) \][/tex]
[tex]\[ \text{Confidence Interval} = 127 \pm 1.645 \left( \frac{0.80}{\sqrt{8}} \right) \][/tex]
[tex]\[ \text{Confidence Interval} = 127 \pm 1.645 \left( \frac{0.80}{2.828} \right) \][/tex]
[tex]\[ \text{Confidence Interval} = 127 \pm 1.645 \times 0.283 \][/tex]
[tex]\[ \text{Confidence Interval} = 127 \pm 0.466 \][/tex]
The confidence interval is ( (126.534, 127.466) ).
Since 128 ounces is not within the confidence interval, it seems unlikely that the population mean could be exactly 128 ounces.
The Glee Club sold a total of 150 tickets to their spring concert. Student tickets cost $5.00 each and adult tickets cost $8.00 each. If they had $1,020 in ticket sales, how many adult tickets did they sell?
Answer: 60 student tickets were sold
90 adult tickets were sold
Step-by-step explanation:
Let x represent the total number of student tickets sold.
Let y represent the total number of adult tickets sold.
The Glee Club sold a total of 150 tickets to their spring concert. This means that
x + y = 150
x = 150 - y
Student tickets cost $5.00 each and adult tickets cost $8.00 each. If they had $1,020 in ticket sales,then,
5x + 8y = 1020 - - - - - -1
Substituting x = 150 - y into equation 1, it becomes
5(150 - y) + 8y = 1020
750 - 5y + 8y = 1020
- 5y + 8y = 1020 - 750
3y = 270
y = 270/3 = 90
x = 150 - 90 = 60
The graph of function f a shown. Use the zeros and the turning points of the graph to find the rule for f.
Answer:
f(x) = x(x +4)(x -3)
Step-by-step explanation:
Zeros at -4, 0, and 3 tell you the factorization is ...
f(x) = a(x +4)(x)(x -3)
Then f(2) = a(6)(2)(-1) = -12a.
The graph shows f(2) = -12, so a=1. That makes the function rule:
f(x) = x(x +4)(x -3)
__
If you want it multiplied out, it will be
f(x) = x^3 +x^2 -12x
To find the rule for function f(x) using zeros and turning points, analyze the graph. In this case, with a horizontal line between 0 and 20, there are no zeros or turning points as the function doesn't cross the x-axis or change direction.
Explanation:To find the rule for a function f(x) using its zeros and turning points, we analyze the graphical representation of the function. If the graph is a horizontal line, such as when f(x) = 20 for all 0 ≤ x ≤ 20, the function does not have any zeros or turning points within that interval, as it does not cross the x-axis nor does it change direction. Considering this particular function, we conclude that the graph is indeed a horizontal line with no turning points or zeros between x=0 and x=20.
In the general process of graphing, to illustrate the change in f(x) as x varies, we plot specific (x,y) data pairs and use these to determine trends. For functions that are not constant, like the one described above, zeros are the x-values where the function crosses the x-axis (f(x) = 0), and turning points are found where the slope of the function changes sign, which can be determined by examining the first and second derivatives of the function. However, in this scenario, the horizontal nature of the graph precludes the presence of such features.
Let X represent the number on the face that lands up when a fair six-sided number cube is tossed. The expected value of X is 3.5, and the standard deviation of X is approximately 1,708. Two fair six-sided number cubes will be tossed and the numbers appearing on the faces that land up will be added.
Which of the following values is closest to the standard deviation of the resulting sum?
(A) 1.708 (B) 1.848 (C) 2.415 (D) 3.416 (E) 5.835
Answer:
c) 2.415
Step-by-step explanation:
Given that X represent the number on the face that lands up when a fair six-sided number cube is tossed.
The expected value of X is 3.5, and the standard deviation of X is approximately 1.708.
When another die is rolled let Y represent the number on the face that lands up when a fair six-sided number cube is tossed.
The expected value of Y is 3.5, and the standard deviation of Y is approximately 1.708.
