cala used 4 2/3 cups of watermelon and 2 1/6 cups of cherries to make a fruit bowl how many cups of watermelon and cherries were used in all

Anna should use 10 mL of the 72% salt solution and 20 mL of the 54% salt solution to make 30 mL of a 60% salt solution

Let's assume that Anna will use x mL of the 72% salt solution, and therefore she will use (30 - x) mL of the 54% salt solution (since the total volume is 30 mL).

To find out how much of each solution Anna should use, we can set up an equation based on the amount of salt in each solution.

The amount of salt in x mL of 72% salt solution is

= 0.72x

The amount of salt in (30 - x) mL of 54% salt solution is

= 0.54(30 - x)

To make a 60% salt solution, the total amount of salt in the final solution should be

0.6(30) = 18

So we can set up an equation

0.72x + 0.54(30 - x) = 1

Simplifying the equation

0.72x + 16.2 - 0.54x = 18

0.18x = 1.8

x = 10 ml

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The parabolic trajectory of the falling prey can be described by the equation y = 228 – x2/57, where y is the height above the ground and x is the horizontal distance traveled in meters. In this case, the prey was dropped at a height of 228 m and flying at 19 m/s. To calculate the total distance traveled by the prey, we can use the equation for the parabola to solve for x.

We can rearrange the equation y = 228 – x2/57 to solve for x, which gives us[tex]x = √(57*(228 – y))[/tex]. When the prey hits the ground, the height (y) is 0. Plugging this into the equation for x, we can calculate that the total distance traveled by the prey is[tex]x = √(57*(228 - 0)) = √(57*228) = 84.9 m.\\[/tex] Expressing this answer to the nearest tenth of a meter gives us the final answer of 84.9 m.

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Answer:

We can simplify the expression as follows:

5^(2000) + 5^(1999)

5^(1999) - 5^(1997)

= 5^(1999) * (1 + 1/5)

5^(1997) * (1 - 1/25)

= (5/4) * (25/24) * 5^(1999)

= (125/96) * 5^(1999)

Therefore, the value of the expression is (125/96) * 5^(1999).

Step-by-step explanation:

The parking lot has a width of around [tex]0.937[/tex] meters.

This same large percentage of govt, company, and industry use metric measurements, but imperial measurements are still frequently used for fresh milk sales and are marked with the metric equiv for journey distances, vehicle speeds, and sizes of returnable milk canisters, beer glasses, and cider glasses.

100 centimeters make up one meter. Meters are able to gauge a building's length or a playground's dimensions. 1000 meters make up one kilometer.

[tex]3760 cm = 37.6 m[/tex]

Solve for the width,

[tex]area = (1/2) * (base1 + base2) * height[/tex]

where,

base1 [tex]= 73.4 m[/tex]

base2 [tex]= 37.6 m[/tex]

area [tex]= 2,930.4[/tex] square meters

Let's solve for the height first,

[tex]height = 2 * area / (base1 + base2)[/tex]

[tex]height = 2 * 2,930.4 / (73.4 + 37.6)[/tex]

[tex]height = 2 * 2,930.4 / 111[/tex]

[tex]height = 56.16 m[/tex]

We nowadays can apply the algorithm to determine the width.

[tex]width = (area * 2) / (base1 + base2) * height[/tex]

[tex]width = (2 * 2,930.4) / (73.4 + 37.6) * 56.16[/tex]

[tex]width = 5856.8 / 111 * 56.16[/tex]

[tex]width = 5856.8 / 6239.76[/tex]

[tex]width = 0.937[/tex]

Therefore, the width of the parking lot is approximately [tex]0.937[/tex] meters.

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The time difference of 15 minutes, or 25 percent (15 divided by 60 times 100), is called the time saved.

An experiment is a controlled study in which a scientist manipulates a variable in order to determine its effects. An experiment must have a testable hypothesis, be replicable, and produce empirical evidence.

Discussing the time difference in an experiment. In an experiment, it takes one hour to memorize all of the words on a list, and two years later, they are relearned in 45 minutes.

The time difference of 15 minutes, or 25 percent (15 divided by 60 times 100), is referred to as the time saved.

Time saved is the difference between the total time it takes to finish a process with a particular method and the total time it would take to complete the same process without that method.

