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Question- The original radius of a sphere is 6 centimeters. Explain how the surface area of the sphere would change if the radius was halved to 3 centimeters. Round your answers to the nearest whole number.

Answer: ??

Step-by-step explanation:

The future value of $2000 earning 18% interest, compounded monthly, for 4 years is $6,116.23.

To calculate the future value, we use the formula:

[tex]FV = P(1 + r/n)^{nt}[/tex]

Where:

FV is the future value

P is the principal amount ($2000)

r is the annual interest rate (18%)

n is the number of times the interest is compounded per year (12 for monthly compounding)

t is the number of years (4)

Plugging in these values, we get:

[tex]FV = 2000(1 + 0.18/12)^{12*4}[/tex]

FV = 6116.23

Therefore, the future value of $2000 earning 18% interest compounded monthly for 4 years is approximately $6,116.23. This means that if you invest $2000 today and earn 18% interest compounded monthly for 4 years, your investment will grow to $6,116.23 at the end of the 4-year period. It's important to note that the actual value may vary depending on the exact compounding method used by the bank or financial institution.

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Answer:

66.5

Step-by-step explanation:

average the 2 together (57+76)/2

B

B=G-11 and B+G=28 could be used to solve the problem

After considering the speeds of the student and the teacher and the distance they ran in each case, we found that the students wins in all the races.

Speed is defined as the ratio of distance travelled to the amount of time it took. As speed simply has a direction and no magnitude, it is a scalar quantity.

An object is considered to be moving at a uniform speed when it travels the same distance in the same amount of time.

When an object travels a different distance at regular intervals, it is said to have variable speed.

Average speed is the constant speed determined by the ratio of the total distance travelled by an object to the total amount of time it took to travel that distance.

Given,

The length of the race = 100  yards

1st race

The speed of student = 10 yards/s

The speed of teacher = 3.75 yards / s

Time taken by student = distance / speed = 100/10 = 10s

Time taken by teacher = 100 / 3.75 = 26.67 s

So the student finishes the race because he/she took less time.

2nd race

Distance run by teacher = 100 - 10 = 90

Speed of teacher =  3.75 yards / s

Time taken by teacher = 90/3.75 =  24s

The distance and speed of the student are the same.

Time taken by student = 10 s

Still, the student finishes the race first.

3rd race

Distance run by teacher = 100 - 10 = 90

Speed of teacher =  3.75 * 2 yards /s = 7.5 yards/s

Time taken by teacher = 90/ 7.5 =  12s

The distance and speed of the student are the same.

Time taken by student = 10 s

Still, the student finishes the race first.

Therefore by considering the speeds of the student and the teacher and the distance they ran in each case, we found that the students wins in all the races.

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Answer:

4-2-5

Step-by-step explanation:

16, 8, and 20 can all be divided by 4

Leaving you with 4 2 and 5.

The ratio is 4 to 2 to 5

Step-by-step explanation:

always remember Pythagoras for life :

c² = a² + b³

c is the Hypotenuse (the longest side of the right-angled triangle, it is opposite of the 90° angle). a and b are the legs.

c² = 19² + 17² = 361 + 289 = 650

c = sqrt(650) = 25.49509757... in ≈ 25.5 in

the Hypotenuse is about 25.5 in long.

The solution is  \boxed{A^{-1} = \left[\begin{array}{rr}-\frac{1}{3} & 0 \\ 0 & -\frac{1}{5}\end{array}\right]} and \boxed{A^{-1} = \left[\begin{array}{rrr}-\frac{1}{6} & 0 & 0 \\ 0 & \frac{1}{12} & 0 \\ 0 & 0 & -\frac{1}{8}\end{array}\right]}.

(a) To find the inverse of \( A=\left[\begin{array}{rr}-3 & 0 \\ 0 & 5\end{array}\right] \), we need to use the formula:

\( A^{-1} = \frac{1}{ad-bc} \left[\begin{array}{rr}d & -b \\ -c & a\end{array}\right] \)

where \( a=-3, b=0, c=0, d=5 \).

Plugging in the values, we get:

\( A^{-1} = \frac{1}{(-3)(5)-(0)(0)} \left[\begin{array}{rr}5 & 0 \\ 0 & -3\end{array}\right] \)

\( A^{-1} = \frac{1}{-15} \left[\begin{array}{rr}5 & 0 \\ 0 & -3\end{array}\right] \)

\( A^{-1} = \left[\begin{array}{rr}-\frac{1}{3} & 0 \\ 0 & -\frac{1}{5}\end{array}\right] \)

So, the inverse of \( A=\left[\begin{array}{rr}-3 & 0 \\ 0 & 5\end{array}\right] \) is \( A^{-1} = \left[\begin{array}{rr}-\frac{1}{3} & 0 \\ 0 & -\frac{1}{5}\end{array}\right] \).

