The required selling price of each water bottle is $554.06 such that Michelle's robotics club will achieve 25% increase in the purchase price.
Given that number of water bottles purchased is 75 and cost price (C,P) of the 75 water bottles is $443.25. The increase % on the purchase price is profit % = 25%.
To calculate the selling price when cost price and profit % is given by following steps:
Step 1 - Calculate the 25% of the cost price which gives profit.
Step 2 - Calculate SP, Selling price = Cost price + profit.
That implies, Profit = 25% of C.P.
Profit = 25/100 × 443.25.
Therefore, Profit = $110.81.
That implies, Selling price(S.P) = Cost price + profit.
S.P = 443.25 + 110.81.
Thus, S.P = $554.06.
Hence, the required selling price of each water bottle is $554.06 such that Michelle's robotics club will achieve 25% increase in the purchase price.
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HELP!!!!!!!!!!!!!!!!!!
Answer:
D 0.89275
Step-by-step explanation:
help please, i have a test tomorrow
The value of variables in the figure is,
⇒ x = 80
⇒ y = 70
We have to given that;
By using given figure we have to find the value of each variable.
Now, We can formulate;
⇒ 55 = 1/2 (180 - y)
Solve for y;
⇒ 55 × 2 = 180 - y
⇒ 110 = 180 - y
⇒ y = 180 - 110
⇒ y = 70
And, The value of x is,
⇒ x = 180 - (60 + 40)
⇒ x = 180 - 100
⇒ x = 80
Thus, The value of variables in the figure is,
⇒ x = 80
⇒ y = 70
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Picture included! Find the unknowns in the graph below:
The value of x in the triangle is 28.3 degrees., y is 7.837 and z is 7.837.
We can find the value of x by using angle sum property.
It states that the sum of three angles in a triangle is 180 degrees.
61.7+90+x=180
151.7+x=180
Subtract 151.7 from both sides:
x=180-151.7
x=28.3 degrees.
Now let us find the value of y.
We know that tan function is a ratio of opposite side and adjacent side.
tanx=y/14.76
tan(28.3) = y/ 14.76
0.531 = y/14.76
Now apply cross multiplication:
y=0.531×14.76
y=7.837
Now let us find z:
sin 90=y/z
1=7.837/z
z=7.837
Hence, the value of x in the triangle is 28.3 degrees., y is 7.837 and z is 7.837.
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The product of two numbers is 1536.
If the HCF of the two numbers is 16.
find the LCM of these two numbers.
Work Shown:
LCM = (product of two numbers)/(HCF of the two numbers)
LCM = 1536/16
LCM = 96
An employee started a new job and must enroll in a new family health insurance plan. One of the plans involves prescription drug coverage. The employee estimates that the entire family will fill 10 prescriptions per month, totaling $1,250. The employee has two options to choose from:
Option A: $94 monthly premium; 80% coverage for all prescription costs
Option B: $42 monthly premium; 75% coverage for first $500 in prescription costs, then 85% coverage for all prescription costs over $500
Which option would result in the highest overall cost for the employee, and by how much?
A) Option A has the highest overall cost by $64.50.
B)Option B has the highest overall cost by $64.50.
C) Option A has the highest overall cost by $106.50.
D) Option B has the highest overall cost by $106.50.
Option B has the highest overall cost by $64.50.
To determine the option that would result in the highest overall cost for the employee
we need to compare the costs of both options based on the estimated prescription drug expenses.
Option A:
Monthly premium: $94
Prescription coverage: 80%
Option B:
Monthly premium: $42
Prescription coverage: 75% for the first $500, 85% for costs over $500
Let's calculate the costs for each option:
Option A:
Total prescription drug cost: $1,250
Employee's share (20%): 20% × $1,250 = $250
Monthly premium: $94
Total cost for Option A: $250 + $94 = $344
Option B:
Total prescription drug cost: $1,250
Employee's share for the first $500 (25%): 25% × $500 = $125
Employee's share for costs over $500 (15%): 15% × ($1,250 - $500) = $112.50
Monthly premium: $42
Total cost for Option B: $125 + $112.50 + $42 = $279.50
Comparing the total costs for each option, we see that Option B has a lower overall cost for the employee.
