Answer:
48%
Step-by-step explanation:
In order to find the percentage we need to divide the sold cans by total cans and multiply the result by 100.
Total cans = 150
Sold cans = 72
→ 72/150 = 0.48
→ 0.48 * 100 = 48
The percentage of the cans of popcorn stocked were sold that weekend is 48%
The given parameters are:
Total can of popcorn = 150
Sold can of popcorn = 72
The percentage of can sold is then calculated as:
[tex]\%Sold = \frac{72}{150} *100\%[/tex]
Multiply 72 and 100
[tex]\%Sold = \frac{7200}{150}\%[/tex]
Divide 7200 by 150
[tex]\%Sold = \%48[/tex]
Hence, the percentage of the cans of popcorn stocked were sold that weekend is 48%
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A publishing company is going to have 24000 books printed. There are between 3 and 4 books out of every 3000 printed that will have a printing error. At this rate, which number could be the number of books that will have a printing error in the 24000
Given:
24,000 books
Between 3 and 4 books will have a printing error in every 3000 printed books
Find: in the 24000 books, the total number of books that will have a printing error
Solution:
Based from the given, we need to know how many sets of 3000 we
have in the 24000 books so:
24000 / 3000 = 8
Now, in each set of 3000 we have between 3 and 4 errors and we have 8 sets of 3000 books:
minimum errors 3 * 8 = 24
maximum errors 4 * 8 = 32
Therefore, at this rate, the number of books that will have a printing error in the 24000
will be between 24 and 32 or 24 < E < 32.
In 24,000 books, the number of printing errors could be between 24 and 32. This is calculated based on an error rate of 3 to 4 errors per 3,000 books.
To determine the number of printing errors in 24,000 books, we need to understand the error rate. The problem states that there are between 3 and 4 books with errors per 3,000 printed books.
First, find the range of error rates per 3,000 books:
Minimum errors: 3 errors per 3,000 booksMaximum errors: 4 errors per 3,000 booksNext, scale this up to 24,000 books:
Minimum errors: (3 errors/3,000 books) x 24,000 books = 24 errorsMaximum errors: (4 errors/3,000 books) x 24,000 books = 32 errorsTherefore, the number of books with printing errors in 24,000 books will be between 24 and 32.
The Rectangles are similar. Find the value of the variable (Picture Included)
Sam took his family to the zoo. An adult's ticket is two times the cost of a child's ticket. The total cost for two adults' tickets and three children's tickets was $28. How much do the tickets cost? A. Child's ticket = $5.60, adult's ticket = $11.20 B. Child's ticket = $2, adult's ticket = $4 C. Child's ticket = $4, adult's ticket = $8 D. Child's ticket = $7, adult's ticket = $14 \
Sarah bought a lawnmower for $320. She signed up for the buy now pay later plan at the store with the following conditions: $100 down and payments of $25 for the next 12 months. The extra cost paid by taking this plan is equivalent to what actual yearly rate of interest?
Answer:
25%
Step-by-step explanation:
Just here to help cause im doing this too lol
Find f. (use c for the constant of the first antiderivative and d for the constant of the second antiderivative.) f ''(x) = 12x + sin x
You received 1⁄3 pound of candy from your grandmother, 1⁄2 pound of candy from your sister, but your best friend ate 1⁄5 pound of candy. How much candy do you have left?
The total amount of candy left after adding the candy received from the grandmother and the sister, and subtracting the candy eaten by the friend, is approximately 0.63 pounds.
Explanation:First, we add up the amounts of candy you received. You started with 1/3 pound from your grandmother and received an additional 1/2 pound from your sister, for a total of 5/6 pound of candy. However, because your friend ate some, we subtract 1/5 pound from this total. To do this, we need to convert all fractions to have a common denominator, which is 30 in this case. Therefore, 5/6 becomes 25/30, and 1/5 becomes 6/30. Subtraction gives us (25-6)/30 = 19/30 or approximately 0.63 pounds of candy left.
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An employee earns $36 per hour and 1.5 times that rate for all hours in excess of 40 hours per week. assume that the employee worked 60 hours during the week, and that the gross pay prior to the current week totaled $52,200. assume further that the social security tax rate was 6.0%, the medicare tax rate was 1.5%, and federal income tax to be withheld was $605.