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By using differentials to estimate the maximum error when measuring the volume of the sphere if the possible error in measuring the radius is 0.5. It will be 32,572.03 in³. Which is option (d).

The possible error in measuring the radius of the sphere is 0.5 in

The formula for the volume of a sphere is given by V(r) = 4/3πr³

The volume of the sphere when r=72 in is given by V(72) = 4/3π(72)³

When r= 72 + 0.5 in= 72.5 in, the volume of the sphere can be calculated using the formula:

V(72.5) = 4/3π(72.5)³

The difference between these two volumes, V(72) and V(72.5), gives us the maximum error while measuring the volume of a sphere. It can be calculated as follows:

V(72.5) - V(72) = 4/3π(72.5)³ - 4/3π(72)³= 4/3π [ (72.5)³ - (72)³ ]= 4/3π [ (72 + 0.5)³ - 72³ ]= (4/3)π [ 3(72²)(0.5) + 3(72)(0.5²) + 0.5³ ]≈ (4/3)π [ 777.5 ]= 3.28 × 10⁴ in³

Therefore, the maximum error while measuring the volume of a sphere with a radius of 72 in, where the possible error in measuring the radius is 0.5 in, is approximately 3.28 × 10⁴ in³ or 32,572.03 in³. Therefore coorect option is (D).

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(a) The probability that exactly two cars finish the race is 0.0512.

(b) The probability that at most two cars finish the race is 0.05792.

(c) The probability that at least three cars finish the race is 0.94208.

(a) To determine the probability that exactly two cars finish the race, we have to use binomial distribution. In this case, we have n = 5 trials, and p = 0.8 is the probability that a car finishes the race (1 - 0.2). Using the binomial distribution formula:

P(X = k) = (nCk)(p^k)(1 - p)^(n - k)

Where X is the number of cars that finish the race, we get:

P(X = 2) = (5C2)(0.8²)(0.2)³= (10)(0.64)(0.008)= 0.0512

Therefore, the probability that exactly two cars finish the race is 0.0512.

(b) To determine the probability that at most two cars finish the race, we have to calculate the probabilities of 0, 1, and 2 cars finishing the race and add them up.

P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)= (5C0)(0.8⁰)(0.2)⁵ + (5C1)(0.8¹)(0.2)⁴ + (5C2)(0.8²)(0.2)³= 0.00032 + 0.0064 + 0.0512= 0.05792

Therefore, the probability that at most two cars finish the race is 0.05792.

(c) To determine the probability that at least three cars finish the race, we can calculate the probability of 0, 1, and 2 cars finishing the race and subtract it from 1, which gives us the probability of at least three cars finishing the race.

P(X ≥ 3) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)]= 1 - (0.00032 + 0.0064 + 0.0512)= 0.94208

Therefore, the probability that at least three cars finish the race is 0.94208.

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Answer:

If the GM between √2 and 2√2 is a find the value of a.​

Step-by-step explanation:

To find the geometric mean between two numbers, we simply take the square root of their product.

In this case, we want to find the geometric mean between √2 and 2√2.

Their product is:

√2 * 2√2 = 2√4 = 2*2 = 4

So, the geometric mean between √2 and 2√2 is the square root of 4, which is:

√4 = 2

Therefore, the value of a is 2.

The probability of getting tails in the three coins would be 0.125 or 12.5%.

To calculate the probability of an event happening, first, we need to identify the rate of the desired outcome versus the total possible outcomes. Moreover, to determine the total probability of two or more events happening we need to calculate the probability of each event and then multiply the results.

Probability of getting tails in any of the three coins:

1 / 2 = 0.5

Total probabilityy:

0.5 x 0.5 x 0.5 = 0.125 or 12.5%

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The rate of change of the distance for limousine is less than the rate of change of the convertible.

How much a quantity changes over a specific time period or interval is the subject of the mathematical notion of rate of change. Several real-world occurrences are described using this basic calculus notion.

In mathematics, the ratio of a quantity change to a time change or other independent variable is used to indicate the rate of change. For instance, the rate at which a location changes in relation to time is called velocity, and the rate at which a velocity changes in relation to time is called acceleration.

The equation of the distance travelled by the convertible is given as:

y = 35x

The equation of the limousine can be calculated using the coordinates of the graph (1, 30) and (2, 60).