Hence, Option B has the highest overall cost by $64.50.
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All else being equal, if you cut the sample size in half, how does this affect the margin of error when using the sam
to make a statistical inference about the mean of the normally distributed population from which it was drawn?
ME-
Z.S
O The margin of error is multiplied by √0.5.
O The margin of error is multiplied by √√2-
O The margin of error is multiplied by 0.5.
O The margin of error is multiplied by 2.
Answer:
When you cut the sample size in half while making a statistical inference about the mean of a normally distributed population, the effect on the margin of error depends on the relationship between the sample size and the margin of error. Generally, the margin of error is inversely proportional to the square root of the sample size.
So, if you reduce the sample size by half, it means you are taking a smaller sample, which will result in a larger margin of error. In other words, the margin of error is multiplied by a factor greater than 1.
Among the given options, the correct answer is:
D. The margin of error is multiplied by 2.
This option correctly reflects the relationship between reducing the sample size by half and the resulting increase in the margin of error.
During a normal day, there are 782 passengers in average that
are late for their plane each day. However, during the
Christmas holidays, there are 1,835 passengers that are late for
their planes each day which caused delays of 14 planes. How
many more passengers are late for their planes in each day
during the Christmas holidays?
i picked 4th grade lol
During the Christmas holidays, there are 1,053 more passengers who are late for their planes each day compared to a normal day.
To solve this problemCalculating the difference between the quantity of late passengers on a typical day and the quantity of late passengers on holidays is necessary.
1,835 travellers were late over the Christmas break.
On an average day, there are 782 travelers who are late.
Difference = Number of late passengers during the Christmas holidays - Number of late passengers on a normal day
Difference = 1,835 - 782
Difference = 1,053
Therefore, during the Christmas holidays, there are 1,053 more passengers who are late for their planes each day compared to a normal day.
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Answer:
Step-by-step explanation:
During the Christmas holidays, there is an increase of 1,053 passengers who are late for their planes each day compared to the average daily number of 782 passengers who are late. This surge in late passengers during the holiday period contributes to delays in 14 planes.
The higher volume of travelers, combined with potential factors like weather conditions, congestion, and heightened travel demand, likely leads to a greater number of individuals encountering delays and difficulties reaching their departure gates on time.
Hope it helps! :)
PLSSA HELP ME
Describe a way to transform to . Be specific
The way to transform A to B is (1,-1).
We are given that;
The coordinates (-1,1),(-3,1),(-1,3) and (1,-1),(3,-1),(1,-3).
Now,
To transform the coordinates (-1,1),(-3,1),(-1,3) to (1,-1),(3,-1),(1,-3), you can apply a reflection about the origin. This transformation can be represented by the matrix [-1 0; 0 -1]. When you multiply this matrix with the original coordinates, you will get the transformed coordinates.
First, represent the point (-1,1) as a column vector:
[-1]
[ 1]
Then, multiply this vector by the reflection matrix [-1 0; 0 -1]:
[-1 0; 0 -1] * [-1] = [ 1]
[ 1] [-1]
Therefore, by transformation the answer will be (1,-1).
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Carl's class is collecting canned food for a
food drive. If the class collects 33 pounds
on the first day and 41 pounds on the
second day, how many pounds of food
have they collected so far?
Give your answer as a mixed number in simplest form.
Enter the number that belongs in the green box.
[2] pounds
The class has collected a total of 74 pounds of food so far.
The class collected 33 pounds on the first day and 41 pounds on the second day.
To find the total pounds of food collected, we add these two amounts together:
33 pounds + 41 pounds = 74 pounds
On the first day, the students gathered 33 pounds, and on the second day, they collected 41 pounds.