Also we find that X and Y are independent
Let U = X+Y
Then we have U as the random variable representing the sum shown by two dice
Since X and Y are independent
[tex]Var(x+y) = Var(x) +Var(y)\\= 1.708^2 *2\\= 5.83333[/tex]
Std dev for sum
= [tex]\sqrt{5.8333} \\=2.4152[/tex]
Hence option C 2.415 is correct
Answer:
2.415
Step-by-step explanation:
Its not as complicated as that other response. You never subtract standard deviations, you always add them, and you don't add them directly, you have to square them to make them variances, add them, and find the square root of it.
In this problem, you do [tex]\sqrt{(1.708^{2}) + (1.708^{2})}[/tex]
Put it in the calculator and you get 2.415
A certain company currently has how many employees?
(1) If 3 additional employees are hired by the company and all of the present employees remain, there will be at least 20 employees in the company.
(2) If no additional employees are hired by the company and 3 of the present employees resign, there will be fewer than 15 employees in the company.
Answer: The company currently has 17 employees
Step-by-step explanation:
Let X represent the number of current employees in the company
From the first information, it can be expressed mathematically as
X + 3 ≥ 20
X ≥ 20 - 3
X ≥ 17
From the second information, it can be expressed mathematically as
X - 3 < 15
X < 15 + 3
X < 18
From the above solutions, it can be deduced that
17 ≤ X < 18
The only number that fulfils this criteria is 17.
Therefore, X = 17
Design a rectangular milk carton box of width $$w, length $$l, and height $$h, which holds $$128 cubic cm of milk. The sides of the box cost $$1 cent per square cm and the top and bottom cost $$2 cents per square cm. Find the dimensions of the box that minimize the total cost of materials used.
To minimize the cost of a rectangular milk carton that holds 128 cubic cm, we need to calculate the dimensions that minimize the surface area cost. By setting up an optimization problem and using calculus, we can find the values of length, width, and height that satisfy the volume constraint and result in the lowest cost.
Minimizing Cost for a Rectangular Milk Carton
To find the dimensions of a rectangular milk carton with a given volume that minimize the cost of materials used, we can set up an optimization problem using calculus. First, we know the volume of the milk carton must be $$128 cm^3$, which gives us the constraint:
V = lwh = 128
Next, we need to express the cost function in terms of the dimensions of the box. The sides of the box cost $$1 cent per square cm, while the top and bottom cost $2 cents per square cm. Thus, the total cost, C, in cents, is:
C = 2lw + 2wh + 2lh + (4 * l * w)
To minimize the cost, we would take the partial derivatives with respect to l, w, and h, set them equal to zero, and solve the system of equations while taking into account the volume constraint. This involves the method of Lagrange multipliers or directly substituting the volume constraint into the cost function to eliminate one variable and then taking the derivative with respect to the other variables.
By finding the derivative of the cost function and setting it to zero, you can determine the values of l, w, and h that will result in the minimum cost while respecting the volume constraint. Since this is an applied problem, it is important to check that the resulting values are practical, meaning they should be positive and make sense for a milk carton.
Carmen received a $100 bill as a birthday gift. She bought a book online for $24.95. Then she bought a backpack for $39.75 at a variety store. How much money did she have left after making her purchases
Answer:35.3
Step-by-step explanation:
100-24.95-39.75=35.3
Final answer:
Carmen spent a total of $64.70 on a book and a backpack and had $35.30 left from her original $100 after her purchases.
Explanation:
To calculate how much money Carmen had left after making her purchases, we first need to add the cost of the book and the backpack to find the total amount spent. She spent $24.95 on the book and $39.75 on the backpack. Add these two amounts together to find the total spent:
Book: $24.95Backpack: $39.75Total spent: $24.95 + $39.75 = $64.70Next, subtract the total spent from the original $100 bill:
Original amount: $100.00Total spent: $64.70Money left: $100.00 - $64.70 = $35.30Therefore, Carmen has $35.30 left after her purchases.
Just took pictures to make it easier.
Answer:
8. [tex]\displaystyle \frac{9[x + 5]}{x - 14}[/tex]
7. [tex]\displaystyle -\frac{2x - 1}{2[3x - 5]}[/tex]
6. [tex]\displaystyle \frac{2[x - 4]}{5[x + 3]}[/tex]
5. [tex]\displaystyle \frac{2x + 7}{x + 3}[/tex]
4. [tex]\displaystyle 3x^{-1}[/tex]
Step-by-step explanation:
All work is shown above from 8 − 4.
