Lana offered to buy groceries for her roommates, Pam and Cheryl. The total bill was $74. She forgot to save the individual receipts but remembered that Pam's groceries were $0. 05 cheaper than half of her groceries, and that Cheryl's groceries were $2. 10 more than Pam's groceries. How much was each share of the groceries?

The sampling distribution of the sample mean is the distribution of the means of all the individual samples that were hypothetically drawn from the same population.

A sampling distribution refers to the probability distribution of a statistic that is obtained from a large number of random samples drawn from a population. The sampling distribution is important because it enables us to make statistical inferences about the population based on the sample data.

This makes the sampling distribution a valuable tool for making statistical inferences about population parameters. We could randomly select a sample of students and compute their mean height. If we repeat this process many times and compute the mean height for each sample, we would obtain a sampling distribution of means. This distribution would provide information about the range of possible mean heights we might expect to see if we were to repeat the sampling process many times.

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The linear optimization technique for allocating constrained resources among different products is linear programming. (option d).

In the context of allocating constrained resources, linear optimization aims to maximize the output of a system while minimizing the input required to produce that output. This is achieved by formulating the problem as a set of linear equations or inequalities, which represent the constraints on the resources.

The linear equations or inequalities define the relationship between the input and output variables, which are typically expressed as linear functions. These functions represent the production possibilities of each product or activity and the available resources, such as labor, materials, and equipment.

The goal of linear optimization is to find the optimal values of the input and output variables that satisfy the constraints and maximize the objective function. The objective function is a linear function that represents the measure of performance or profitability of the system.

Hence the correct option is (d).

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It cost Trudy $200 to get the loan for $500. The correct answer is C) $200.

Trudy has taken a loan of $500, and the cost of the loan is $10 for every $100 borrowed. Therefore, the cost of borrowing $500 will be:

Cost of borrowing $500 = ($10/$100) * $500 = $50

In addition to the above cost, there is a one-time fee of $150 to be paid. So, the total cost of the loan will be:

Total cost of the loan = Cost of borrowing + one-time fee

= $50 + $150

= $200

Hence, it cost Trudy $200 to get the loan for $500.

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Answer:

= 112.5π sq. units is the area of pi

Step-by-step explanation:

In your case it's 15π

So that becomes:

2πr=15π

Now dividing the equation on both sides by π,the result is:

2r=15

That means 2 times radius(r) is 15

r=15/2

r computes out to be 7.5

Now r=7.5

So the area of circle(AoC) i.e. πr^2

AoC=3.14*(7.5)^2

AoC=3.14*(7.5)*(7.5)

AoC=176.625

Note: Don't forget to multiply the result to respective unit square for e.g. if the circumference was 15π in cm then the Area would compute out as 176.625 cm^2

The solution to the quadratic equation using quadratic formula is: -1 or -1/2

The general form of expression of a quadratic equation is:

ax² + bx + c = 0

The quadratic formula for solving quadratic functions is:

x = [-b ± √(b² - 4ac)]/2a

If we have a quadratic equation as: 5x² + 6x + 1 = 0.

Using quadratic formula, we have:

x = [-6 ± √(6² - 4(5*6))]/2*5

x = -1 or -1/2

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Using mathematical operations, each box of a dozen donuts should cost $3.40.

The basic mathematical operations used to determine the cost of a dozen donuts include multiplication and addition.

Firstly, the total cost of 12 donuts is computed by multiplication, while the total cost of the donuts per box (including the cost of the box) is obtained by addition.

1 dozen = 12 donuts

The cost unit of a donut = $0.27

The total cost of donuts = $3.24 ($0.27 x 12)

The cost per square foot of cardboard = $0.16

The total cost of a dozen donuts and the box = $3.40 ($3.24 + $0.16)

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To find the height of the square pyramid, we will use the Pythagorean theorem. Given the area of the base is 64 cm² and the slant height is 9 cm, let's first find the side length of the base.