We combine these two figures together to get the total number of pounds of food collected:
33 lbs. plus 41 lbs. equals 74 lbs.
On the first day, the class brought in 33 pounds, and on the second, 41 pounds.
We combine these two figures to get the total weight in pounds of food collected:
44 pounds plus 33 pounds equals 74 pounds.
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How many significant digits are in 26.04813?
7 significant figures.
Zeros that come at the start do not count as significant figures.
Eg. 0.0001 would only be 1 significant figure.
Zeros that come after is counted as a significant figure.
Eg. 2.20 Is three significant figures.
s = the amount of snowfall
t = the length of time it takes to plow all of the streets
which is the dependent variable?
Use the information above to answer the questions that follow.
2.2.1
2.2.2
2.2.3
2.2.4
sanitation in Johannesburg if a property is 175 m².
Write down, to the nearest ten cents and excluding VAT, the cost for
Calculate the cost for 4,1 ke sanitation in Cape Town before the increase.
Mr Jones lives in Johannesburg and Ms Brown lives in Cape Town. They
both own a property with an area of 550 m² and each was billed for 22 kl
sanitation.
Use the table above to determine the difference in the cost of sanitation
for the two properties.
Explain how the tariff system used in Johannesburg is beneficial to
home owners in terms of water usage.
(2)
(8)
(2)
[34]
mor
MO
ud.
0
0
Mr. Jones in Johannesburg is billed R9,767.12 and Ms. Brown in Cape Town is billed R680.24 for their respective properties.
From the provided information for Cape Town's sanitation tariffs:
0-4.2 kl: R16.03 per kl
Since 4.1 ke is equivalent to 4100 liters, which is less than 4.2 kl, we can use the tariff rate for the 0-4.2 kl range.
Cost for 4.1 ke of sanitation
= 4.1 ke x R16.03 per kl
= 4100 liters x R16.03 per kl
= R65.83
For Mr. Jones and Ms. Brown, who own properties with an area of 550 m² each and were billed for 22 kl of sanitation.
we need to determine the applicable tariff rate based on the property size and calculate the cost.
In Johannesburg, based on the provided information, the tariff rate for properties larger than 300 m² to 1,000 m² is R443.96.
Cost of sanitation for Mr. Jones in Johannesburg:
= 22 kl x R443.96 per kl
= R9,767.12
Cost of sanitation for Ms. Brown in Cape Town:
= 22 kl x R30.92 per kl = R680.24
Therefore, Mr. Jones in Johannesburg is billed R9,767.12 and Ms. Brown in Cape Town is billed R680.24 for their respective properties.
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For which situations would it be appropriate to calculate a probability about the difference in sample means?
1) Both population shapes are unknown. n1 = 50 and n2 = 100.
2) Population 1 is skewed right and population 2 is approximately Normal. n1 = 50 and n2 = 10.
3) Both populations are skewed right. n1 = 5 and n2 = 10.
4) Population 1 is skewed right and population 2 is approximately Normal. n1 = 10 and n2 = 50.
5) Both populations have unknown shapes. n1 = 50 and n2 = 100.
6) Both populations are skewed left. n1 = 5 and n2 = 40.
It is generally appropriate to calculate a probability about the difference in sample means when the sample sizes are large (typically n ≥ 30) and the populations are approximately normally distributed.
How did we arrive at this assertion?To determine if it is appropriate to calculate a probability about the difference in sample means, consider the assumptions and conditions for conducting a hypothesis test or constructing a confidence interval. The appropriateness of calculating a probability about the difference in sample means depends on the following factors:
1) Both population shapes are unknown. n1 = 50 and n2 = 100:
- It is generally appropriate to calculate a probability about the difference in sample means when the sample sizes are large (typically n ≥ 30) and the populations are approximately normally distributed. Since the population shapes are unknown in this case, it is difficult to assess this condition. However, the large sample sizes (n1 = 50 and n2 = 100) may suggest that it is reasonable to approximate the population distributions as normal. Therefore, calculating a probability about the difference in sample means could be considered.
