Fiona cleans offices. She is allowed 5 seconds per square foot. She cleans building A, which is 3,000 square feet, and building B, which is 2,460 square feet. Will she finish these two buildings in an 8-hour shift?

Lydia's claim is incorrect, and the true statement is (c) No; the slopes of segment EF and segment DF are not opposite reciprocals.

Right triangles have a pair of perpendicular lines

Coordinates

The coordinates are given as:

Start by calculating the slopes of DF and EF using:

\(m = \frac{y_2 -y_1}{x_2 -x_1}\)

So, we have:

\(m_{DF} = \frac{0 + 1}{0 +2}\)

\(m_{DF} = \frac{1}{2}\)

Also, we have:

\(m_{EF} = \frac{0 -2 }{0+2}\)

\(m_{EF} = \frac{-2 }{2}\)

\(m_{EF} = -1\)

Notice that the slopes of both lines are not opposite reciprocals.

Hence, the Lydia's claim is incorrect

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Answer:

option c

Step-by-step explanation:

i did the test

The truth set of the inequality is all values of x greater than -9. In interval notation, this can be written as (-9, ∞)

Multiply both sides by -2 (and reversing the direction of the inequality because we are multiplying by a negative number) gives:

x > -2*(5/2 + 2)

x > -5 - 4

x > -9

Therefore, the truth set of the inequality is all values of x greater than -9. In interval notation, this can be written as:

(-9, ∞)

Below is the complete question:

Find the truth set of x*(-1/2) < 5/2 + 2

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radius= 14 inches

the area of the circle.

\(area = \pi {r}^{2} \)

\(a = \pi \times {14}^{2} \)

\(a = 615.75216\)

\(a = 615.75 \: {in}^{2} \)

therefore, the area of the circle is 615.75 square inches.

The area is:

615.44 in²

Work/explanation:

The formula for the area of a circle is:

\(\sf{A=\pi r^2}}\)

Where

A = area

π = 3.14

r = radius

Diagram:

\(\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier(2.3,0)(2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,-2.121)(0,-2.3)\qbezier(2.3,0)(2.121,-2.121)(-0,-2.3)\put(0,0){\line(1,0){2.3}}\put(0.5,0.3){\bf\large 14\ in}\end{picture}\)

Plug in the data:

\(\sf{A=3.14\times14^2}\\\\\sf{A=3.14\times196}\\\\\sf{A=615.44\:in^2}\)

9514 1404 393

Answer:

Step-by-step explanation:

The sum of angles in any triangle is 180°.

  (x +45) +(2x +95) +(3x +48) = 180

  6x +188 = 180 . . . . . . . . collect terms

  6x = -8 . . . . . . . . . . . subtract 188

  x = -4/3 . . . . . . . . divide by 6

Then the angles are ...

  x +45 = -4/3 +45 = 43 2/3°

  2x +95 = 2(-4/3) +95 = 92 1/3° . . . . the obtuse angle

  3x +48 = 3(-4/3) +48 = 44°

Answer:

b + 32 = 76

Step-by-step explanation:

Answer:

One time

Step-by-step explanation:

The probability of picking a purple marble once is 1/6, so if I pick a marble 6  times, then (1/6)*6=1

The probability values for the questions posed are :

A.)

Number of students Taking a public speaking class majoring in business administration.

P(PS or BA) = 55/100 = 11/20

Therefore, the probability of PS or BA is 11/20

B.)

For the survey described, P(taken a public speaking class | majoring in Business Administration) represents the probability that a selected student has taken a public speaking class given that the student is majoring in business administration.

Hence, as inferred from P(A|B) ; probability of A given B.

C.)

P(PS|BA) = n(PSnBA) / n(BA)

n(PS) = 10+20 = 30

n(PSnBA) = 10

P(PS|BA) = 10/30 = 1/3

Therefore , the probability of PS|BA is 1/3.

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Answer:

-----------------------------

Given sequence:

If t(n) = 200, we can try to find the value of n:

There is no integer solution, since 32 < 40 < 64 or 2⁵ < 40 < 2⁶, the value of n should be between 5 and 6.

The sequence should include integer numbers, so there is no solution.

Given function:

Solve for x if f(x) is 200:

Answer:

1.  No

\(\textsf{2.} \quad x=\dfrac{\ln 40}{\ln 2} \approx5.32\;(\sf 2\;d.p.)\)

Step-by-step explanation:

Given sequence:

\(t(n)=5 \cdot 2^n\)

To determine if the sequence has a term with a value of 200, substitute t(n)=200 into the equation and solve for n:

\(\implies 5 \cdot 2^n=200\)

\(\implies 2^n=40\)

\(\implies \ln 2^n=\ln 40\)

\(\implies n\ln 2=\ln 40\)

\(\implies n=\dfrac{\ln 40}{\ln 2}\)

\(\implies n=5.3219280...\)

In a sequence, n is a positive integer.  Therefore, it is not possible for the sequence to have a term with the value of 200, as when t(n)=200, n is not a positive integer.