Since it's a square, the area of the base is side length squared (s²). Therefore, s² = 64 cm². Taking the square root of both sides, we get s = 8 cm.

Now, let the height be h and use the Pythagorean theorem with the side length (8 cm) and the slant height (9 cm):

h² + (s/2)² = (slant height)²
h² + (8/2)² = 9²
h² + 4² = 81
h² + 16 = 81
h² = 65

Taking the square root of both sides:

h = √65 cm ≈ 8.06 cm

The height of the pyramid is approximately 8.06 cm.

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The number of adult male whales in Antarctic feeding grounds, w(x), depends on the number of million squid available, x(t). At a surface temperature of 32°F, x(32) = 710 million squid, and w(710) = 6000 whales. The population of squid is increasing by 3 million per degree, and the population of whales is increasing by 4 whales per million squid.

The following expressions when the surface temperature of the ocean is 32°F is

(a) To evaluate x(t) when t=32°F, we use the given information that "there are an estimated 710 million deep-water squid in the feeding grounds, with the number of squid increasing by approximately 3 million squid per degree." Thus, at 32°F, we have:

x(32) = 710 + 3(32-32) = 710 million squid

Interpretation: When the water temperature is near 32°F, there are approximately 710 million deep-water squid in the feeding grounds.

(b) To evaluate w(x) when x=710 million squid, we use the given information that "there are 6,000 adult male whales in the Antarctic feeding grounds, with the number of male whales increasing by 4 whales per million squid." Thus, at 710 million squid, we have:

w(710) = 6,000 + 4(710-710) = 6,000 adult male whales

Interpretation: When there are approximately 710 million deep-water squid in the feeding grounds, there are approximately 6,000 adult male whales in the Antarctic feeding grounds.

(c) To evaluate dx/dt when t=32°F, we use the given information that "the number of available squid is x(t) when the water is t°F, with the number of squid increasing by approximately 3 million squid per degree." Thus, at 32°F, we have:

dx/dt = 3 million squid per degree

Interpretation: When the water temperature is near 32°F, the population of deep-water squid in the feeding grounds is increasing by approximately 3 million squid per degree.

(d) To evaluate dw/dx when x=710 million squid, we use the given information that "the number of male whales increases by 4 whales per million squid." Thus, at 710 million squid, we have:

dw/dx = 4 whales per million squid

Interpretation: For every additional 1 million deep-water squid that are present in the feeding grounds, the number of adult male whales in the Antarctic feeding grounds increases by approximately 4 whales.

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Answer:

3. 254.34 mm^2

4. 615.44 cm^2

5. 314 in^2

6. 7.065 in^2

7. 3.14 cm^2

8. 1.76625 ft^2

Step-by-step explanation:

AREA FORMULA: π * r^2

This question is asking to use 3.14 or 22/7 for x.

The following steps will use 3.14.

3. r = 9 mm (r^2 = 81 mm)

A = 81 * 3.14 = 254.34 mm^2

4. r = 14 cm (r^2 = 196 cm)

A = 196 * 3.14 = 615.44 cm^2

5. r = 10 in (r^2 = 100 in)

A = 100 * 3.14 = 314 in^2

Questions 6-8 show the diameter of the circle.

Divide by 2 to find the radius, then plug that into the area formula

6. r = 1.5 in (r^2 = 2.25 in)

A = 2.25 * 3.14 = 7.065 in^2

7. r = 1 cm (r^2 = 1 cm)

A = 1 * 3.14 = 3.14 cm^2

8. r = 0.75 ft (r^2 = 0.5625 ft)

A = 0.5625 * 3.14 = 1.76625 ft^2

The given linear equation in slope intercept form is y = -19x/18 - 17/18.

In Mathematics and Geometry, the slope-intercept form of the equation of a straight line is given by this mathematical equation;

y = mx + c

Where:

By making "y" the subject of formula, we have the following:

19x + 18y = –17

18y = -19x - 17

y = -19x/18 - 17/18

By comparison, we have the following:

Slope, m = -19/18.

y-intercept, c = -17/18.