2) Population 1 is skewed right and population 2 is approximately Normal. n1 = 50 and n2 = 10:
- In this case, the assumption of approximately normally distributed populations is violated for population 1, which is skewed right. When the population distributions are not approximately normal, it may not be appropriate to calculate a probability about the difference in sample means. The small sample size for population 2 (n2 = 10) may also limit the accuracy of any inference made based on this sample.
3) Both populations are skewed right. n1 = 5 and n2 = 10:
- Similar to the previous case, the assumption of approximately normally distributed populations is violated for both populations. Additionally, the small sample sizes (n1 = 5 and n2 = 10) may not provide sufficient information for reliable inferences. Therefore, it is generally not appropriate to calculate a probability about the difference in sample means in this situation.
4) Population 1 is skewed right and population 2 is approximately Normal. n1 = 10 and n2 = 50:
- Similar to case 2, the assumption of approximately normally distributed populations is violated for population 1, which is skewed right. In this case, the sample size for population 1 (n1 = 10) is also small, which may limit the accuracy of any inference made based on this sample. The larger sample size for population 2 (n2 = 50) might make it more reasonable to approximate the population distribution as normal. However, the violation of the assumption for population 1 suggests caution when interpreting the results. It is not generally appropriate to calculate a probability about the difference in sample means in this situation.
5) Both populations have unknown shapes. n1 = 50 and n2 = 100:
- Similar to case 1, the population shapes are unknown. However, the large sample sizes (n1 = 50 and n2 = 100) might suggest that it is reasonable to approximate the population distributions as normal. As mentioned before, calculating a probability about the difference in sample means could be considered in this case.
6) Both populations are skewed left. n1 = 5 and n2 = 40:
The assumption of approximately normally distributed populations is violated for both populations, as they are skewed left. Additionally, the small sample sizes (n1 = 5 and n2 = 40) may not provide sufficient information for reliable inferences. Therefore, it is generally not appropriate to calculate a probability about the difference in sample means in this situation.
Summarily, it is generally appropriate to calculate a probability about the difference in sample means when the sample sizes are large (typically n ≥ 30) and the populations are approximately normally distributed. However, when the population distributions are not approximately normal or when the sample sizes are small, it is generally not appropriate to calculate a probability about the difference in sample means.
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find the first derivative with respect to x of the following function: f(x)=(7x+3)(4-3x)
You have the derivative of a product:
first limit (7x+3)
second limit (4-3x)
The Rule for the derivative of a product is:
(first limit × derivative of second limit) + (second limit × derivative of first limit)
f(x) = (7x+3)(4-3x)
f'(x)=(7x+3)×-3 + (4-3x)×7 = -21x- 9 + 28-21x
f'(x) = -42x + 19
In Boston, a popular restaurant recently added a delivery option. In reviewing the sales for this new delivery service, the restaurant found the mean delivery orders in a random sample of Friday nights was x¯=48 orders, with a margin of error of 12 orders.
The 95% confidence interval for the mean number of delivery orders on a Friday night is 36 to 60 orders.
What is the confidence interval?The confidence interval for the mean number of delivery orders on a Friday night is constructed assuming a 95% confidence level.
Given data:
Sample mean (x) = 48 orders
The margin of error (E) = 12 orders
To calculate the confidence interval, we'll use the formula:
Confidence Interval = x ± E
Substituting the values:
Confidence Interval = 48 ± 12
The lower limit of the confidence interval:
Lower Limit = 48 - 12 = 36
The upper limit of the confidence interval:
Upper Limit = 48 + 12 = 60
Therefore, the 95% confidence interval for the mean number of delivery orders on a Friday night is 36 to 60 orders.
