Find the exact volume of a cylinder with radius 4 ft. and height 10 ft.

The range, difference of maximum and minimum values, is larger for Class C with a range of 10 compared to Class D with a range of 3. The data provided is not sufficient to calculate the IQR (Interquartile Range) for either class.

To begin, let's understand what the terms 'range' and 'IQR' (Interquartile Range) refer to in statistics. The range is simply the difference between the largest and smallest data values. The IQR is the range of the middle 50% of the values when ordered from lowest to highest.

For Class D, the lowest absence number is 0 and the highest is 3, thus the range is 3-0 = 3. For Class C, the lowest absence number is 0 and the highest is 10, so the range is 10-0 = 10. Thus, the range for the distribution of absences is larger for Class C.

Calculating IQR requires determining the 1st Quartile (Q1) and 3rd Quartile (Q3) values and subtracting Q1 from Q3. The data given doesn't include enough information to directly calculate these values and hence, it is impossible to definitively say which class has a larger IQR based on the provided information.

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Log₁₆(256b) in terms of the logarithm of b, which is the factor that was multiplied by 256 inside the logarithm is  2 + log₁₆(b)

We can use the product rule of logarithms, which states that the logarithm of a product is equal to the sum of the logarithms of the factors.

Therefore, we can write

log₁₆(256b) = log₁₆(256) + log₁₆(b)

We can simplify log₁₆(256) as follows:

log₁₆(256) = log₁₆(16^2) = 2

Therefore, we have:

log₁₆(256b) = log₁₆(256) + log₁₆(b)

= 2 + log₁₆(b)

So, we have rewritten log₁₆(256b) in terms of the logarithm of b, which is the factor that was multiplied by 256 inside the logarithm.

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The statement that could be true for h is h(2) = 16.

Option A is the correct answer.

We have,

The function h has a domain of -3 ≤ x ≤ 11 and a range of 1 ≤ h(x) ≤ 25, and we know that h(8) = 19 and h(-2) = 2.

To determine which statement could be true for h, we can use the given domain and range, along with the two known function values, to narrow down the possible values of h(x) for different values of x.

h(2) = 16

We do not have enough information to determine whether this statement could be true or not.

It is possible that h(2) = 16, but it is also possible that h(2) could be a different value within the range of 1 ≤ h(x) ≤ 25.

h(8) = 21

This statement cannot be true, as we already know that h(8) = 19.

h(13) = 18

This statement cannot be true, as 13 is outside the given domain of -3 ≤ x ≤ 11.

h(-3) = -1

This statement cannot be true, as -1 is outside the given range of 1 ≤ h(x) ≤ 25.

Therefore,

The statement that could be true for h is h(2) = 16.

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The probability of the task taking between 48 and 49.9 minutes is 0.1357 or 13.57% (rounded to four decimal places).

To calculate this probability, we first need to find the total probability of the entire range between 43 and 57 minutes. This can be found by subtracting the lower bound from the upper bound and dividing by the total range:

P(43 < X < 57) = (57 - 43) / (57 - 43) = 1

Since the probability of the entire range is 1, we can find the probability of any sub-interval by dividing the length of the sub-interval by the length of the total range. Therefore, the probability of the task taking between 48 and 49.9 minutes is:

P(48 < X < 49.9) = (49.9 - 48) / (57 - 43) = 1.9 / 14 = 0.1357 or 13.57%

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The number of cans brought by Joes team is J = 299 cans

Given data ,

Let the number of cans brought by Joes team be J

Now , sage team brought 224 cans

And , sage team brought 3 times cans as asifs team

And , joes team brought 4 times as asifs team

On simplifying the equation , we get

Asif's team brought 224/3 = 74.67 (rounded to the nearest whole number) cans.

Joe's team brought 74.67 x 4 = 298.68

So , J = 299 cans

Hence , the equation is solved and J = 299 cans

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The system of equations in the context of this problem is given as follows:

The solution is given as follows:

h = 35, c = 55.