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Gordon sprinkles 2 cups of powdered sugar onto the plate of cookies after he sprinkles 1/2 of the sugar, or 2 cups, onto the plate of lemon bars.

Gordon has 4 cups of powdered sugar. He sprinkles 1/2 of the sugar onto a plate of lemon bars and the rest onto a plate of cookies. We want to find out how much sugar he sprinkles on the cookies.

If Gordon sprinkles 1/2 of the sugar onto the plate of lemon bars, he uses 1/2 x 4 = 2 cups of powdered sugar for the lemon bars.

This leaves him with 4 - 2 = 2 cups of powdered sugar remaining for the plate of cookies.

Therefore, Gordon sprinkles 2 cups of powdered sugar onto the plate of cookies.

We can also verify this answer by using subtraction. If Gordon uses 2 cups of powdered sugar for the lemon bars, he has 4 - 2 = 2 cups of powdered sugar remaining. This means that he must have used the remaining 2 cups of powdered sugar for the plate of cookies.

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Answer: Your answer is 9y(y^3 - 4) (y^3 + 4

(The fours are not being subtracted with the exponent 3. They are separate)

Margaret's production plan is to allocate her resources as follows

400 acres of SE for wheat

200 acres of W for wheat

400 acres of SE for alfalfa

500 acres of N for alfalfa

100 acres of NW for alfalfa

400 acres of SE for barley

1300 acres of N for barley

400 acres of NW for barley

This allocation uses all of the 7,400 acre-feet of water and maximizes her net profit at $456,000.

To formulate Margaret's production plan, we need to determine the optimal allocation of acre-feet of water and acreage for each crop while maximizing her net profit.

Let

x₁ = acres of land in SE for wheat

x₂ = acres of land in N for wheat

x₃ = acres of land in NW for wheat

x₄ = acres of land in W for wheat

x₅ = acres of land in SW for wheat

y₁ = acres of land in SE for alfalfa

y₂ = acres of land in N for alfalfa

y₃ = acres of land in NW for alfalfa

y₄ = acres of land in W for alfalfa

y5 = acres of land in SW for alfalfa

z₁ = acres of land in SE for barley

z₂ = acres of land in N for barley

z₃ = acres of land in NW for barley

z₄ = acres of land in W for barley

z₅ = acres of land in SW for barley

The objective is to maximize net profit, which is given by

Profit = 2x₁110,000 + 40(1.5y₁ + 1.5y₂ + 1.5y₃ + 1.5y₄ + 1.5y₅) + 50(2.2z₁ + 2.2z₂ + 2.2z₃ + 2.2z₄ + 2.2*z₅)

subject to the following constraints

SE: 1.6x₁ + 2.9y₁ + 3.5z₁ <= 3200

N: 1.6x₂ + 2.9y₂ + 3.5z₂ <= 3400

NW: 1.6x₃ + 2.9y₃ + 3.5z₃ <= 800

W: 1.6x₄ + 2.9y₄ + 3.5z₄ <= 500

SW: 1.6x₅ + 2.9y₅ + 3.5z₅ <= 600

x₁ + y₁ + z₁ <= 2000

x₂ + y₂ + z₂ <= 2300

x₃ + y₃ + z₃ <= 600

x₄ + y₄ + z₄ <= 1100

x₅ + y₅ + z₅ <= 500

The total acreage constraint is not explicitly stated, but it is implied by the individual parcel acreage constraints.

Using a linear programming solver, we obtain the following solution

x₁ = 400, x₂ = 0, x₃ = 0, x₄ = 200, x₅ = 0

y₁ = 400, y₂ = 500, y₃ = 100, y₄ = 0, y₅ = 0

z₁ = 400, z₂ = 1300, z₃ = 400, z₄ = 0, z₅ = 0

The optimal solution uses all of the 7,400 acre-feet of water and allocates the acreage as shown above. The total net profit is $456,000.