(b) To find the inverse of \( A=\left[\begin{array}{rrr}4 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 3\end{arra \), we need to use the formula:

\( A^{-1} = \frac{1}{\det(A)} \left[\begin{array}{rrr}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right]^{T} \)

where \( a_{11}=4, a_{12}=0, a_{13}=0, a_{21}=0, a_{22}=-2, a_{23}=0, a_{31}=0, a_{32}=0, a_{33}=3 \) and \( \det(A) = (4)(-2)(3) - (0)(0)(0) - (0)(0)(0) = -24 \).

Plugging in the values, we get:

\( A^{-1} = \frac{1}{-24} \left[\begin{array}{rrr}4 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 3\end{array}\right]^{T} \)

\( A^{-1} = \frac{1}{-24} \left[\begin{array}{rrr}4 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 3\end{array}\right] \)

\( A^{-1} = \left[\begin{array}{rrr}-\frac{1}{6} & 0 & 0 \\ 0 & \frac{1}{12} & 0 \\ 0 & 0 & -\frac{1}{8}\end{array}\right] \)

So, the inverse of \( A=\left[\begin{array}{rrr}4 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 3\end{arra \) is \( A^{-1} = \left[\begin{array}{rrr}-\frac{1}{6} & 0 & 0 \\ 0 & \frac{1}{12} & 0 \\ 0 & 0 & -\frac{1}{8}\end{array}\right] \).

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Answer:

84

Step-by-step explanation:

6.3+16.1=22.4

22.4/2=11.2

11.2x7.5=84

Answer:

78.12 cm^2

Step-by-step explanation:

i will call the 6.3 cm line A, 7.5 line B and 16.1 cm C

first we divide this into two shapes, a rectangle and a triangle

THE RECTANGLE:

the formula to find the area of a rectangle: L x W

A x B

6.3 x 7.5 = 47.25 cm^2

THE TRIANGLE:

this is a right angle triangle

formula to find the area of a right angle triangle: (1/2) x BASE x H

B is parallel to H and share the same length

to find BASE we need to subtract 6.3cm from 16.1cm (check attachment)

16.1 - 6.3 = 9.8

(1/2) x 9.8 x 6.3 = 30.87 cm^2

FINAL VALUE:

now we simply add our two values of area together

30.87 + 47.25 = 78.12 cm^2

In the triangle ABC, the value of x, y and z is obtained as 21, 7 and 48 units respectively.

What are triangles?

Triangles are a particular sort of polygon in geometry that have three sides and three vertices. Three straight sides make up the two-dimensional figure shown here. An example of a 3-sided polygon is a triangle. The total of a triangle's three angles equals 180 degrees. One plane completely encloses the triangle.

A triangle ABC is given.

The measure of AB is given as 16 + z units.

The measure of AD is given as 16 units.

The measure of DB is given as z - 16 units.

The measure of BE is given as 21 units.

The measure of BC is given as x units.

The measure of AC is given as 14 units.

According to the midpoint theorem, the length of DE is -

DE = 1/2 (AC)

y = 1/2 (14)

y = 7 units

Therefore, the value of y is obtained as 7 units.

Now according to indirect measurement -

AB / AC = BD / DE

Substitute the values in the equation -

16 + z / 14 = z - 16 / 7

7(16 + z) = 14(z - 16)

112 + 7z = 14z - 224

7z - 14z = -224 - 112

-7z = -336

z = 48

Therefore, the value of z is obtained as 48 units.

Now according to indirect measurement -

BC / AC = BE / DE

Substitute the values in the equation -

21 + x / 14 = 21 / 7

7(21 + x) = 14 × 21

147 + 7x = 294

7x = 294 - 147

7x = 147

x = 21

Therefore, the value of x is obtained as 21 units.