The slope is given as:

slope = (change in y) / (change in x) = (60 - 30) / (2 - 1) = 30

Using the point slope form:

y - 30 = 30(x - 1)

y = 30x

So the equation of the limousine is y = 30x.

Comparing the rates, that is the slope we observe that, the rate of change of the limousine is lower than the rate of change of the convertible.

Hence, the rate of change of the limousine is less than the rate of change of the convertible.

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The result (recurrent value), A = sum j=1 to 89 ln(j), is true for every n. This is the desired result.

As in T(n) = T(n/2) + n, T(0) = T(1) = 1, a recurrence or recurrence relation specifies an infinite sequence by explaining how to calculate the nth element of the sequence given the values of smaller members.

We can start by proving the base case in order to demonstrate the first portion through recurrence. Let n = 1. Next, we have:

Being true, ln(a1) = ln(a1). If n = k, let's suppose the formula is accurate:

Sum j=1 to k ln = ln(prod j=1 to k aj) (aj)

Prod j=1 to k aj * ak+1 = ln(prod j=1 to k+1 aj)

(Using the logarithmic scale) = ln(prod j=1 to k aj) + ln(ak+1)

Using the inductive hypothesis, the property ln(ab) = ln(a) + ln(b)) = sum j=1 to k ln(aj) + ln(ak+1) = sum j=1 to k+1 ln (aj)

(b), we can use the just-proven formula:

A = ln(1, 2,...) + ln + ln (89)

= ln(j=1 to 89) prod

sum j=1 to 89 ln = ln(prod j=1 to 89 j) (j).

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Answer:

Step-by-step explanation:

kernel composition rules 2 1 point possible (graded) let and be two vectors of the same dimension. use the the definition of kernels and the kernel composition rules from the video above to decide which of the following are kernels. (note that you can use feature vectors that are not polynomial.) (choose all those apply. )

Answer:

(-2,9)

Step-by-step explanation:

when moving it 5 units left on the x axis it would be 5-7

So in turn you would be given (-2,9)

Because the y stays the same you would still have (?,9)

Answer:

142 sq cm

Step-by-step explanation:

A= 2(lh + wh + lw)

2(7*3+5*3+7*5)

2(21+15+35)

2(71)

A= 142 sq cm

Answer:

12

Step-by-step explanation:

i think its right

The Trapezoidal rule and Simpson's rule are two methods used to approximate the value of a definite integral. The Trapezoidal rule approximates the integral by dividing the region between the lower and upper limits of the integral into n trapezoids, each with a width h. The approximate value of the integral is then calculated as the sum of the areas of the trapezoids. The Simpson's rule is similar, except the region is divided into n/2 trapezoids and then the integral is approximated using the weighted sum of the area of the trapezoids.

For the given integral 4 x x2 1 0 dx, with n = 200, the Trapezoidal rule and Simpson's rule approximate the integral to be 7.4528 and 7.4485 respectively, rounded to four decimal places. The exact value of the integral is 7.4527. The difference between the exact and approximate values is very small, thus indicating that both the Trapezoidal rule and Simpson's rule are accurate approximations.

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A bin can hold 28 pounds. each toy car weighs 7 ounces., so the bin can hold 64 toy cars.

To determine the number of toy cars the bin can hold, we must first convert the weight limit of the bin and the weight of the toy cars to a uniform unit of measure.

We'll then divide the weight limit of the bin by the weight of one toy car. After that, we'll multiply the resulting value by the number of ounces in one pound (16).

Here's how to solve the problem:

1 pound = 16 ounces

Therefore, a bin that can hold 28 pounds can hold:28 × 16 = 448 Ounces

The weight of one toy car is 7 ounces.

Divide the weight limit of the bin (448 ounces) by the weight of one toy car (7 ounces):

448 ÷ 7 = 64

Therefore, the bin can hold 64 toy cars.

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Answer:

[tex] |x - 300| \leqslant 10[/tex]

a. S = {(1, -1), (2, 1)}Let's begin by calculating the determinant of the matrix composed of the vectors of S, and checking if it is equal to 0. Because the two vectors are not colinear, they should span R2.|1 -1||2 1| determinant is not 0, therefore S spans R2. No geometric description is required for this example.

b. S = {(1, 1)} The set S contains one vector. A set containing only one vector cannot span a plane because it only spans a line. Therefore, S does not span R2. Geometric description: S spans a line that passes through the origin (0, 0) and the point (1, 1).c. S = {(0, 2), (1, 4)} Let's again begin by calculating the determinant of the matrix composed of the vectors of S, and checking if it is equal to 0.|0 2||1 4| determinant is 0, thus S does not span R2. In this scenario, S only spans the line that contains both vectors, which is the line with the equation y = 2x.