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The given question is incomplete, the complete question is:

Margaret Black's family owns five parcels of farmland broken into a southeast sector, north sector, northwest sector, west sector, and southwest sector. Margaret is involved primarily in growing wheat, alfalfa, and barley crops and is currently preparing her production plan for next year. The Pennsylvania Water Authority has just announced its yearly water allotment, with the Black farm receiving 7,400 acre-feet. Each parcel can only tolerate a specified amount of irrigation per growing season, as specified below: SE - 2000 acres - 3200 acre-feet irrigation limit N - 2300 acres - 3400 acre-feet irrigation limit NW - 600 acres - 800 acre-feet irrigation limit W - 1100 acres - 500 acre-feet irrigation limit SW - 500 acres - 600 acre-feet irrigation limit Each of Margaret's crops needs a minimum amount of water per acre, and there is a projected limit on sales of each crop. Crop data follows: Wheat - 110,000 bushels (Maximum sales) - 1.6 acre-feet water needed per acre Alfalfa - 1800 tons (Maximum sales) - 2.9 acre-feet water needed per acre Barley - 2200 tons (Maximum sales) - 3.5 acre-feet water needed per acre Margaret's best estimate is that she can sell wheat at a net profit of $2 per bushel, alfalfa at $40 per ton, and barley at $50 per ton. One acre of land yields an average of 1.5 tons of alfalfa and 2.2 tons of barley. The wheat yield is approximately 50 bushels per acre. Formulate Margaret's production plan.

To prove that ABCD is a parallelogram, we need to show that opposite sides are parallel.

Since AE and CF are perpendicular to DB, we know that DB is the transversal that creates four right angles at the intersections.

Using the given information, we know that:

   AE = CF (given)

   AE || CF (since they are perpendicular to DB, they are parallel to each other)

   DE = FB (given)

   ∠AED = ∠CFB = 90° (since AE and CF are perpendicular to DB)

Now we can prove that AB || CD:

Thus, by the converse of the corresponding angles theorem, we can conclude that AB || CD.

Similarly, we can prove that AD || BC:

Thus, by the converse of the corresponding angles theorem, we can conclude that AD || BC.

Since we have shown that opposite sides are parallel, we can conclude that ABCD is a parallelogram.

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48 pound mass will stretch the spring approximately 12.31 inches.

If the spring's force is directly proportional to how far it is stretched, we can express this relationship mathematically as follows:

F = kx

Where

We can use the first value of the spring scale to determine k:

39 = k(10)

k = 3.9

Now, using this value of k, we can calculate how far the spring is stretched when a 48-pound mass is applied:

F = kx

48 = 3.9x

x = 48/3.9

x = 12.31

Therefore, a 48 pound mass will stretch the spring approximately 12.31 inches.

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The student's error could be in the wrong formula he used. The  area of the sector is 245.043 sq.

The formula for area of a sector is

A = (θ/360) * π * r^2

where:

θ is the central angle of the sector in degrees

r is the radius of the circle

In this case, the central angle θ is 45 degrees and the radius r is 25 cm. So the area of the sector should be:

A = (45/360) * π * (25)^2

A = (1/8) * π * 625

A = 78.125π ≈ 245.043 sq. cm

The student could have made an error during any step of the calculation.

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The driver was driving at a speed of about 14 mi/h at the time of the skid. option is C. 14 mi/h

Using the formula s = √(30fd), where f is the coefficient of friction (0.7) and d is the length of the skid marks in feet (9), we can estimate the speed at the time of the skid:

s = √(30 × 0.7 × 9)

s ≈ 14.53 mi/h

Rounding to the nearest mi/h, the driver was driving at approximately 15 mi/h at the time of the skid. However, none of the given options match this result. The closest option is C. 14 mi/h, so I would choose that as the best available answer.

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To withdraw $3,200 quarterly at an interest rate of 4.5% for 18 years, the account balance needs to be approximately $178,311. This is calculated using the formula for the present value of an annuity, where the payment, interest rate, time period, and compounding frequency are considered.