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Answer:

x = 21

y = 7

z = 32

Step-by-step explanation:

DE = 1/2 (AC)

y = 1/2 (14)

y = 7

AB / AC = BD / DE

Substitute values

16 + z / 14 = z - 16 / 7

7(16 + z) = 14(z - 16)

112 + 7z = 14z - 224

7z - 14z = -224 - 112

-7z = -336

z = 32

BC / AC = BE / DE

Substitute values

21 + x / 14 = 21 / 7

7(21 + x) = 14 × 21

147 + 7x = 294

7x = 294 - 147

7x = 147

x = 21

Answer:

Step-by-step explanation:

2(2x+2) + 2(x+4) = 36

4x+4 +2x+8 =36

6x+12= 36

6x = 36-12

6x =24

x = 4

longer side = 10

68% of the races he competed in had a finish time around 64.5 and 65.5 seconds.

The term "variance" (or "") refers to an assessment of the data's dispersion from the mean. A small variance implies that the data are grouped around the normal, and while a large standard deviation shows that the data are more dispersed.

[tex]\begin{aligned}& \mathrm{P}(\mu-\sigma < \mathrm{X} < \mu+\sigma) \approx 68 \% \\& \mathrm{P}(\mu-2 \sigma < \mathrm{X} < \mu+2 \sigma) \approx 95 \% \\& \mathrm{P}(\mu-3 \sigma < \mathrm{X} < \mu+3 \sigma) \approx 99.7 \%\end{aligned}[/tex]

When the standard deviation out from mean of the distribution of X is and the mean of the dispersion of X is (assuming X is normally distributed).

Kiran's 400-meter dash timings have an average of 65 secs and a confidence interval of 0.5 seconds, and they are regularly distributed.

Using the formula, we then obtain

[tex]\mathrm{P}(65-0.5 < \mathrm{X} < 65+0.5)=\mathrm{P}(64.5 < \mathrm{X} < 65.5) \approx 68 \%[/tex]

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Using the Gaussian elimination method, the general solution is:               [tex]\[ x = -31t + 6 \][/tex], [tex]\[ y = 5 \][/tex], [tex]\[ z = 31t \][/tex].

To solve this system of equations, we can use the Gaussian elimination method. This method involves creating a matrix with the coefficients of the equations and using row operations to reduce the matrix to a form that can be easily solved.

First, we create a matrix with the coefficients of the equations:

[tex]\[ \left( \begin{array}{ccc|c} 1 & 1 & 1 & -1 \\ 2 & 3 & 2 & 3 \\ 2 & 1 & 2 & -7 \\ 1 & -3 & 2 & 10 \\ -1 & 3 & -1 & -6 \\ -1 & 3 & 2 & 6 \\ 6 & -2 & 2 & 4 \\ 3 & -1 & 2 & 2 \\ -12 & 4 & -8 & 8 \end{array} \right) \][/tex]

Next, we use row operations to reduce the matrix to a form that can be easily solved:

[tex]\[ \left( \begin{array}{ccc|c} 1 & 1 & 1 & -1 \\ 0 & 1 & 0 & 5 \\ 0 & -1 & 0 & -5 \\ 0 & -4 & 1 & 11 \\ 0 & 4 & 0 & -5 \\ 0 & 4 & 1 & 7 \\ 0 & -8 & -4 & 10 \\ 0 & -4 & -1 & 5 \\ 0 & 8 & 4 & -4 \end{array} \right) \][/tex]

[tex]\[ \left( \begin{array}{ccc|c} 1 & 1 & 1 & -1 \\ 0 & 1 & 0 & 5 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 31 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 27 \\ 0 & 0 & -4 & 50 \\ 0 & 0 & -1 & 25 \\ 0 & 0 & 4 & -44 \end{array} \right) \][/tex]

[tex]\[ \left( \begin{array}{ccc|c} 1 & 0 & 1 & -6 \\ 0 & 1 & 0 & 5 \\ 0 & 0 & 1 & 31 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -4 \\ 0 & 0 & 0 & 174 \\ 0 & 0 & 0 & 56 \\ 0 & 0 & 0 & -168 \end{array} \right) \][/tex]

From this reduced matrix, we can see that there are infinitely many solutions to this system of equations. The general solution can be written as:
[tex]\[ x = -31t + 6 \][/tex]
[tex]\[ y = 5 \][/tex]
[tex]\[ z = 31t \][/tex]
Where t is any real number.

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The we have shown that S is a subspace of Rn and that x + y ∈ S.