Answer:
An employee’s rate of pay is $36 per hour, with time and a half for all hours worked in excess of 40 during a week. The employee worked 48 hours during the week. The amount of the employee’s gross pay for the week is:
Step-by-step explanation:
A golden rectangle is to be constructed such that the longest side is 18 inches long. How long is the other side? (Round your answer to the nearest tenth of an inch.)
An item is regularly priced at
$80
. It is now priced at a discount of
85%
off the regular price. What is the price now?
85% = 0.85
1-0.85 = 0.15
80 x 0.15 = 12
the price now is $12
Crestwood Paint Supply had a beginning inventory of 10 cans of paint at $25.00 per can. They purchased 20 cans during the month at $30.00 per can. They had an ending inventory valued at $500. How much paint in dollars was used for the month? A. $250 B. $1,350 C. $850 D. $350
On which number line do the points represent negative seven and one over two and +1?
Answer:
d
Step-by-step explanation:
greens theorem. find the max value of the line integral where f=(13x^2y+3y^3-y)i-12x^3j and C is any positively oriented closed curve. max=?
Final answer:
To find the max value of a line integral over a closed curve using Green's Theorem, consider the curl of the given vector field and apply the theorem to express the result. The maximum value of the line integral is -2y²dy, determined through vector calculus and Green's Theorem application.
Explanation:
Green's Theorem states that for a vector field f in the form given, the max value of the line integral over any positively oriented closed curve C can be found by considering the curl of f.
By applying Green's Theorem, we can find that the maximum value of the line integral is -2 y²dy.
This computation involves utilizing vector calculus and understanding how to apply Green's Theorem to find the extremum of the line integral.
A copy machine makes 28 copies per minute. How long does it take to make 154 copies?
k friends evenly divided up a 12-slice pizza. One of the friends, Harris, ate 1 fewer slice than he was given. How many slices of pizza did Harris eat? Write your answer as an expression.
Final answer:
Harris ate 12/k - 1 slices of pizza after a 12-slice pizza was divided evenly among k friends and he ate one less than he was given.
Explanation:
To find out how many slices of pizza Harris ate, we initially need to determine how many slices each person would get if the 12-slice pizza is divided evenly among k friends.
Each friend would get 12/k slices.
Since Harris ate 1 fewer slice than he was given, we subtract 1 from the number of slices he was supposed to get.
Therefore, the expression for the number of slices Harris ate is 12/k - 1.
Assume that two fair dice are rolled. First compute P(F) and then P(F|E). Explain why one would expect the probability of F to change as it did when we added the condition that E had occurred.
F: the total is two
E: an even
total shows on the dice
Compute P(F).
P(F)equals=
nothing
(Simplify your answer.)
In the process of calculating probabilities of events on a pair of dice, we found P(F), the probability of rolling a total of two, to be 1/36. P(E), the probability of rolling an even total, to be 1/2. However, when determining P(F|E), the probability of rolling a two given we've rolled an even total, the probability changes to 1/18 due to the reduced sample space.
Explanation:The concepts involved in this question are related to probability, specifically the principles governing dice rolls. In this particular scenario, the events are rolling two dice and getting a total of two (Event F), and rolling an even total on the dice (Event E).
In this specific scenario, event F (the total is two) can only occur in one way - when both dice show 1. Since there are 36 potential outcomes when two dice are rolled (6 possibilities for the first die and 6 for the second), the probability of event F, P(F), is 1/36.
Event E (an even total) can occur in 18 ways (2,4,6 for the first die and 1,3,5 for the second or 1,3,5 for the first die and 2,4,6 for the second). So, P(E) = 18/36 = 1/2. However, when considering P(F|E) (the probability of event F given that event E has occurred), you need to adjust your consideration of 'total possibilities' based only on event E. Since P(E) = 1/2, your total possibilities now become 18. From these 18, only one will result in a total of two. Therefore, P(F|E) = 1/18.
Of course, there's different perspectives to consider how adding the condition that E had occurred would change the probability of event F. Essentially, by narrowing down the potential outcomes to only those that involve event E, you're working with a reduced sample space. This in turn affects the likelihood of event F occurring, hence the alteration in probability.
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Find the sum of the series. 1 + z/5 + z^2/25 + z^3.125
The City Housing Authority has received 75 applications from qualified applicants for ten low-income apartments. Five of the apartments are on the north side of town, and five are on the south side. If the apartments are to be assigned by means of a lottery, find the following probabilities. (a) A specific qualified applicant will be selected for one of these apartments. (Round your answer to three decimal places.) (b) Two specific qualified applicants will be selected for apartments on the same side of town. (Round your answer to five decimal places.)