I am joyous to assist you anytime.
Let xij = gallons of component i used in gasoline j. Assume that we have two components and two types of gasoline. There are 8,000 gallons of component 1 available, and the demand gasoline types 1 and 2 are 11,000 and 14,000 gallons respectively. Write the supply constraint for component 1.
a) x21 + x22 = 8000
b) x12 + x22 = 8000
c) x11 + x12 = 8000
Answer:
c) x11 + x12 = 8000
Step-by-step explanation:
Xij = Gallons of the ith component used in the jth Gasoline type
This invariably tells us what component is in which gasoline type.
The gasoline types are:
Gasoline 1 = 11,000 gallons
Gasoline 2 = 14,000 gallons
Assuming two(2) components types:
Component 1
Component 2
The possible combinations Xij are :
Gasoline 1 Gasoline 2
Component 1 X11 X12
Component 2 X21 X22
From the above , it is clear that the supply constraint for component 1 across the gasoline types is given by X11 & X12
Mathematically, since there are 8,000 gallons of Component 1, the supply constraint is given by:
X11 + X12 = 8000
The supply constraint for component 1, given the available gallons and the usage in two types of gasoline, is represented by the equation x11 + x12 = 8000.
Explanation:The supply constraint for component 1 when formulating an equation that represents the total usage of component 1 in both types of gasoline should reflect the total availability of component 1. Since there are 8,000 gallons of component 1 available, the correct mathematical representation of the supply constraint for component 1 is:
x11 + x12 = 8000
This equation means that the sum of gallons of component 1 used in gasoline type 1 (x11) and gasoline type 2 (x12) must equal the total available gallons of component 1, which is 8,000 gallons.
Paper depot is swnding out 28 trucks to deliver paper to customers today. Each truck is being loaded with 3 cases of paper. 283 of all the cases are plain white paper. How many cases of all types of paper will be delivered today?
Answer:
84 cases
Step-by-step explanation:
Given that:
Number of trucks: 28
Paper cases each truck can load = 3
Total cases of white paper = 283
So the cases delivered will be = 28 *3 = 84 cases will be delivered today
i hope it will help you!
In 2010, a town's population was 83 thousand. By 2015 the population had grown to 105 thousand.a) Find an exponential equation for the town's population.b) Determine in what year the population will exceed 135 thousand. As always, show your work for finding the equation and solving for the year algebraically.
Answer:
(a) [tex]y = 83 (1.048)^x[/tex]
(b) 2020
Step-by-step explanation:
(a) Let the exponential equation that shows the population in thousand after x years,
[tex]y = ab^x[/tex]
Also, suppose the population is estimated since 2010,
So, x = 0, y = 83 thousands,
[tex]83 = ab^0[/tex]
[tex]\implies a = 83[/tex]
Again by 2015 the population had grown to 105 thousand,
i.e. y = 105, if x = 5,
[tex]\implies 105 = ab^5[/tex]
[tex]\implies 105 = 83 b^5[/tex]
[tex]\implies b = (\frac{105}{83})^\frac{1}{5}=1.0481471103\approx 1.048[/tex]
Hence, the required function,
[tex]y = 83 (1.048)^x[/tex]
(b) if y = 135,
[tex]135 = 83(1.048)^x[/tex]
[tex]\implies x = 10.375\approx 10[/tex]
Hence, after approximately 10 years since 2010 i.e. in 2020 the population would be 135.
Final answer:
By assuming exponential growth, we derived the equation P = 83,000 * e^(0.048t) and determined that the town's population would exceed 135,000 around the year 2021.
Explanation:
To find an exponential equation for the town's population and determine in what year the population will exceed 135 thousand, we start by assuming the population growth follows the form P = P0 * e^(rt), where P is the final population, P0 is the initial population, r is the rate of growth, and t is the time in years. For this town, in 2010 (t=0), the population was 83,000 (P0=83,000), and by 2015 (t=5), the population grew to 105,000.
Substituting these values into the formula, we have 105,000 = 83,000 * e^(5r). Solving for r, we find that r ≈ 0.048. Thus, the exponential growth equation is roughly P = 83,000 * e^(0.048t).