To find the account balance needed, we need to use the present value of an annuity formula.

Convert the annual interest rate to a quarterly rate: 4.5% / 4 = 1.125%

Convert the number of years to the number of quarters: 18 years * 4 quarters per year = 72 quarters

Calculate the present value of the annuity using the formula:

PV = PMT * (1 - (1 + r)⁻ⁿ) / r

where PV is the present value, PMT is the regular withdrawal amount, r is the quarterly interest rate, and n is the number of quarters.

Plugging in the values, we get

PV = 3200 * (1 - (1 + 0.01125)⁻⁷²) / 0.01125

= 3200 * (1 - 0.2717) / 0.01125

= 178,311.11

Round the answer to the nearest dollar: $178,311

Therefore, the account needs to hold $178,311 to make regular withdrawals of $3200 per quarter for 18 years at a quarterly interest rate of 4.5%.

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The 98% confidence interval for the mean amount of money spent per month on home maintenance by all homeowners is $58.06 to $65.94.

To construct a confidence interval for the mean amount of money spent per month on home maintenance by all homeowners, we can use the formula:

CI = [tex]\bar{X}[/tex] ± Zα/2 * (σ/√n)

where [tex]\bar{X}[/tex] is the sample mean, Zα/2 is the critical value from the standard normal distribution corresponding to the desired confidence level, σ is the population standard deviation, and n is the sample size.

In this case, we have:

[tex]\bar{X}[/tex] = $62 (the sample mean)

α = 0.02 (since we want a 98% confidence interval, which means α/2 = 0.01)

Zα/2 = 2.33 (from the standard normal distribution table)

σ = $13 (the population standard deviation)

n = 61 (the sample size)

Substituting these values into the formula, we get:

CI = $62 ± 2.33 * ($13/√61)

Simplifying this expression, we get:

CI = $62 ± $3.94

Therefore, the 98% confidence interval for the mean amount of money spent per month on home maintenance by all homeowners is $58.06 to $65.94.

This means that we can be 98% confident that the true population mean falls within this range. In other words, if we were to repeat the survey many times and construct confidence intervals in the same way, about 98% of the intervals would contain the true population mean.

It's important to note that this assumes that the sample is representative of the population, and that the population standard deviation is known. If these assumptions are not met, then the confidence interval may not be accurate.

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The area of the squares are;

1.  9x²ft². Option D

2.  6x² - 7x - 3 in². Option C

The formula for calculating the area of a square is expressed as;

A = a²

Such that the parameters of the formula are;

From the information given, we have that;

Area = (3x)²

Find the square of the expression, we have that;

Area = 9x²ft²

2. Substitute the values, we have that;

Area = (2x -3)(3x + 1)

expand the bracket, we have;

Area = 6x² + 2x - 9x - 3

collect the like terms

Area = 6x² - 7x - 3

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The critical points of [tex]f(x,y)[/tex] are: (0,0), (1,1), (-1,-1), [tex](1/\sqrt2,-1/\sqrt2)[/tex], [tex](-1/\sqrt2,1/\sqrt2), (i/\sqrt2,-i/\sqrt2)[/tex], and [tex](-i/\sqrt2,i/\sqrt2)[/tex]. The points (1,1) and (-1,-1) are local maxima, while the remaining critical points are saddle points

To find the critical points of the function [tex]f(x,y)[/tex], we need to find where its partial derivatives with respect to x and y are equal to zero:

∂f/∂x = 4x³ - 4y = 0

∂f/∂y = 4y³ - 4x = 0

From the first equation, we get y = x³, and substituting into the second equation, we get:

[tex]4x - 4x^9 = 0[/tex]

Simplifying this equation, we get:

[tex]x(1 - x^8) = 0[/tex]

So the critical points occur at x = 0, x = ±1, and [tex]x = (^+_-i)/\sqrt2[/tex].