Given function:

\(f(x)=5 \cdot 2^x\)

To determine if the function has an output of 200, substitute f(x)=200 into the function and solve for x:

\(\implies 5 \cdot 2^x=200\)

\(\implies 2^x=40\)

\(\implies \ln 2^x=\ln 40\)

\(\implies x=\dfrac{\ln 40}{\ln 2}\)

\(\implies x=5.3219280...\)

Therefore, it is possible for the function to have an output of 200 when:

\(x=\dfrac{\ln 40}{\ln 2}\)

Answer:

5.4 = 7.56

1.5 = 2.1

Step-by-step explanation:

Just find the answer for 40% of the given meters.

Answer:

option A is your answer okk

Answer:

y = 2\(\sqrt{x}\) - 3

Step-by-step explanation:

Answer:

x = 4

Step-by-step explanation:

if a line is parallel to a side of a triangle and it intersects the other two sides then id divides those sides proportionally.

DE is such a line , then

\(\frac{BD}{AD}\) = \(\frac{BE}{EC}\)  ( substitute values )

\(\frac{x+2}{x}\) = \(\frac{3}{2}\) ( cross- multiply )

3x = 2(x + 2)

3x = 2x + 4 ( subtract 2x from both sides )

x = 4

Answer:

a. Possible outcomes for this chance experiment are:

(Here, D means a defective battery, and N means a nondefective battery.)

b. The number of outcomes in the sample space is infinite, as the experiment could potentially continue indefinitely. However, in practice, we can define a maximum number of batteries that could be examined before the experiment is stopped.

c. The event E, that the number of batteries examined is a number, would include outcomes where the experiment stopped at a particular number of batteries. For example, if the experiment stopped after examining three batteries (i.e., the fourth battery was nondefective), then the outcome would be DDDN, and it would be included in event E. However, outcomes where the experiment continued indefinitely (e.g., DDDDD...) would not be included in event E.

Answer:

A

Step-by-step explanation:

A. This is true and occurred during the industrial revolution

B. The government had not started managing the economy (yet)

C. This was happening even in the 1700s

D. This was also more apparent in the 1700s with the 13 colonies, especially with cash crops such as tobacco.

Answer:5.5

Step-by-step explanation:

The key is to divide by 2.54. It was not terminating so I rounded.

Step-by-step explanation:

=> 16/7 x ⅝ x 41/10

=> 41 x 2 ÷ 7 x 2

=> 41 ÷ 7

=> 5 6/7

Answer:

C.

Step-by-step explanation:

because x value cannot be repeated in an ordered pair

the one on the bottom left

Answer:

the large boxes=70

the small boxes= 60

Step-by-step explanation:

large boxes) 60 x 70 = 4,200

small boxes) 25 x 60 = 1,500

                                      5,700

The function notation of 9x + 3y = 12 is given as follows:

f(x) = 4 - 3x.

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

9x + 3y = 12.

The format for the function notation is given as follows:

Hence we must isolate the variable y, as follows:

3y = 12 - 9x

y = 4 - 3x (each term of the expression is divided by 3).

f(x) = 4 - 3x.

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Answer:

5AB + 40 = +2

Step-by-step explanation:

IN FIRST BLANK (5AB)

SECOND BLANK (40)

a. The distribution of X is X ~ N(68, 14).

b. The corresponding area to the right of 0.9286, which is approximately 0.1772.

c. The longest amount of time a patron can spend and still receive the discount is approximately 48.5654 minutes.

d. The Inter Quartile Range (IQR) for time spent at the hot springs is approximately 21.373 minutes.

a. The distribution of X is X ~ N(68, 14), where X represents the amount of time a person spends at Grover Hot Springs, 68 is the mean, and 14 is the standard deviation.

b. To find the probability that a randomly selected person stays longer than 81 minutes, we need to calculate the area under the normal curve to the right of 81.

Using the z-score formula: z = (x - μ) / σ, where x is the value (81), μ is the mean (68), and σ is the standard deviation (14).

Plugging in the values, we have z = (81 - 68) / 14 = 0.9286.

Using a standard normal distribution table or a calculator, we can find the corresponding area to the right of 0.9286, which is approximately 0.1772.

c. To find the longest amount of time a patron can spend at the hot springs and still receive the discount, we need to find the value that corresponds to the lowest 8% of the distribution.

Using a standard normal distribution table or a calculator, we can find the z-score that corresponds to the 8th percentile, which is approximately -1.4051.

Using the z-score formula, we can calculate the longest amount of time: x = μ + z \(\times\) σ = 68 + (-1.4051) \(\times\) 14 = 48.5654 minutes.