To determine when the population will exceed 135,000, we set P > 135,000 and solve for t. This gives us the inequality 135,000 < 83,000 * e^(0.048t). Solving this, we find that t ≈ 10.24 years after 2010, which rounds up to the year 2021 when the population will exceed 135 thousand.
the volume of a cube with side length x is V(x)=x^3. The volume of a cylinder with radius x and height 0.5x is shown in the graph. When x=1, which volume is greater?
Answer:
the cylinder volume is greater
Step-by-step explanation:
The volume of a cube with x=1 is ...
V(1) = 1^3 = 1
The graph shows y ≈ 1.5 for x=1. Since 1.5 > 1, the volume of the cylinder is greater.
When x = 1, the volume of the cylinder is greater than the volume of the cube.
To determine which volume is greater when x = 1, we can calculate the volume of the cube and the volume of the cylinder at x = 1 and compare them.
For the cube with side length x, the volume is given by V(x) = x^3. So, when x = 1:
V(cube) = (1)^3 = 1
For the cylinder with radius x and height 0.5x, the volume is given by the formula for the volume of a cylinder: V(cylinder) = πr^2h, where r is the radius and h is the height.
When x = 1:
r = 1 (radius)
h = 0.5(1) = 0.5 (height)
V(cylinder) = π(1^2)(0.5) = π(0.5) = 0.5π
Now, we need to compare the volumes.
V(cube) = 1
V(cylinder) = 0.5π ≈ 1.57 (rounded to two decimal places)
So, when x = 1, the volume of the cylinder is greater than the volume of the cube.
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Mary scored a total of 27 points in a basketball game. During the game, all her points came from 3-point shots and 2-point shots. The number of 3-point shots she made is 4 more than her 2-point shots. How many 3-point shots and 2-point shots did she make?
A) 2 three-point shots and 6 two-point shots
B) 3 three-point shots and 7 two-point shots
C) 6 three=point shots and 2 two-point shots
D) 7 three-point shots and 3 two-point shots
Answer:
D
Step-by-step explanation:
7x3=21 and 3x2=6
21+6=27
D) The number of three point shots is 7 and the number of two points shots is 3.
What are Linear Equations?Linear equations are equation involving one or more expressions including variables and constants and the variables are having no exponents or the exponent of the variable is 1.
Given that,
Mary scored a total of 27 points in a basketball game, all from 3-point shots and 2-point shots.
Let x be the number of 3-point shots and y be the number of 2-point shots.
3x + 2y = 27 [Equation 1]
The number of 3-point shots she made is 4 more than her 2-point shots.
x = y + 4 [Equation 2]
Substitute [Equation 2] in [Equation 1].
3(y + 4) + 2y = 27
3y + 12 + 2y = 27
5y = 15
y = 3
x = y + 4 = 3 + 4 = 7
Hence the option is D) 7 three-point shots and 3 two-point shots.
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Help me!!! 100 points !!
Answer:
200000 times larger, its not in scientifc notation it should be 1.3*10^9, and the answer is 1.824*10^15,or182400000000000
Step-by-step explanation:
Answer:
Step-by-step explanation:
To find how many times larger, divide the CEO's salary by the teachers to get
5 000 000 / 25 000 = 200
this in scientific notation is 2.00 * 10^2
d - is not in scientific notation because there is more than one digit before the decimal place
1.3 * 10 ^9
e - 1.748 * 10^15
determine the intervals on which the function is increasing, decreasing, and constant
PLEASE HELP!!!
A total of 517 tickets were sold for the school play. They were either adult tickets or student tickets. There were 67 more student tickets sold than adult tickets. How many adult tickets were sold?
Answer:
225 adult tickets were sold
Step-by-step explanation:
You are asked to find the number of adult tickets sold. It is convenient to let a variable represent that quantity. We can call it "a" to remind us it is the number of adult tickets (not student tickets).
The number of student tickets is 67 more, so can be represented by (a+67). The total number of tickets sold is the sum of the numbers of adult tickets and student tickets:
(a) + (a+67) = 517
2a + 67 = 517 . . . . . collect terms
2a = 450 . . . . . . . . . subtract 67
a = 225 . . . . . . . . . . divide by 2
There were 225 adult tickets sold.