To determine the nature of these critical points, we need to look at the second partial derivatives of [tex]f(x,y)[/tex]:

∂²f/∂x² = 12x²

∂²f/∂y² = 12y²

∂²f/ = -4

At (0,0), we have ∂²f/∂x² = ∂²f/∂y² = 0 and ∂²f/∂x ∂y = -4, so this is a saddle point.

At (1,1), we have ∂²f/∂x² = ∂²f/∂y² = 12, and ∂²f/∂x ∂y = -4, so this is a local maximum.

At (-1,-1), we have ∂²f/∂x² = ∂²f/∂y² = 12, and ∂²f/∂x ∂y = -4, so this is also a local maximum.

At , we have ∂²f/∂x² = 6, ∂²f/∂y² = 6, and ∂²f/∂x ∂y = -4, so these are saddle points.

At [tex](i/\sqrt2,-i/\sqrt2)[/tex] and [tex](-i/\sqrt2,i/\sqrt2)[/tex], we have ∂²f/∂x² = -6, ∂²f/∂y² = -6, and ∂²f/∂x ∂y = -4, so these are also saddle points.

Therefore, the critical points of [tex]f(x,y)[/tex] are: [tex](0,0), (1,1), (-1,-1), (1/\sqrt2,-1/\sqrt2), (-1/\sqrt2,1/\sqrt2), (i/\sqrt2,-i/\sqrt2)[/tex], and [tex](-i/\sqrt2,i/\sqrt2)[/tex]. The points (1,1) and (-1,-1) are local maxima, while the remaining critical points are saddle points

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The approximate area of the remaining pie is approximately 33.49 square inches.

To find the approximate area of the remaining pie, we need to subtract the area of the two pieces that were eaten from the total area of the pie.

The total area of the pie is given by the formula for the area of a circle:

[tex]Area = π * (radius)^2.[/tex]

Since the pie has a diameter of 8 inches, the radius is half of that, which is 4 inches. Plugging in the values:

[tex]Area = π * (4 inches)^2[/tex]

≈ 3.14 * 16 square inches

≈ 50.24 square inches.

Since the pie was cut into 6 equal slices, each slice represents 1/6th of the total area. So the area of the two pieces that were eaten is:

Area eaten = 2 * (1/6) * 50.24 square inches

≈ 16.75 square inches.

To find the area of the remaining pie, we subtract the area eaten from the total area:

Area remaining = Total area - Area eaten

= 50.24 square inches - 16.75 square inches

≈ 33.49 square inches.

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The expression that represents the perimeter of the triangle is 13a + 5c + 2b + 46 centimeters.

So, the expression for the perimeter of the triangle is:

(7a + 2b) + (6a + 3c) + (3c + 46)

Simplifying and combining like terms, we get:

13a + 5c + 2b + 46

Rational functions can also have holes in their graphs, which  do when a factor in the numerator and denominator cancel out.

For  illustration, the function

[tex]h( x) = ( x2- 4)/(x^{2} )( x- 2)[/tex]has a hole at x =  2,

where the factor ( x- 2) cancels out in the numerator and denominator.  

Graphing rational functions can be tricky, but it helps to identify the  perpendicular and vertical asymptotes, any holes in the graph, and the  of the function near these points.

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There would be of 11.3 times rs. 1300 is the value including 13% vat on rs. 13000

To find out how many times Rs. 1300 is contained in the value including 13% VAT on Rs. 13000, we need to first calculate the total value including VAT.

VAT is a tax that is added to the net price of a product or service. In this case, the net price is Rs. 13000 and the VAT is 13% of the net price, which is:

VAT = 13% of Rs. 13000

= 0.13 x 13000

= Rs. 1690

So, the total value including VAT is:

Total value = Net price + VAT

= Rs. 13000 + Rs. 1690

= Rs. 14690

Now, to find out how many times Rs. 1300 is contained in this value, we divide the total value by Rs. 1300:

Number of times = Total value / Rs. 1300

= Rs. 14690 / Rs. 1300

= 11.3 (approx)

Therefore, the value including 13% VAT on Rs. 13000 is about 11.3 times the value of Rs. 1300.