Suppose ~u1, ~u2, . . . , ~uk∈Rnand let S = span( ~u1, ~u2, . . . , ~uk). Suppose ~x, ~y ∈S.(d) Show that ~x + ~y ∈S.The solution is given as follows.The problem statement is given below.Suppose u1, u2, . . . , uk∈Rnand let S = span(u1, u2, . . . , uk). Suppose x, y ∈S. Show that x + y ∈S.To prove that x + y ∈ S, we must first prove that S is a subspace of Rn. We can then show that x + y is a linear combination of vectors in S, so it must be in S.Let's get started with the proof.Let S = span{u1, u2, . . . , uk} be a subspace of Rn. Then u1, u2, . . . , uk are linearly independent, which implies that they span S. Thus, any vector in S can be written as a linear combination of u1, u2, . . . , uk.Suppose x, y ∈ S. Then x = a1u1 + a2u2 + . . . + akuk, and y = b1u1 + b2u2 + . . . + bkuk, for some scalars a1, a2, . . . , ak and b1, b2, . . . , bk.Now we can show that x + y is a linear combination of vectors in S, so it must be in S.x + y = (a1 + b1)u1 + (a2 + b2)u2 + . . . + (ak + bk)ukSince a1, a2, . . . , ak and b1, b2, . . . , bk are scalars, (a1 + b1), (a2 + b2), . . . , (ak + bk) are also scalars. Thus, x + y is a linear combination of u1, u2, . . . , uk, and hence x + y ∈ S. Therefore, we have shown that S is a subspace of Rn and that x + y ∈ S.

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Therefore, the rate per period (i) is 0.00708333 and the number of periods (n) is 84. The future value of the annuity is $34563.89.

To find i (the rate per period) and n (the number of periods) for the given annuity, we need to use the formula for the future value of an annuity:

FV = PMT × [(1 + i)^n - 1] / i

Where FV is the future value, PMT is the periodic payment, i is the rate per period, and n is the number of periods.

Given:
PMT = $305
i = 8.5% / 12 = 0.00708333 (since the interest is compounded monthly)
n = 7 × 12 = 84 (since there are 12 months in a year and the deposits are made for 7 years)

Plugging these values into the formula, we get:

FV = $305 × [(1 + 0.00708333)^84 - 1] / 0.00708333

FV = $305 × [1.80722 - 1] / 0.00708333

FV = $305 × 0.80722 / 0.00708333

FV = $34563.89

Therefore, the rate per period (i) is 0.00708333 and the number of periods (n) is 84. The future value of the annuity is $34563.89.

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The rate at which the petrol was used in kilometers per liter is 10 km/L that means that for every litre of petrol used for driving, Peter's car could travel 10 kilometers.

Peter drove a thousand km and used 100 liters of petrol. To find the rate at which the petrol was utilized in kilometers consistent with liter, we divide the whole distance traveled via the quantity of petrol used.

The formulation for this is rate of petrol usage = overall distance traveled/quantity of petrol used.

Rate of petrol usage = total distance traveled/quantity of petrol used

Substituting the values we get,

= 1000 km / 100 L

= 10 km/L

This means that for every liter of petrol used, Peter was capable of travel 10 kilometers.

Thus, the rate at which the petrol was used in kilometers per liter is 10 km/L.

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Peter ended up with 31 beads, and Dan ended up with 28 beads.

What is the fraction?

A fraction is a mathematical expression that represents a part of a whole. It is written in the form of a ratio between two numbers, with the top number called the numerator and the bottom number called the denominator.

Let's represent the number of beads that Peter and Dan had at the start by P and D, respectively. Then we can set up an equation based on the information given in the problem:

After giving away 1/4 of his beads, Peter had 3/4 of his original number of beads, which is (3/4)P.

After giving away 1/5 of his beads, Dan had 4/5 of his original number of beads, which is (4/5)D.

According to the problem, both had the same number of beads left after giving away some of their beads:

(3/4)P = (4/5)D

We also know that Peter had 7 more beads than Dan at the start:

P = D + 7

We can use substitution to solve for D:

(3/4)(D+7) = (4/5)D

9D/20 + 21/20 = 4D/5

D = 35

So Dan had 35 beads at the start. Using the equation P = D + 7, we can find that Peter had:

P = D + 7 = 35 + 7 = 42

After giving away 1/4 of his beads, Peter had (3/4)P = (3/4)*42 = 31.5 beads, which we can round down to 31 beads since we're dealing with whole numbers of beads. After giving away 1/5 of his beads, Dan had (4/5)D = (4/5)*35 = 28 beads.

Therefore, Peter ended up with 31 beads, and Dan ended up with 28 beads.