Geometric description: S spans a line that passes through the origin (0, 0) and the point (1, 2).

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The reason why the terms a and m have to be relatively prime integers is that it is the only way to make sure that ax≡1 (mod m) is solvable for x within the integers modulo m.

Theorem:"If a and m are relatively prime integers and m > 1, then an inverse of a modulo m exists. Furthermore, this inverse is unique modulo m. (That is, there is a unique positive integer a less than m that is an inverse of a modulo m and every other inverse of a modulo m is congruent to a modulo m.)"If a and m are relatively prime integers and m > 1, then an inverse of a modulo m exists. Furthermore, this inverse is unique modulo m. (That is, there is a unique positive integer a less than m that is an inverse of a modulo m and every other inverse of a modulo m is congruent to a modulo m.)The inverse of a modulo m is another integer, x, such that ax≡1 (mod m).

This theorem has an interesting explanation: if a and m are not co-prime, then there is no guarantee that ax≡1 (mod m) has a solution in Zm. The reason for this is that if a and m have a common factor, then m “absorbs” some of the factors of a. When this happens, we lose information about the congruence class of a, and so it becomes harder (if not impossible) to undo the multiplication by .This is the reason why the terms a and m have to be relatively prime integers.

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An inequality representing the amount that the island of Martinique can received for hurricane relief efforts, where x is the number of days is y ≤ 32,000 + 4,500x.

Inequality is an algebraic statement that two or more mathematical expressions are unequal.

Inequalities can be represented as:

The total amount received by the island = $32,000

The daily receipt of donations = $4,500

Let the number of days = x

Let the funds raised for aid = y

y ≤ 32,000 + 4,500x

Thus, the inequality for the funds that the island can fundraise for hurricane relief aid by the end of the month is y ≤ 32,000 + 4,500x.

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Answer:

[tex]{ \rm{s = 12( v_{0} + v_{1} )t}} \\ \\{ \boxed { \rm{t = \frac{s}{12(v_{0} + v_{1})} \: \: }}}[/tex]

Answer: x=211/1296

Step-by-step explanation:

Answer:

The 2nd equation is false.

Step-by-step explanation:

You don't even have to solve. DE is not 58, it's 40.

The 2nd equation is false.

Solve the equation [tex]7/11=18/x+1[/tex] we find the solution is [tex]x = 27.2857[/tex]

Your example is an equation since an equation is any statement with an equals sign. Equations are frequently utilized for mathematical equations since mathematicians like equal signs. A set of instructions for achieving a certain result is called an equation.

A number, a constant, or a mix of numbers, variables, and operation symbols make up an expression. Two expressions joined by such an assignment operator form an equation.

we can cross-multiply,

[tex]7(x+1) = 11(18)[/tex]

Expanding the left side,

[tex]7x + 7 = 198[/tex]

Subtracting [tex]7[/tex] from both sides,

[tex]7x = 191[/tex]

Dividing both sides by [tex]7[/tex],

[tex]x = 191/7[/tex]

Therefore, the solution to the proportion is

[tex]x = 27.2857[/tex]

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Answer:

3x, 4x | 5, 3

Step-by-step explanation:

a block slides along a track that descends through distance h. The track is frictionless except for the lower section. There the block slides to a stop in a certain distance d because of friction. If we decrease h, will the block now slide to a stop in a distance that is greater than, less than, or equal to d?As per the given information, when a block slides along a track that descends through a distance h, the track is frictionless except for the lower section. There the block slides to a stop in a certain distance d because of friction. Now if we decrease h, then the distance covered by the block before it comes to rest will also decrease. So the block will slide to a stop in a distance that is less than d. Hence the answer is less than d.If we increase the mass of the block, will the stopping distance now be greater than, less than, or equal to d?

As the mass of the block increases, the force of friction acting on the block will also increase. Hence the stopping distance will also increase. So the stopping distance now will be greater than d. Hence the answer is greater than d.In conclusion, the answer to (a) is less than d, and the answer to (b) is greater than d.