_____
Check
The number of student tickets sold is ...
a+67 = 225 +67 = 292
And the total number of tickets sold is ...
225 + 292 = 517 . . . . . answer checks OK
1. In a survey sample of 83 respondents, about 30.1 percent of the sample work less than 40 hours per week. Calculate a 68 percent confidence interval for the proportion of persons who work less than 40 hours per week.
Answer:
A 68 percent confidence interval for the proportion of persons who work less than 40 hours per week is (0.251, 0.351), or equivalently (25.1%, 35.1%)
Step-by-step explanation:
We have a large sample size of n = 83 respondents. Let p be the true proportion of persons who work less than 40 hours per week. A point estimate of p is [tex]\hat{p} = 0.301[/tex] because about 30.1 percent of the sample work less than 40 hours per week. We can estimate the standard deviation of [tex]\hat{p}[/tex] as [tex]\sqrt{\hat{p}(1-\hat{p})/n}=\sqrt{0.301(1-0.301)/83} = 0.0503[/tex]. A [tex]100(1-\alpha)%[/tex] confidence interval is given by [tex]\hat{p}\pm z_{\alpha/2}\sqrt{\hat{p}(1-\hat{p})/n[/tex], then, a 68% confidence interval is [tex]0.301\pm z_{0.32/2}0.0503[/tex], i.e., [tex]0.301\pm (0.9944)(0.0503)[/tex], i.e., (0.251, 0.351). [tex]z_{0.16} = 0.9944[/tex] is the value that satisfies that there is an area of 0.16 above this and under the standard normal curve.
Final answer:
To calculate the 68 percent confidence interval for the proportion of persons working less than 40 hours per week from a sample of 83 respondents with a sample proportion of 30.1 percent, we use the formula for the confidence interval for a proportion. The resulting interval is approximately 25.06% to 35.14%.
Explanation:
To calculate a 68 percent confidence interval for the proportion of persons who work less than 40 hours per week from a sample of 83 respondents, where 30.1 percent work less than 40 hours, we use the formula for a confidence interval for a proportion:
In this formula:
p is the sample proportion (0.301 in this case).z* is the z-value corresponding to the desired confidence level (for 68 percent confidence, use the z-value corresponding to one standard deviation from the mean in a standard normal distribution, which is approximately 1).n is the sample size (83).Plugging the values into the formula we get:
0.301±1*sqrt((0.301(0.699)/83))
Calculating the square root part, we have:
0.301±1*sqrt((0.301*0.699)/83)
= 0.301±1*sqrt(0.210699/83)
= 0.301±1*sqrt(0.002539)
= 0.301±1*0.05039
= 0.301±0.05039
The confidence interval is thus:
0.301-0.05039 to 0.301+0.05039
= 0.25061 to 0.35139
Hence, with a 68 percent confidence level, we can say that the true proportion of the population that works less than 40 hours per week is estimated to be between 25.06% and 35.14%.
PLEASE HELP! Will make correct answer brainliest
Answer:
C. g(0)=8; g(1)=12
Step-by-step explanation:
Answer:
The answer to your question is g(0) = 12; g(1) = 8. Letter D
Step-by-step explanation:
g(x) = 12(2/3)[tex]^{x}[/tex]
a) Substitute 0 in the equation and simplify
g(0) = 12(2/3) ⁰
g(0) = 12(1)
g(0) = 12
b) Substitute 1 in the equation and simplify
g(1) = 12(2/3)¹
g(1) = 12(2/3)
g(1) = 24/3
g(1) = 8
If A, B, and C are the measures of the angles of any triangle and if a, b, and c are the lengths of the sides opposite the corresponding angles, then which of the following expressions does not represent the area of the triangle?
A. (1/2)bcsinA
B. (1/2)acsinB
C. (1/2)acsinA
D. (1/2)absinC
Answer:
C. (1/2)acsinA
Step-by-step explanation:
Given is that, A, B, and C are the measures of the angles of a triangle and a, b, and c are the lengths of the sides opposite the corresponding angles.
So, the expression that does not represent the area of the triangle is :
C. (1/2)acsinA
i have a couple questions with my geometry homework which is due tomorrow, could someone try answering them? I've been trying to solve them for the past hour.