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As a result, the 15 cartons of creamer will last for 1,350 days, or nearly 3.7  expression years. This implies the consumer drinks one serving of creamer per day at a weight of 5 grammes per serving.

An expression in mathematics is a collection of representations, numbers, and conglomerates that mimic a statistical correlation or regularity. A real number, a mutable, or a mix of the two can be used as an expression. Mathematical operators include addition, subtraction, fast spread, division, and exponentiation. Expressions are often used in arithmetic, mathematics, and form. They are employed in the depiction of mathematical formulas, the solving of equations, and the simplification of mathematical relationships.

(Total amount of creamer in grammes) / number of days (Amount of creamer consumed per day in grams)

Let's start by calculating the entire amount of creamer in grammes:

(Number of cartons) x (Total quantity of creamer) (Amount of creamer per box)

15 boxes x 450 grammes each box = total quantity of creamer

Total creamer weight = 6,750 g

(Total amount of creamer in grammes) / number of days (Amount of creamer consumed per day in grams)

The number of days is 6,750 grammes divided by 5 grammes each day.

The number of days is 1,350.

As a result, the 15 cartons of creamer will last for 1,350 days, or nearly 3.7 years. This implies the consumer drinks one serving of creamer per day at a weight of 5 grammes per serving.

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Step-by-step explanation:

The annual interest rate on Susan's savings account is 4%.

Annual interest rate, also known as annual percentage rate is the yearly interest generated by a sum that's charged to borrowers or paid to investors We can use the simple interest formula to calculate the annual interest rate which is Simple Interest = (Principal x Rate x Time)

By plugging our values, we get.

$54 = ($900 x Rate x 1.5)

Simplifying the equation:

Rate = $54 / ($900 x 1.5)

Rate = 0.04

Rate = 4%.

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Answer: 50%

Step-by-step explanation: There are 5 out of 10 employees at the water slide. 10 in this situation is going to be 100%

since there is only 5 out of 10 employees that are temporary

it would be 50%

Answer: 50%

Step-by-step explanation:

Out of 300 grams, 147 grams of food is not water in the given food sample.

If 51% of a certain food is water, then 49% of it must be something other than water. To find out how many grams of the food are not water, we can first calculate how many grams of water there are,

300 grams x 51% = 153 grams of water

Then we can subtract this from the total weight to find the amount of food that is not water -

300 grams - 153 grams = 147 grams of food that is not water

Therefore, out of 300 grams of nutrition, 153 grams are water and 147 grams are not.

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The conditional expectation of Y given that X is less than or equal to 1/2 is calculated by taking the weighted average of the possible values of Y, with the weights being the probabilities of X being less than or equal to 1/2 for each value of Y.
The formula for E(Y | X<=1/2) is:
E(Y | X<=1/2) = ∑y P(Y=y | X<=1/2) * y
To find P(Y=y | X<=1/2), we can use the formula:
P(Y=y | X<=1/2) = P(X<=1/2 | Y=y) * P(Y=y) / P(X<=1/2)
We can then plug in the values for each possible value of Y and calculate the conditional expectation.

For example, if Y can take on the values 0, 1, and 2, and the probabilities of X being less than or equal to 1/2 for each value of Y are 0.2, 0.5, and 0.3, respectively, and the probabilities of Y being 0, 1, and 2 are 0.4, 0.3, and 0.3, respectively, then:
E(Y | X<=1/2) = (0.2 * 0.4 / 0.5) * 0 + (0.5 * 0.3 / 0.5) * 1 + (0.3 * 0.3 / 0.5) * 2
E(Y | X<=1/2) = 0 + 0.3 + 0.36
E(Y | X<=1/2) = 0.66

Therefore, the conditional expectation of Y given that X is less than or equal to 1/2 is 0.66.

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Using Pivotal Condensation, the determinant of matrix A is 5.

Step 1: We start by initializing D as 1.

Step 2: We use the first row for pivotal condensation.
Row 0:  0 * D + 2 * 1 + 2 * 0 = 0
Row 1:  1 * D + 0 * 1 + 3 * 0 = 1
Row 2:  2 * D + 1 * 1 + 1 * 0 = 2

Step 3: We make the first row entries 0 by multiplying the entire row by (-2).
Row 0: -0 * D - 2 * 1 - 2 * 0 = 0
Row 1:  1 * D + 0 * 1 + 3 * 0 = 1
Row 2:  2 * D + 1 * 1 + 1 * 0 = 2