The variables used for the system of equations are given as follows:

Considering the Monday's earnings, the equation is given as follows:

5h + 3c = 340.

Considering the Wednesday's earnings, the equation is given as follows:

2h + 6c = 400.

Hence the system is:

Simplifying the second equation by 2, we have that:

h + 3c = 200.

Subtracting the first equation by the second, we can obtain the value of h as follows:

4h = 140

h = 140/4

h = 35.

Hence the value of c is given as follows:

c = (200 - 35)/3

c = 55.

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Answer:

B) 0.08 and 0.1

The answer was 0.09, which is between 0.08 and 0.1

Answer:

660,430 gallons AKA 2499.9995045071 kilograms

you would proably round it to 2500 kilograms

Step-by-step explanation:

looked it up, had the same measurements as you gave

Note that the difference in the radii of the two circular tracks is about 13.68 ft

recall that the circumference of a circle is denoted by the expression

C = 2πr

In this case, C = Circumference
and r   radius

Since we have two tracks

Let ra = radius of the purple track and

rb = radius of the yello track

So

265 = 2πra

170 = 2πrb

making ra and rb subject of the expression we have

ra = 256/(2π)  =40.7436654315 ≈ 40.74

rb = 170 / (2π) = 27.0563403256 ≈ 27.06

Hence, the difference is

40.74 - 27.06 = 13.68ft

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Full Question:

at a fair Daniel and Claire went on a ride that has two separate circular tracks. Daniel rode in a purple car that travels a total distance of 265 feet around the track . Ciara rode in a yellow car that travels a total distance of 170 feet around the track. They drew drew a sketch of the ride. What is the difference ofthe radii of the two circle tracks

The area of the shaded region for the normal distribution is approximately 0.6985.

We have,
To find the area of the shaded region under the standard normal distribution curve between z = -0.87 and z = 1.24, we can use a standard normal distribution table or a calculator with a standard normal distribution function.

Using a standard normal distribution table, we look up the area to the left of z = -0.87 and the area to the left of z = 1.24 and then subtract the smaller area from the larger area to find the area between z = -0.87 and z = 1.24.

The table value for z = -0.87 is 0.1949, and the table value for z = 1.24 is 0.8934.

The area between z = -0.87 and z = 1.24 is:

= 0.8934 - 0.1949

= 0.6985

So the area of the shaded region is approximately 0.6985.

Thus,

The area of the shaded region for the normal distribution is approximately 0.6985.

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Anna will make  $1602.71 in all over 10 weeks.

Here, the first week salary of Anna = $116

i.e., the initial salary a =  $116

She was guaranteed a raise of 7% each week.

so, her salary in the next week would be,

116 + 7% of 116

7 percent of 116 is:

116 × 7/100 = 8.12

so, her salary will be $124.12

So, the equation for the salary after 'm' weeks would be,

[tex]n = 116\times (1 + 0.07)^{m - 1}\\\\n = 116\times (1 .07)^{m - 1}[/tex]

Using this equation , the salary in the second week m = 2 would be,

n = 124.12

Salary in the third week m = 3 would be,

n = 116 × [tex](1.07)^{3-1}[/tex]

n = 132.81

Salary in the fourth week m = 4 would be,

n = 116 × [tex](1.07)^{4-1}[/tex]

n = 142.11

Salary in the fifth week m = 5 would be,

n = 116 × [tex](1.07)^{5-1}[/tex]

n = 152.05

Salary in the sixth week m = 6 would be,

n = 116 × (1.07)⁶⁻¹

n = 162.7

Salary in the seventh week m = 7 would be,

n = 116 × (1.07)⁷⁻¹

n = 174.08

Salary in the eighth week m = 8 would be,

n = 116 × (1.07)⁸⁻¹

n = 186.27

Salary in the nineth week m = 9 would be,

n = 116 × (1.07)⁹⁻¹

n = 199.31

And the salary after the tenth week m = 10 would be,

n = 116 × (1.07)¹⁰⁻¹

n = 213.26

The total money after 10 weeks would be,

T = 116 + 124.12 + 132.81+ 142.11 + 152.05 + 162.7 + 174.08 + 186.27 + 199.31 + 213.26

T = $1602.71

This is the required amount.

