Look at image that is attached

The key number, ac, is -104.

Key number is the product of the coefficients of the first and last terms. The right binomial factors for a quadratic expression can be determined with its assistance.

We must multiply the coefficient of the quadratic term by the constant term to obtain the key value, ac, which results in:

ac = 13(-8) = -104

Therefore, The key number, ac, is -104.

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The graph is a function: yes

The domain of this function is: x ≥ -3.

The range of this function: all real numbers.

In Mathematics and Geometry, a domain is the set of all real numbers for which a particular function is defined.

Additionally, the vertical extent of any graph of a function represents all range values and they are always read and written from smaller to larger numerical values, and from the bottom of the graph to the top.

By critically observing the graph shown in the image attached above, we can reasonably and logically deduce the following domain and range:

Domain = {-3, ∞}, x ≥ -3, or -3 ≤ x ≤ ∞.

Range = {-∞, ∞} or all real numbers.

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Answer:

E. 108 in.²

Step-by-step explanation:

The piece of wood has the shape of a triangular prism.

SA = lateral area + area of the bases

SA = perimeter × height + 2 × bh/2

SA = (5 + 3 + 4) in. × 8 in. + 3 in. × 4 in.

SA = 108 in.²

The linear regression equation is y = 22.6x + 52.33

The profit would reach 234 thousand dollars during the calendar year of 2020.

Replacing the means, the intercept a is given by:

108.8 = 22.6(2.5) + a.

a = 108.8 - 22.6(2.5)

a = 52.33

Therefore, the linear regression equation is y = 22.6x + 52.33

The profit would reach 234 thousand dollars during the calendar year of 2020.

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Answer:

Step-by-step explanation:

The missing parts of the triangle are

BD = 8

AD = 8 sqrt(2)

The missing part BD of the figure is solved using similar triangles

The expression is as follows

h / 8 = 8 / h

h^2 = 8 x 8

h^2 = 64

h sqrt = (64)

h = 8

Solving for AB we use Pythagoras theorem

AB^2 = h^2 + 8^2

AB^2 = 8^2 + 8^2

AB = sqrt(64 + 64)

AB = sqrt (128)

AB = 8 sqrt (2)

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Option C is the true statement from the data that has been gathered and shown in the image

The interquartile range exemplifies the "dispersion" or width of a set [1] by determining the difference between the top quartile (the 25% highest) and lower quartile (the 25% lowest). In reference to the provided picture:

- The bus drivers' interquartile range is 10, represented by subtracting their distance's lowest point (10) from the highest (20).

20 - 10 = 20

- Similarly, the teachers' interquartile range is 15, which stems from finding the difference between their lowest distance (15) and highest distance (30).

30 - 15 = 15

Therefore, comparing both ranges reflects that the bus drivers have an interquartile range of distances that is 5 miles smaller than the one for the teachers. Consequently, we opt for option C.

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Note that the estimated population mean is 2.2

The mean absolute deviation is 1

The average distance each data value is from the mean is 1 hours per night

The distribution has _Low_ variability

Where the data graphed is:

Hours Per Night       Students

0.5                                       0

1                                            2

1.5                                           3

2                                              4

2.5                                        2

3                                             2

3.5                                          1

4                                             1

The estimate of the population mean is

0.5 (0 ) + 1 ( 2) +  1.5( 3) + 2(4) + 2.5(2) + 3( 2 ) + 3.5( 1) + 4( 1) =3 3

⇒ 33 / 15

population Mean = 2.2



To find the Mean Absolute Deviation:

0.5 - 2.2 = - 1.7

1 -  2.2 = -1 .2

1.5 - 2.2   = - 0.7

2 -  2.2  = - 0.2

2.5   - 2.2  =0.3

3  -  2.2 = 0.8

3.5 - 2.2 = 1.3

4 - 2.2 =   1.8

The absolute deviations from the sample mean are

1.7 ,  1.2, 0.7, 0.2 , 0.3, 0.8 , 1.3,1.8

Thus

(1.7 + 1.2 + 0.7 + 0.2+0.3 + 0.8 + 1.3 +1.8) / 8 = 8/8 = 1

Therefore, the M.A. D is 1 hours per nigh


For average distance each data value is from the mean...

|(-1.7)| + |( -1.2) | + | ( -0.7)| + |( -0.2 | + | 0.3| + |  0.8| + | 1.3 | + | 1.8 | = 8

8/ 8 = 1


So average distance of each data value form the mean is 1 hours per night.