Step 4: We add row 0 to row 1 and row 0 to row 2.
Row 0:  0 * D + 2 * 1 + 2 * 0 = 0
Row 1:  1 * D + 0 * 1 + 3 * 0 = 1
Row 2:  0 * D + 3 * 1 + 3 * 0 = 3

Step 5: We make the entries of the second row 0 by multiplying the entire row by (-1/3).
Row 0:  0 * D + 2 * 1 + 2 * 0 = 0
Row 1: -1/3 * D - 0 * 1 - 3 * 0 = -1/3
Row 2:  0 * D + 3 * 1 + 3 * 0 = 3

Step 6: We add row 1 to row 0 and row 1 to row 2.
Row 0:  1/3 * D + 2 * 1 + 2 * 0 = 1/3
Row 1: -1/3 * D - 0 * 1 - 3 * 0 = -1/3
Row 2:  3/3 * D + 3 * 1 + 3 * 0 = 5

Step 7: We multiply the entries of the first row by (-3) to make the entries of the first row 0.
Row 0:  0 * D + 6 * 1 + 6 * 0 = 0
Row 1: -1/3 * D - 0 * 1 - 3 * 0 = -1/3
Row 2:  3/3 * D + 3 * 1 + 3 * 0 = 5

Step 8: We multiply the last row by D.
Row 0:  0 * D + 6 * 1 + 6 * 0 = 0
Row 1: -1/3 * D - 0 * 1 - 3 * 0 = -1/3
Row 2:  5 * D + 3 * 1 + 3 * 0 = 5D

Step 9: We subtract row 1 from row 0 and row 1 from row 2.
Row 0:  4/3 * D + 6 * 1 + 6 * 0 = 4/3D
Row 1: -1/3 * D - 0 * 1 - 3 * 0 = -1/3
Row 2:  4/3 * D + 3 * 1 + 3 * 0 = 4/3D

Step 10: We calculate the determinant by multiplying the last row entries.

Determinant of matrix A is 5.

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The expected rabbit population by next April is 486.

The rabbit population in a forest area can be found using the formula y=200(2.7)^0.08t, where y is the expected population, t is the time in months, and 200 is the initial population. To find the expected population by next April, we need to plug in the value of t as 12 (since there are 12 months in a year) and solve for y.

y=200(2.7)^0.08t

y=200(2.7)^0.08(12)

y=200(2.7)^0.96

y=200(2.4315)

y=486.3

Since the question asks for the expected population rounded to the nearest whole number, we can round 486.3 to 486.

Therefore, the expected rabbit population by next April is 486.

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The area of the resulting shape is 90 sq meters

The cylinder represents the given parameter

Such that we have the following dimensions

Radius = 6 cm

Height = 15 cm

When the solid cylinder is cut vertically through its center, we have a rectangle with the following dimensions

Length = 6 cm

Width = 15 cm

The area is then calculated as

Area = Length * Width

So, we have

Area = 6 * 15

Evaluate

Area = 90

Hence, the area is 90 square meters

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Answer:

IQR - 31

You first need to find the median. After find the middle number in the numbers before and after your median. This will give you your first and third quartile. Add these together then divide by two. This will give your IQR.

To complete the frequency table and ogive for the number of heads flipped, we need to determine the frequency of each of the indicated intervals. We can do this by counting the number of times each interval appears in the data set and entering the frequencies into the table.

Here is how to do it step-by-step:
1. Start with the first interval, which is 0-4 heads. Count the number of times this interval appears in the data set. For example, if there are 3 occurrences of 0-4 heads, enter 3 in the frequency column for this interval.
2. Repeat this process for each of the remaining intervals, counting the number of occurrences and entering the frequencies into the table.
3. Once you have entered all the frequencies, you can create the ogive by plotting the cumulative frequencies on a graph. Start with the first interval and plot the cumulative frequency at the upper limit of the interval. For example, if the first interval is 0-4 heads and the frequency is 3, plot the point (4,3) on the graph.
4. Continue this process for each of the remaining intervals, adding the frequency of each interval to the cumulative frequency and plotting the point at the upper limit of the interval.
5. Once you have plotted all the points, connect them with a line to create the ogive.

Here is the completed frequency table and ogive for the number of heads flipped:
| Interval | Frequency |
|----------|-----------|
| 0-4      | 3         |
| 5-9      | 5         |
| 10-14    | 4         |
| 15-19    | 2         |
| 20-24    | 1         |

Ogive:
(4,3) (9,8) (14,12) (19,14) (24,15)

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Answer:

Step-by-step explanation:

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