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The greatest of the given integers is 64.

Given that, the sum of four consecutive integers is 250. We need to find what is the greatest of these integers.

Let the consecutive integers be = x, x+1, x+2, x+3

Therefore,

x + x+1 + x+2 + x+3 = 250

4x + 6 = 250

2x + 3 = 125

2x = 122

x = 61

Therefore, the greatest integer = 61 + 3 = 64

Hence, the greatest of the given integers is 64.

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Answer:

B. The candidate received at least 250 votes in half of the 6 towns.

This is because the median number of votes is the middle value when all the vote counts are put in order. This means that at least three towns gave the candidate more than 250 votes, and at least three towns gave the candidate fewer than 250 votes. So, the candidate received at least 250 votes in half of the 6 towns. The total number of votes the candidate received in the election cannot be determined from this information. Similarly, the number of votes received in individual towns cannot be determined.

The expression 16x⁹y¹⁸√3xy is equivalent to √769x¹⁹y³⁷

If a > O and y > 0, which expression is equivalent to √769x¹⁹y³⁷

Factor and rewrite the radicand in exponential form:

√16²×3x¹⁸×xy³⁶×y

Simplify the radical expression:

16x⁹y¹⁸ √2²×3xy

Factor and rewrite the radicand in exponential form:

8x⁹y¹⁸√2²×3xy

8x⁹y¹⁸.2√3xy

Multiply the monomials:

16x⁹y¹⁸√3xy

Hence, the expression 16x⁹y¹⁸√3xy is equivalent to √769x¹⁹y³⁷

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The total height of the three jumps is A = 20.744 feet

Given data ,

1st high jump score: 7.016 feet

2nd high jump score: 5.42 feet

3rd high jump score: 8.308 feet

On adding the scores , we get

7.016 + 5.42 + 8.308 = 20.744 feet

On simplifying the equation , we get

A = 20.744 feet

Hence , Oliver jumped a total of 20.744 feet in the three high jumps

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Answer:

Step-by-step explanation:The system of equations represented by the given points is:1x + 0y = 1   (equation 1)

1x + 6y = 1   (equation 2)

8x + 26y = 1  (equation 3)

8x - 22y = 1  (equation 4)

We can use the first equation to solve for x in terms of y:1x + 0y = 1

1x = 1

x = 1

Then we can substitute x=1 into equations 2-4 to obtain three equations in terms of y:1(1) + 6y = 1    =>   6y = 0   =>   y = 0

8(1) + 26y = 1   =>   26y = -7/8   =>   y = -7/208

8(1) - 22y = 1   =>   -22y = -7/8 - 7   =>   y = 15/44

Therefore, the solution to the system is x = 1 and y = -7/208 or y = 0 or y = 15/44.

However, we are given that x = -2, which contradicts our solution of x = 1. Therefore, there is no solution to the system with the given values.

The area of triangle ABC is :  33.44 cm²

The area of a triangle is the region enclosed within the sides of the triangle. The area of a triangle varies from one triangle to another depending on the length of the sides and the internal angles. The area of a triangle is expressed in square units, like, m2, cm2, in2, and so on.

We have the information from the question is:

In triangle ABC :

AB=6 cm,

AC=15 cm, and

m ∠A=48°.

We have to find the area of Triangle ABC

Area of triangle is : 1/2 × bc × (SinA)

Area of Triangle = 1/2 × 6 × 15 × (Sin 48° )

We know the value of Sin 48° is 0.7431

Area of Triangle = (1/2) × 66.88

Area of triangle = 33.44 cm²

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Answer: The brother is 16 and she is 19.