We discovered that the M.A. D in this scenario is 1hr/ night. We may claim that the distribution has low variability because this number is quite tiny.

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The rate of change for the interval between 0 and 2 for the quadratic equation is 10.

A quadratic equation is a type of polynomial equation of the second degree, which means it has one or more terms that are raised to the power of two.

The general form is;

ax² + bx + c = 0

Where;

The rate of change for the interval between 0 and 2 for the quadratic equation is calculated as;

f(2) - f(0) = 7 - (-3)

             = 7 + 3

             = 10

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Answer: Discover how a 3D solid shape can be made up from a 2D net. Understand how nets are formed with examples of common 3D polygons and prisms.

Step-by-step explanation:

The walkway around the swimming pool should be 8 feet wide.

From the question, we are given:

Length of the pool (l) = 15 feet

Width of the pool (w) = 7 feet

width of walkway (X) = X feet

Total Area =  Area of pool and walkway = 713 square feet

Let's derive equation for the pool and walkaway combined:

length of the pool and walkway = 15 + 2X [walkway is on both sides]

width of the pool and walkway = 7 + 2X

Total Area = (length x width) of the pool and walkway

713 = (15 + 2X) x (7 + 2X)

713 = 15(7 + 2X) + 2X(7 + 2X)

713 = 105 + 30X + 14X + 4X²

713 = 105 + 44X + 4X²

Rearrange the equation

4X² + 44X + 105 = 713

Collect like terms and express in a proper quadratic equation

4X² + 44X - 608 = 0

Dividing both sides by 4, we get

X² + 11X - 152 = 0

Solve this equation quadratically

X =  [tex]\frac{-b \± \sqrt{b^{2} - 4ac}}{2a}[/tex]

where

a = 1

b = 11

c = -152

Plug in the value into the equation

X = [tex]\frac{-11 \± \sqrt{11^{2} - 4(1)(-152)}}{2(1)}[/tex]

X = [tex]\frac{-11 \± \sqrt{121 + 608}}{2(1)}[/tex]

X = [tex]\frac{-11 \± \sqrt{729}}{2}[/tex]

X = [tex]\frac{-11 \± 27}{2}[/tex]

X = [tex]\frac{-11 + 27}{2}[/tex] or [tex]\frac{-11 - 27}{2}[/tex]

X = [tex]\frac{16}{2}[/tex] or [tex]\frac{-38}{2}[/tex]

X = 8 or -19

But since the width of the walkway cannot be negative, the only valid solution is therefore:

X = 8

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Answer:

The correlation coefficient of a data set can be significantly altered by removing an influential point from the data set.

Step-by-step explanation:

Please give the brainliest, really appreciated. Thank you

Answer:

[tex] {( \frac{1}{5}) }^{3} (312) = \frac{312}{125} = 2.496[/tex]

The volume of this rectangular prism is 2.496 cubic centimeters.

The volume of the rectangular prism is 2.496 cubic centimeters.

If the rectangular prism has 312 cubes with an edge length of 1/5 cm, we may calculate the total volume by multiplying the number of cubes by the volume of each cube.

V = x3 gives the volume of a cube with edge length x. Because the edge length, in this case, is 1/5 cm, the volume of each cube is (1/5)3 = 1/125 cm3.

We multiply the number of cubes by the volume of each cube to determine the volume of the rectangular prism:

312 cubes (1/125 cm3/cube) = 2.496cm3 volume

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The complement and the supplement of an angle of 68º are given as follows:

Hence the complement and the supplement of an angle of 68º are obtained as follows:

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Manuel bought a car for 65% (percentage) of the original price of $7,000, he must have paid $4,550 for the car.