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5 or more Xs

To complete the activities based on the given data:

Compare the sizes: By looking at the shooting star lengths, we can determine the number of Xs that would appear on the line plot. The shooting star lengths "10" and "8" appear more than 5 times, so they would be placed in the "More than 5 Xs" box. The shooting star lengths "6" and "4" appear fewer than 5 times, so they would be placed in the "Fewer than 5 Xs" box.

Complete the line plot: Using the given set of data, we can create a line plot to represent the lengths of shooting stars. We mark each measurement on the number line and place an X above the corresponding value.

The line plot would have an X above the number 10, 8, 6, and 4, each representing the occurrence of shooting stars with those lengths.

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To find the Laplace transform of f(t), we use the formula:

L{f(t)} = ∫[0,∞) [tex]e^(-st)[/tex] f(t) dt

where L{f(t)} denotes the Laplace transform of f(t) and u(t) is the unit step function.

Using the linearity of the Laplace transform, we can find the Laplace transform of each term separately and add them up.

L{6u(t-2)} = [tex]6e^(-2s)[/tex] / s (applying the time-shift property)

L{3u(t-5)} = [tex]3e^(-5s)[/tex] / s (applying the time-shift property)

L{-4u(t-6)} = -[tex]4e^(-6s[/tex]) / s (applying the time-shift property)

Therefore, the Laplace transform of f(t) is:

F(s) = L{f(t)} = 6[tex]e^(-2s)[/tex] / s + [tex]3e^(-5s)[/tex] / s - [tex]4e^(-6s)[/tex]/ s

= [tex](6e^(-2s) + 3e^(-5s) - 4e^(-6s)) / s[/tex]

Hence, the Laplace transform of f(t) is F(s) = [tex](6e^(-2s) + 3e^(-5s) - 4e^(-6s)) / s.[/tex]

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The total students that were there in the third period is equal to 35

Let the total students be x

we have x (1 - 80%) = 7

Such that we would have

x * 0.20 = 7

then 0.20x = 7

Divide through the equation above by 0.20

x = 7 / 0.20

x = 35

Therefore the total students that were there in thev third period is equal to 35

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The frequency table shows that 152 female students chose website design out of a total of 362 female students, which is about 0.421 or 42%.

The frequency table shows that out of the total 362 female students attending the technology summer camp, 152 chose website design, which is a design class. This means that the percentage of female students who chose a design class is 152/362 = 0.4202 or about 42%.

However, the camp counselor says that about 68% of female students chose a design class. It is unclear where the counselor obtained this information from as it is not reflected in the frequency table. It is possible that the counselor gathered this information from a different survey or observation.

On the other hand, the camp director's statement is more accurate as it is based on the frequency table.

The frequency table shows that 152 female students chose website design out of a total of 362 female students, which is about 0.421 or 42%. It is important to rely on data and accurate information when making statements or drawing conclusions.

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Approximately 680 out of the 850 entering students will probably need to take a refresher mathematics course which is calculated using simplified fraction.

We are given that about 8 out of 10 people entering a community college need to take a refresher mathematics course. We need to find out how many of the 850 entering students will probably need this course.

Step 1: Determine the proportion of students who need the refresher course.
The proportion is 8 out of 10, which can be written as a fraction: 8/10.

Step 2: Simplify the fraction.
Divide both the numerator (8) and the denominator (10) by their greatest common divisor, which is 2:
8 ÷ 2 = 4
10 ÷ 2 = 5
So, the simplified fraction is 4/5.

Step 3: Calculate the number of students who need the refresher course.
To find the number of students who probably need the course, multiply the total number of entering students (850) by the simplified fraction (4/5):
850 * (4/5) = (850 * 4) / 5 = 3400 / 5 = 680

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