Answer:
2. 100° 22. 7
18. 6 23. 9
19. 8 24. 65°
20. 55° 25. AB = (1/2)DF
21. 6 26. AB ║ DF
Step-by-step explanation:
Please be aware that the triangle measurements shown for problems 18–20 and 22–24 cannot exist. In the first case, the angle is closer to 48.6° (not 35°), and in the second case, the angle is closer to 51.1°, not 25°. So, you have to take the numbers at face value and not think too deeply about them. (This state of affairs is all too common in geometry problems these days.)
_____
2. You have done yourself no favors by marking the drawing the way you have. Look again at the given conditions. You will find that x+2x must total a right angle, so x=30°. Angle P is the complement of 40°, so is 50°. Then the sum of x and angle P is 30° +50° = 80°, and the angle of interest is the supplement of that, 100°.
__
18–20. Perpendicular bisector NO means ∆NOL ≅ ΔNOM. Corresponding parts have the same measures, and angle L is the complement of the marked angle.
__
21. ∆ODA ≅ ∆ODB by hypotenuse-angle congruence (HA), so corresponding parts are the same measure. DB = DA = 6.
__
22–24. ∆VPT ≅ ∆VPR by LL congruence, so corresponding measures are the same. Once again, the angle in question is the complement of the given angle.
__
25–26. You observe that A is the midpoint of DE, and B is the midpoint of FE, so AB is what is called a "midsegment." The features of a midsegment are that it is ...
half the length of the base (DF)parallel to the baseLet X, Y , Z be three random variables which satisfy the following conditions: Var(X) = 4, Var(Y ) = 9, Var(Z) = 16. Cov(X, Y ) = −2, Cov(Z, X) = 3, and Y and Z are independent. Find: (a) Cov(X + 2Y, Y − Z). (b) Var(3X − Y ). (c) Var(X + Y + Z)
Answer:
13,57,31
Step-by-step explanation:
Given that X, Y , Z be three random variables which satisfy the following conditions:
Var(X) = 4, Var(Y ) = 9, Var(Z) = 16. Cov(X, Y ) = −2, Cov(Z, X) = 3,
Var(y,z) =0 since given as independent
To find
[tex]a) Cov (x+2y, y-z)\\ \\= cov (x,y) +cov (2y,y) -cov (x,z) -cov(2y,z)\\= cov (x,y) +2cov (y,y) -cov (x,z) -2cov(y,z)\\=-2+2 var(y) -3-0\\= -2+18-3\\=13[/tex]
b) [tex]Var(3X − Y ).\\= 9Var(x)+var(y) -6 covar (x,y)\\= 36 +9+12\\= 57[/tex]
c) Var(X + Y + Z)[tex]=Var(x) = Var(Y) +Var(z) +2cov (x,y) +2cov (y,z) +2cov (x,z)\\= 4+9+16+(-4) +6\\= 31[/tex]
Note:
Var(x+y) = var(x) + Var(Y) +2cov (x,y)
Var(x+2y) = Var(x) +4Var(y)+4cov (x,y)
18 - 24i write in trigonometric form
[tex]\boxed{18-24i=30(cos(5.36)+isin(5.36))}[/tex]
Explanation:Unlike 0, we can write any complex number in the trigonometric form:
[tex]z=r(cos\alpha+isin\alpha)[/tex]
We have the complex number:
[tex]18-24i[/tex]
So [tex]r[/tex] can be found as:
[tex]r=\sqrt{x^2+y^2} \\ \\ \\ Where: \\ \\ x=18 \\ \\ y=-24 \\ \\ r=\sqrt{18^2+(-24)^2} \\ \\ r=\sqrt{324+576} \\ \\ r=\sqrt{900} \\ \\ r=30[/tex]
Now for α:
[tex]\alpha=arctan(\frac{y}{x}) \\ \\ Since \ the \ complex \ number \ lies \ on \ the \ fourth \ quadrant: \\ \\ \alpha=arctan(\frac{-24}{18})=-53.13^{\circ} \ or \ 360-53.13=306.87^{\circ}[/tex]
Finally:
[tex]Convert \ into \ radian: \\ \\ 360^{\circ}\times \frac{\pi}{180}=5.36rad \\ \\ \\ Hence: \\ \\ \boxed{18-24i=30(cos(5.36)+isin(5.36))}[/tex]
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Charles and Lisa were having a apple eating contest. They ate eighteen apples between the two of them. Lisa ate two more apples than Charles. How many apples did Lisa eat?