You attend an amusement park with your family. Your parents buy you an all-ride pass for $20, shown as fx. Instead of getting a pass, your parents decide to pay $4 for each ride they take, shown as gx. What function shows the correct combination of these two functions to represent the total cost to them of attending the amusement park that day, shown as hx?
A. fx = 20x, gx = 4, hx = 20x + 4
B. fx = 20, gx = 4, hx = 4 + 20
C. fx = 20, gx) = 4x, hx = 4x + 20
D. fx = 20x, gx = 4x, hx = 20x + 4x
A $33$-gon $P_1$ is drawn in the Cartesian plane. The sum of the $x$-coordinates of the $33$ vertices equals $99$. The midpoints of the sides of $P_1$ form a second $33$-gon, $P_2$. Finally, the midpoints of the sides of $P_2$ form a third $33$-gon, $P_3$. Find the sum of the $x$-coordinates of the vertices of $P_3$.
1.
The midpoint MPQ of PQ is given by (a + c /
2, b + d / 2)
2.
Let the x coordinates of the vertices of P_1 be :
x1, x2, x3,…x33
the x coordinates of P_2 be :
z1, x2, x3,…z33
and the x coordinates of P_3 be:
w1, w2, w3,…w33
3.
We are given with:
X1 + x2 + x3… + x33 = 99
We also want to find the value of w1 + w2 + w3… + w33.
4.
Now, based from the midpoint formula:
Z1 = (x1 + x2) / 2
Z2 = (x2 + x3) / 2
Z3 = (x3 + x4) / 2
Z33 = (x33 + x1) / 2
and
W1
= (z1 + z1) / 2
W2 = (z2 + z3) / 2
W3
= (z3 + z4) / 2
W13 = (z33 + z1) / 2
.
.
5.
W1 + w1 + w3… + w33 = (z1 + z1) / 2 + (z2 + z3) / 2 + (z33 + z1) / 2 = 2 (z1 + z2 + z3… + z33) / 2
Z1
+ z1 + z3… + z33 = (x1 + x2) / 2 + (x2 + x3) / 2
+ (x33 + x1) / 2
2 (x1 + x2 + x3… + x33) / 2 = (x1 + x2 +
x3… + x33 = 99
Answer: 99
If 3✖️/4 =7 ➖x/3,then x=
Using rectangles whose height is given by the value of the function at the midpoint of the rectangle's base, estimate the area under the graph using first two and then four rectangles. f(x)equals=x squared2 between xequals=1 and xequals=2
The area under the graph by using the first two and then four rectangles is [tex]2.958[/tex] units square.
For reference use the below-given graph.
Given function is
[tex]f(x)=x^{2}[/tex] when [tex]x=1[/tex] to [tex]x=2[/tex] .
The first rectangle of the first part graph goes from [tex]1.0[/tex] to [tex]1.6[/tex], so the width will be [tex]0.6[/tex] units. And the height measured from the middle point i.e. [tex]1.3[/tex] is
[tex]f(1.3)=(1.3)^{2}[/tex]
[tex]=1.69[/tex] units.
Then the area of the first rectangle is [tex]0.6\times1.69=1.014[/tex] units square.
Similarly, the second rectangle of the first part graph goes from [tex]1.6[/tex] to [tex]2.0[/tex], so the width will be [tex]0.4[/tex] units. And the height measured from the middle point i.e. [tex]1.8[/tex] is
[tex]f(1.8)=(1.8)^{2}[/tex]
[tex]=3.24[/tex] units.
So, the area of the second rectangle is [tex]0.6\times3.24=1.944[/tex] units square.
Hence, the final area under the graph will be [tex]1.014+1.944=2.958[/tex] units square.
Further, we can do the same for another part of the graph to find the area under the graph by using four rectangles.
For example, the first rectangle of the four has a width of [tex]0.6[/tex] units and a height of [tex]f(1.1)=(1.1)^{2}[/tex]
[tex]=1.21[/tex] units.
Therefore, the area under the graph by using the first two and then four rectangles is [tex]2.958[/tex] units square.
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The estimated areas under the curve of the function f(x)=x^2 between x = 1 and x = 2 are 2.3125 using two rectangles and 2.3281 using four rectangles
Explanation:To estimate the area under the graph of the function f(x)=x^2 between x = 1 and x = 2 using rectangles, we use the method of midpoint Riemann sums. For this question, let's use 2 rectangles and then 4 rectangles.