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The volume of the solid whose base is the region inside the circle with radius 3 if the cross sections taken parallel to one of the diameters are equilateral triangles is 81/2*\sqrt3 by using the slicing method.

To find the volume of the solid whose base is the region inside the circle with radius 3, we need to integrate the area of the cross sections taken parallel to one of the diameters, which are equilateral triangles.

Let's consider a cross section of the solid taken at a distance x from the center of the circle.

Since the cross section is an equilateral triangle, all its sides have the same length.

Let  this length be y. Since the triangle is equilateral, its height can be found using the Pythagorean theorem as follows:

[tex]height = \sqrt{(y^2 - (y/2)^2)} = \sqrt{(3/4y^2)}= \sqrt{3/2y}[/tex]

Therefore, the area of the cross section at a distance x from the center of the circle is:

[tex]A(x) = (1/2)y\sqrt{3/2y} = \sqrt{3/4y^2}[/tex]

Now, we need to integrate this area over the range of x from -3 to 3 (since the circle has radius 3):

[tex]V = \int\ [-3,3]\sqrt{3/4*y^2} dx[/tex]

To find the limits of integration for y, we need to consider the equation of the circle:

[tex]x^2 + y^2= 3^2[/tex]

Solving for y, we get:

[tex]y =\pm\sqrt{(3^2 - x^2)}=\pm\sqrt{(9^2 - x^2)}[/tex]

Since we want the cross sections to be equilateral triangles, we know that y is equal to the height of an equilateral triangle with side length equal to the diameter of the circle, which is 2*3 = 6. Therefore, we can write:

[tex]y = 3*\sqrt{3}[/tex]

Substituting this into the integral, we get:

[tex]V = \int\ [-3,3] \sqrt{3/4*(3\sqrt3)^2} dx[/tex]

[tex]= \int\ [-3,3] 27/4*\sqrt{3} dx[/tex]

Integrating, we get:

[tex]V = [27/4\sqrt{3x}]*[-3,3][/tex]

[tex]= 81/2*\sqrt{3}[/tex]

Therefore, the volume of the solid is [tex]81/2*\sqrt3[/tex]cubic units

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The average number of hours of sleep for working college students is between 6.28 and 6.72 hours of sleep each night

According to the given data,

Sample size n = 101

Sample mean x = 6.5

Standard deviation s = 2.14

Level of confidence C = 96%

Using a 96% confidence level, the value of t* for 100 degrees of freedom is 2.081, as given in the question.

Now, the formula for the confidence interval is:x ± (t* × s/√n)Here, x = 6.5, s = 2.14, n = 101, and t* = 2.081

Substituting the values in the above formula, we get:

Lower limit = x - (t* × s/√n) = 6.5 - (2.081 × 2.14/√101) = 6.28

Upper limit = x + (t* × s/√n) = 6.5 + (2.081 × 2.14/√101) = 6.72

Therefore, the 96% confidence interval for the average number of hours of sleep for working college students is between 6.28 and 6.72 hours of sleep each night.

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The value of the double integral using the easier order, ydA bounded by y = x − 6; x = y² is 125/12.

The double integral, indicated by ', is mostly used to calculate the surface area of a two-dimensional figure. By using double integration, we may quickly determine the area of a rectangular region. If we understand simple integration, we can easily tackle double integration difficulties. Hence, first and foremost, we will go over some fundamental integration guidelines.

Given, the double integral ∫∫yA and the region y = x-6 and x = y²

y = x-6

x = y²

y² = y +6

y² - y - 6 = 0

y² - 3y +2y - 6 = 0

(y-3) (y+2) = 0

y = 3 and y = -2

[tex]\int\int\limits_\triangle {y} \, dA\\ \\[/tex]

= [tex]\int\limits^3_2 {y(y+6-y^2)} \, dx \\\\\int\limits^3_2 {(y^2+6y-y^3)} \, dx \\\\(\frac{y^3}{3} + 3y^2-\frac{y^4}{4} )_-_2^3\\\\\frac{63}{4} -\frac{16}{3} \\\\\frac{125}{12}[/tex]

The value for the double integral is 125/12.

Integration is an important aspect of calculus, and there are many different forms of integrations, such as basic integration, double integration, and triple integration. We often utilise integral calculus to determine the area and volume on a very big scale that simple formulae or calculations cannot.

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