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Complete question:
In Boston, A Popular Restaurant Recently Added A Delivery Option. In Reviewing The Sales For This New Delivery Service, The Restaurant Found The Mean Delivery Orders In A Random Sample Of Friday Nights Was X¯=48 Orders, With A Margin Of Error Of 12 Orders. Construct A Confidence Interval For The Mean Number Of Delivery Orders On A Friday Night.
What is the approximate volume of a cylinder with a radius of centimeters and a height of 4centimeters?
Use 3.14 for π and round the answer to the nearest tenth.
A. 150.7 cm3
B. 28.3 cm3
c. 113.0 cm3
d. 75.4 cm3
To calculate the volume of a cylinder, we use the formula:
Volume = π * radius^2 * height
Given:
Radius = cm
Height = 4 cm
π (pi) = 3.14
Let's substitute these values into the formula and calculate the volume:
Volume = 3.14 * (cm)^2 * 4 cm
= 3.14 * (cm^2) * 4 cm
= 12.56 * cm^3
Rounding the answer to the nearest tenth, we get:
Volume ≈ 12.6 cm^3
Among the answer choices provided, the closest option to 12.6 cm^3 is:
B. 28.3 cm^3
Therefore, the correct answer is B. 28.3 cm^3.
Consider the equation x^3 - 3x^2 + 2x - 6 = 0. Find all the complex roots of this equation and represent them in the form a + bi.
Answer:
x=i[tex]\sqrt{2}[/tex], x=-i[tex]\sqrt{2}[/tex]
Step-by-step explanation:
This equation can be factored. In order to do so, we can split the expression into two parts: (x^3-3x^2) and (2x-6). The greatest common factor of the first group is x^2, factoring it out gives us x^2(x-3). The greatest common factor of the second group is 2, factoring it out gives us 2(x-3). Since both x^2 and 2 are being multiplied by x-3, we add them into a factored form of (x^2+2)(x-3)=0
x-3 provides us with one real root, so we can disregard it.
x^2+2=0 ---> x^2=-2--->x=±[tex]\sqrt{-2}[/tex]
Factoring out the imaginary number i (equals [tex]\sqrt{-1}[/tex]), we get x=±i[tex]\sqrt{2}[/tex]
A bag contains 19 blue, 28 purple, 21 red, and 29 orange balls. You pick one ball at random. Find the probability that it is not orange. P(not orange) = Simplify your answer completely.
[tex]|\Omega|=19+28+21+29=97\\|A|=19+28+21=68\\\\P(A)=\dfrac{68}{97}[/tex]
Luther bought a snack pack of crackers, and he thinks it's interesting that the cylindrical
tower of crackers came in a box shaped like a square prism. He wonders how much extra
space the box contains around the cracker tower. The box has a height of 8 inches and a side
length of 2 inches. If the tower of crackers has a height of 7.9 inches and a diameter of
1.9 inches, what is the volume of the extra space in the box?
Round your answer to the nearest tenth.
The volume of the extra space in the box is 9.0 cubic inches.
How to estimate the volume of the extra space in the box?To estimate the volume of the extra space in the box, we shall find the difference between the volume of the box and the volume of the cracker tower.
First, volume of the box:
The volume can be calculated as the product of the height, length, and width (we assume the width is the same as the side length since it is a square prism):
Box volume = height * length * width
Given:
height = 8 inches
side length = 2 inches
Box volume = 8 * 2 * 2
= 32 in³
Next, the volume of the cracker tower:
The tower of the cracker is cylindrical, so we shall use the formula for the volume of a cylinder to estimate the volume:
Tower volume = π * (radius²) * height
Given:
height = 7.9 inches
diameter = 1.9 inches
radius = diameter / 2 = 1.9 / 2 = 0.95
Tower volume = π * (0.95)² * 7.9
≈ 22.994 in³ (rounded to three decimal places)
Then, the volume of the extra space in the box:
Volume of the extra space = Tower volume - Box volume:
= 32 in³ - 22.994 in³
≈ 9.006 in³ (rounded to three decimal places)
Therefore, the volume of the extra space in the box = 9.0 in³.