First, for 2 rectangles, the interval from 1 to 2 is divided into 2 equal parts: [1, 1.5] and [1.5, 2]. The midpoints of these intervals are 1.25 and 1.75. The height of each rectangle is given by the function value at these midpoints: [tex]f(1.25) = (1.25)^2 =1.5625, and f(1.75) = (1.75)^2 = 3.0625.[/tex] The total area of the rectangles is thus (0.5 * 1.5625) + (0.5 * 3.0625) = 2.3125.
Next, for 4 rectangles, the interval from 1 to 2 is divided into 4 equal parts: [1, 1.25], [1.25, 1.5], [1.5, 1.75], [1.75, 2]. The midpoints of these intervals are 1.125, 1.375, 1.625, 1.875. The height of each rectangle is given by the function value at these midpoints: [tex]f(1.125) = (1.125)^2 = 1.26562, f(1.375) = (1.375)^2 = 1.8906, f(1.625) = (1.625)^2 = 2.6406[/tex], and f(1.875) = (1.875)^2 = 3.5156. The total area of the rectangles is thus [tex](0.25 * 1.26562) + (0.25 * 1.8906) + (0.25 * 2.6406) + (0.25 * 3.5156) = 2.3281.[/tex]
These are the estimated areas under the curve for 2 rectangles and 4 rectangles respectively. And as you can see, the more rectangles we use, the closer we get to the actual area under the curve.
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BC is tangent to circle A at B and to circle D at C. What is AD to the nearest tenth?
Which set of coordinates, when paired with (-3, -2) and (-5, -2), result in a square?
Do the side lengths of 5, 6, and 8 form a Pythagorean triple?
Yes
No
In a kitchen there are four containers that can hold different quantities of water as shown in the figure below
1-(x-2) liters
2- x liters
3- (x+2)liters
4- (x+4) liters
How many liters of water can the four containers hold in all
X^4+4
2x+4
X^2+2x
4x+4
A cube is packed with decorative pebbles. If the cube has a side length of 6 inches, and each pebble weighs on average 0.5 lb per cubic inch, what is the total weight of the pebbles in the cube?
Answer: 108 lbs.
Step-by-step explanation:
Given : A cube is packed with decorative pebbles. If the cube has a side length of 6 inches.
Volume of cube = [tex](side)^3[/tex]
i.e. Volume of cube = [tex](6)^3=216\text{ cubic inches}[/tex]
Since , each pebble weighs on average 0.5 lb per cubic inch.
Then, the total weight of the pebbles in the cube will be
= 0.5 x Volume of cube
= [tex]0.5\times216=108\text{ lb}[/tex]
Hence, the total weight of the pebbles in the cube =108 lbs.
A test consists of 20 problems and students are told to answer any 10 of these questions. In how many different ways can they choose the 10 questions?
Answer: The required number of ways is 184756.
Step-by-step explanation: Given that a test consists of 20 problems and students are told to answer any 10 of these questions.
We are to find the number of different ways in which the students choose 10 questions.
We know that
the number of ways in which r things can be chosen from n different things is given by
[tex]N=^nC_r.[/tex]
Therefore, the number of ways in which students chose 10 questions from 20 different questions is given by
[tex]N\\\\=^{20}C_r\\\\\\=\dfrac{20!}{10!(20-10)!}\\\\\\=\dfrac{20\times19\times18\times17\times16\times15\times14\times13\times12\times11\times10!}{10!\times 10\times9\times8\times7\times6\times5\times4\times3\times2\times1}\\\\\\=184756.[/tex]
Thus, the required number of ways is 184756.
find the slope of each line 5x-y=-7
You arrive in your history class today only to discover there is a pop quiz! You haven't studied and you aren't at all prepared. Fortunately, the quiz is multiple choice. Each question has five answer choices. You happen to have a die in your pocket. For each question you roll the die and answer A if the die shows 1, B if the die shows 2, etc, leaving the question blank if the die shows a six. For each question you are given one point if you answer it correctly and lose 1/4 point if you answer it incorrectly. You aren't penalized if you leave it blank, you just don't earn a point. What is the expected value for points earned on each question? Enter your answer as a decimal, rounded to two decimal places if necessary
at the beginning of a lesson, a piece of chalk is 4.875 inches long. at the end of the lesson, it is 3.125 inches long. writ the two amounts in expanded form using fractiones.