Therefore, the longest amount of time a patron can spend and still receive the discount is approximately 48.5654 minutes.

d. The Inter Quartile Range (IQR) is a measure of the spread of the data and represents the range between the first quartile (Q1) and the third quartile (Q3).

To find Q1 and Q3, we can use the z-score formula and the standard normal distribution table.

For Q1, we find the z-score corresponding to the 25th percentile, which is approximately -0.6745.

Using the formula Q1 = μ + z \(\times\) σ, we have Q1 = 68 + (-0.6745) \(\times\) 14 = 57.053.

Therefore, Q1 is approximately 57.053 minutes.

For Q3, we find the z-score corresponding to the 75th percentile, which is approximately 0.6745.

Using the formula Q3 = μ + z \(\times\) σ, we have Q3 = 68 + (0.6745) \(\times\) 14 = 78.426.

Therefore, Q3 is approximately 78.426 minutes.

Finally, we can calculate the IQR by subtracting Q1 from Q3: IQR = Q3 - Q1 = 78.426 - 57.053 = 21.373 minutes.

Therefore, the Inter Quartile Range (IQR) for time spent at the hot springs is approximately 21.373 minutes.

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Answer:

0

200

100

50

150

Step-by-step explanation:

0% of 200 = 0

100% of 200

= ( 100 / 100 ) x 200

= 1 x 200

= 200

50% of 200

= ( 100% of 200 ) / 2

= 200 / 2

= 100

25% of 200

= ( 50% of 200 ) / 2

= 100 / 2

= 50

75% of 200

= ( 75 / 100 ) x 200

= ( 75 x 200 ) / 100

= 75 x 2

= 150

Given:

The graph that represents the enrollment for college R between 1950 and 2000.

To find:

The average rate of increase in enrollment per decade between 1950 and 2000?

Solution:

The average rate of change of function f(x) over the interval [a,b] is:

\(m=\dfrac{f(b)-f(a)}{b-a}\)

So, the average rate of increase in enrollment per year between 1950 and 2000 is:

\(m=\dfrac{f(2000)-f(1950)}{2000-1950}\)

\(m=\dfrac{7-4}{50}\)

\(m=\dfrac{3}{50}\)

\(m=0.06\)

It is given average rate of increase in enrollment per year between 1950 and 2000 is 0.06.

We need to find the average rate of increase in enrollment per decade between 1950 and 2000, So, multiply the average rate of increase in enrollment per year by 10.

\(0.06\times 10=0.6\)

Therefore, the average rate of increase in enrollment per decade between 1950 and 2000 is 0.6.

Answer:

The cost function for \(C(x) = 0.06\cdot e^{0.08\cdot x}\) is \(c(x) = 0.75\cdot e^{0.08\cdot x}+10\).

Step-by-step explanation:

The marginal cost function (\(C(x)\)) is the derivative of the cost function (\(c(x)\)), then, we should integrate the marginal cost function to find the resulting expression. That is:

\(c(x) = \int {C(x)} \, dx + C_{f}\)

Where:

\(C_{f}\) - Fixed costs, measured in US dollars.

If we know that \(C(x) = 0.06\cdot e^{0.08\cdot x}\) and \(C_{f} = \$\,10\), then:

\(c(x) = 0.06\int {e^{0.08\cdot x}} \, dx + 10\)

\(c(x) = 0.75\cdot e^{0.08\cdot x}+10\)

The cost function for \(C(x) = 0.06\cdot e^{0.08\cdot x}\) is \(c(x) = 0.75\cdot e^{0.08\cdot x}+10\).

Answer: 22.4402

Step-by-step explanation:

Answer:

84.4% of 23 is 19.412.

Step-by-step explanation:

Let the unknown number be x.

Now, 84.4% of x = 19.412.

∴ (84.4 ÷ 100) × x = 19.412.

By simplifying the equation,

x = (19.412 × 100) ÷ 84.4

x = 1941.2 ÷ 84.4

∴ x = 23

Thus, 84.4% of 23 is 19.412.

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If α is an ordinal and β ∈ α, then β satisfies all three properties of an ordinal. Therefore, β is also an ordinal.

To prove the statement "If α is an ordinal and β ∈ α, then β is an ordinal," we need to demonstrate that if α is an ordinal and β is an element of α, then β satisfies the three properties of an ordinal:

Well-Ordering: Every element of β is strictly well-ordered by the membership relation ∈. This property holds because α is an ordinal and satisfies the well-ordering property, and β being an element of α inherits this property.

Transitivity: For any two elements γ and δ in β, if γ ∈ δ and δ ∈ β, then γ ∈ β. Since β is an element of α and α is transitive, the transitivity property carries over to β.

Trichotomy: For any two elements γ and δ in β, either γ ∈ δ, δ ∈ γ, or γ = δ. Again, this property is inherited from α, as β is an element of α.

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