Step-by-step explanation:

35 - 3 = 32

32 / 2 = 16

16 + 3 = 19

Answer:

$5.50(2/3)π(6.5^3) = $3,163.45

$5.50(2/3)(3.14)(6.5^3) = $3,161.85

$3,161.85 is the correct answer.

The solution to the system of equations is:

x = 3, y = 2, z = 2

To solve for x, y, and z, we can use the elimination method or substitution method. Here, we will use the elimination method.

First, we will eliminate y from the equations by multiplying the first equation by 3 and the second equation by 1, and then adding them:

(3) (2x – y + 3z = 10)

6x – 3y + 9z = 30

(1) (x + 3y – 2z = 5)

x + 3y – 2z = 5

Adding the two equations gives:

7x + 7z = 35

Simplifying, we get

x + z = 5 (Equation 1)

Next, we will eliminate y again by multiplying the second equation by 2 and the third equation by 3, and then adding them:

(2) (x + 3y – 2z = 5)

2x + 6y – 4z = 10

(3) (3x – 2y + 4z = 12)

9x – 6y + 12z = 36

Adding the two equations gives:

11x + 8z = 46

Substituting x + z = 5 from Equation 1 into the above equation, we get:

11(x + z) + 8z = 46

11x + 19z = 46

Solving for z, we get:

z = 2

Substituting z = 2 into x + z = 5 from Equation 1, we get:

x + 2 = 5

x = 3

Finally, substituting x = 3 and z = 2 into any of the original equations, we can solve for y:

2x – y + 3z = 10

2(3) – y + 3(2) = 10

6 – y + 6 = 10

y = 2

Therefore, the solution to the system of equations is:

x = 3

y = 2

z = 2

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The difference of p and z times the sum of p and z is p² - 2pz - z².

An algebraic expression in mathematics is an expression which is made up of variables and constants, along with algebraic operations such as addition, subtraction, division and multiplication.

Hence, the difference of p and z times the sum of p and z can be represented as follows:

Therefore,

(p - z) × (p + z) = (p - z) (p + z)

(p - z) (p + z)  = p² - pz - pz - z²

p² - pz - pz - z² = p² - 2pz - z²

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Answer:

28 hours

Step-by-step explanation:

Let's use "B" to represent the number of hours Bianca practiced.

From the problem, we know that Filipe practiced 12 fewer hours than Bianca, which means:

Filipe's practice time = Bianca's practice time - 12

We are also told that Filipe practiced for 16 hours, so we can substitute that into the equation:

16 = B - 12

To solve for B, we can add 12 to both sides:

16 + 12 = B

So, Bianca practiced for a total of 28 hours.

Evaluating the given function, we can see that the intensity at 35 cm is 0.816

The intensity at a distance d in centimeters is given by the function:

I = 100*d⁻²

Here we want to find the intensity at 35 centimeters from the source, then we need to evaluate the function above at d = 35, doing that we will get:

I = 100*35⁻²

I = 100/(35²)

I = 0.816

The intensity at 35 centimeters from the source is 0.816.

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10: HL

11: AAS

12: SSS

13: SSS

14: HL

15: AAS

16: AAS

17: HL

18: AAS

19: SAS

20 AAS

1: SAS

2: ASA

3: HL

4: HL

5: ASA

6: ASA

7: HL

8: SSS

9: HL

The value of the expression when x = 4 is 12.

The value of the expression when x = -8 is 6.

We have,

To evaluate the expression 10 + x/2 for a given value of the variable x, we simply substitute the value of x into the expression and simplify.

For example:

If x = 4:

10 + x/2 = 10 + 4/2

= 10 + 2

= 12

So when x = 4, the value of the expression is 12.

If x = -8:

10 + x/2 = 10 + (-8)/2

= 10 - 4

= 6

So when x = -8, the value of the expression is 6.

Thus,

The value of the expression when x = 4 is 12.

The value of the expression when x = -8 is 6.

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