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calcula el cuádruple de la parte decimal sumando con el doble de la parte entera del siguiente número decimal : 123,45
The expression that we need to solve is 4*45 + 2*123 , and that is equal to 426.
How to find the number?
Here we want to calculate the quadruple of the decimal part by adding twice the integer part of the following decimal number: 123.45
The decimal part of this number is 45
The integer part of this number is 123.
Then the expression that we need to solve is:
4*45 + 2*123 = 426
That is the value of the mathematical sentence.
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Show (AB)^-1 = B^-1 A^-1
Segment RT has endpoints R(-3,4) & T (-7,-3) what are the coordinates of the midpoint of RT?
Whose solution strategy would work?
Answer:c
Step-by-step explanation:
Alice was provided with the following trinomial: 3x² + 7x-12x - 34 - 2x² + 10 1 Provide Alice with a step-by-step guide on how to factorize the algebraic expression.
Hello!
[tex]3x^{2} + 7x-12x - 34 - 2x^{2} + 10\\\\3x^{2}- 2x^{2} + 7x-12x - 34 + 10\\\\x^{2} - 5x - 24\\\\x^{2} + 3x - 8x - 24\\\\(x^{2} + 3x) + (-8x - 24)\\\\x(x + 3) - 8(x + 3)\\\\\boxed{(x - 8)(x+3)}[/tex]
Urgent please help I am running out of time thank you
Answer:
119
Step-by-step explanation:
The points on the graph represent both an exponential function and a linear function.
Complete this table by reading the values from the graph. Estimate any function values that are less than one.
x -3 -2 -1 0 1 2 3
Exponential function _____ _____ _____ _____ _____ _____ _____
Linear function _____ _____ _____ _____ _____ _____ _____
At approximately what values of x do both the linear and exponential functions have the same value for y?
We are given two curves on a graph paper and we have to find the y values for the given x.
From the graph we can check for a value of x, what is the value of y from the curve.
x -3 -2 -1 0 1 2 3
Exp 8 4 2 1 0.5 0.25 0.125
Lin. 7.5 6 4.5 3 1.5 0 -1.5
We find that the two have the same values where the two curves intersect.
From the graph we find that near to -3 and near to 2 both have the same values.
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The missing graph is attached below.
If KL = x + 8
, LM = 12
, and KM = 3x − 14
, what is KL?
Answer:
x = 17
Step-by-step explanation:
KL + LM = KM
x+8+12 = 3x-14
or, x+20 = 3x-14
or, 3x-x=20+14
or, 2x = 34
x = 17
Answer:
Step-by-step explanation:
3x-14=x+8+12
3x-x=8+12+14
2x=34
x=17
then
KL=(17)+8=25
The cost of 1kg potatoes and 2kg tomatoes was 30 on a certain day. After two days the cost of 2kg potatoes and 4kg tomatoes was found to be 66.
Translate the following phrase into an algebraic expression. Do not simplify. Use the variable names "x" or "y" to describe the unknowns.
the product of two numbers increased by 4
Answer:
The algebraic expression for the phrase "the product of two numbers increased by 4" can be written as `xy + 4`, where `x` and `y` represent the two unknown numbers.
Answer:
Step-by-step explanation:
To translate the phrase “the product of two numbers increased by 4” into an algebraic expression, we can use the variables x and y to represent the two numbers.
The product of two numbers is xy, and if we increase this by 4, we get:
xy + 4
Therefore, the algebraic expression for “the product of two numbers increased by 4” is: xy + 4
I hope this helps! Let me know if you have any other questions.
Triangle JKL has vertices at the points J(2.5,3), K(3.2,-4.7), and L(-6.9,3). Find the slope of side JL.
A. undefined
B. -11
C. -0.76
D. 0