A.
6
B.
10
C.
12
D.
15
Answer:10
Step-by-step explanation:
ok they are 18 apples
and they ate them
one ate two more thatn the other one
so we can say that one of them is x(charles)
and the one who ate two more is x+2(lisa)
so we get this
x+x+2=18
2x+2=18
2x+18-2
2x=16
x=8
so charles ate 8
and lisa 10(8+2)
Answer:
1000% its 10
Step-by-step explanation:
The set of integers is closed under the operation of addition.
A: Which equation illustrates this concept?
B: Which statement correctly explains this concept?
Select one answer for question A and one answer for question B.
A: 2+27=29
A: 34÷4=172
A: 1−3=−2
A: 2⋅6=12
B: The sum of the integers 2 and 27 is the integer 29, which demonstrates that integers are closed under addition.
B: The quotient of the integers 34 and 4 is the integer 172, which demonstrates that integers are closed under addition.
B: The difference of the integers 1 and 3 is not an integer, −2, which does not demonstrate that integers are closed under addition.
B: The product of the integers 2 and 6 is not an integer, 12, which does not demonstrate that integers are closed under addition.
Answer:
A: 2+27=29
B: The sum of the integers 2 and 27 is the integer 29, which demonstrates that integers are closed under addition.
Step-by-step explanation:
Since, closed property of addition for a set A is defined as,
∀ x, y ∈ A ⇒ x + y ∈ A,
∵ Set of integer is closed under multiplication,
If Z represents the set of integer,
Then 2, 27 ∈ Z ⇒ 2 + 27 = 29 ∈ Z,
Hence, the equation illustrates given statement,
2+27 = 29
The statement that correctly explains given statement,
The sum of the integers 2 and 27 is the integer 29, which demonstrates that integers are closed under addition.c
Henry was buying meals for his family. He bought 3 meals that included a chicken, corn, and a drink and spent a total of $20.25. He paid 6% sales tax. How much did each meal cost before tax?
Answer:
The price of each meal cost before tax is $ 6.367
Step-by-step explanation:
Given as :
The total price of three meals = $ 20.25
The sales tax included in the total price = 6 %
So, Let the cost of meal before sales tax = x
Or, x + 6 % of x = $ 20.25
or, x + 0.06 x = $ 20.25
Or, 1.06 x = $ 20.25
∴ x = [tex]\frac{20.25}{1.06}[/tex]
I.e x = $ 19.10
Or, price of three meals before tax = $ 19.10
so, The price of each meal = [tex]\frac{19.10}{3}[/tex] = $ 6.367
Hence The price of each meal cost before tax is $ 6.367 answer
In the figure below, and are right triangles. The measure of is 30°, the measure of is 6 units, the measure of is 3 units, and the measure of is 12 units. Determine the measure of .
To determine the length of an unknown side in a right triangle, we can use the sin, cos, tan ratios or the Pythagorean theorem, depending on which sides and/or angles we already know. However, the specific information about measures and sides in the two right triangles in the question is not clear, hence a specific answer cannot be provided.
Explanation:The subject of this question is trigonometry, a branch of mathematics that studies relationships involving lengths and angles of triangles. As the triangles are right-angled, we can use properties specific to this type of triangle. When given the measure of an angle in a right triangle, we can find the lengths of the sides using trigonometric ratios such as sine, cosine, and tangent.
In this particular problem, we're given the measure of an angle and lengths of some sides but the information about the measures of the triangles and their sides are not clearly mentioned. To provide an answer, we'd need clear information on which triangle contains the given angle and sides.
However, to give you an idea of how this type of problem can be solved, let's consider that the measure of the angle α is 30 degrees, the measure of side a is 6 units, the measure of side b is 3 units, and the measure of side c (the hypotenuse) is 12 units (These are just hypothetical names to represent the sides). If we are asked to find the length of an unknown side, we can use the Sin, Cos or Tan ratios or even the Pythagorean theorem (a² + b² = c²), depending on the sides that we know.
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