The percentage is a number or ratio that represents a fraction of a whole number or the sum of ratios.

The percentage is computed by dividing one number by another and multiplying the result by 100.

The original price of the car = $7,000

The discount factor = 65% (1 - 35%) or 0.65

The discounted price = $4,550 ($7,000 x 0.65)

Thus, by implication, for receiving a discount of 35% (100% - 65%), Manuel paid $4,550 for the car.

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Answer:

C you are so welcome I hope it helped you

The value of the test statistic used in this test is t = -2.168 and the P-value of this test is P-value is 0.034

To test whether the mean time for Dr. McCoy is different from the mean time for Dr. Turk, we can use a two-sample t-test with unequal variances. The test statistic for this test is given by:

t = (X₁ - X₂) / √(s₁²/n₁ + s₂²/n₂)

where X₁ and X₂ are the sample means, s₁ and s₂ are the sample standard deviations, and n₁ and n₂ are the sample sizes.

Plugging in the given values, we get:

t = (78 - 78.7) / √(2.3²/38 + 4.1²/33)

= -2.168

Therefore, the value of the test statistic used in this test is t = -2.168.

b) To find the P-value of this test, we need to use a t-distribution with degrees of freedom given by:

df = (s₁²/n₁+ s₂²/n₂)² / [ (s₁²/n₁)²/(n₁-1) + (s₂²/n₂)²/(n₂-1) ]

Plugging in the given values, we get:

df = (2.3²/38 + 4.1²/33)² / [ (2.3²/38)²/37 + (4.1²/33)²/32 ] ≈ 65.385

Using a t-distribution table or a calculator with t-distribution function, we find that the P-value for a two-tailed test with t = -2.168 and df = 65.385 is approximately 0.034.

Therefore, the P-value of this test is P-value = 0.034

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The value of a in the equation 60°a = [? ]° is dependent on the value of [? ].

To solve for a in the equation 60°a = [? ]°, we need to isolate a on one side of the equation. We can do this by dividing both sides of the equation by 60°:

60°a / 60° = [? ]° / 60°

This simplifies to:

a = [? ]° / 60°

Since we don't have the value of [? ]°, we cannot determine the exact value of a. However, we can simplify the expression by canceling out the ° units:

a = [? ] / 60

Therefore, the value of a is dependent on the value of [? ].

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The missing lengths of the following two triangles:

Case 1: x = 50 km

Case 2: x = 40 km

Case 3: Obtuse

Case 4: Acute

Case 5: Right

Case 6: m = 16, n = 8√3

Case 7: x = 3√2, y = 3

In this problem we must determine all missing lenghts in a triangle, this can be done by law of cosine:

x² = a² + b² - 2 · a · b · cos X

Where X is the angle opposite to side x.

Now we proceed to determine the missing lengths of the following two triangles:

Case 1

x = √[(14 km)² + (48 km)² - 2 · (14 km) · (48 km) · cos 90°]

x = 50 km

Case 2

x = √[(24 km)² + (32 km)² - 2 · (24 km) · (32 km) · cos 90°]

x = 40 km

In addition, we can determine if any triangle is acute, obtuse and right also by law of cosine:

cos X = - (x² - a² - b²) / (2 · a · b)

Case 3

cos X = - [(18 in)² - (12 in)² - (9 in)²] / [2 · (12 in) · (9 in)]

cos X = - 0.458 (Obtuse)

Case 4

cos X = - [(15 ft)² - (12 in)² - (14 in)²] / [2 · (12 in) · (14 in)]

cos X = 0.342 (Acute)

Case 5

cos X = - [(5 ft)² - (3 in)² - (4 in)²] / [2 · (3 in) · (4 in)]

cos X = 0 (Right)

Finaly, we determine the missing lengths of two right triangles by trigonometric functions:

sin θ = y / r

cos θ = x / r

tan θ = y / x

Now we find the missing lengths:

Case 6

m = 8 / cos 60°

m = 16

n = 8 · tan 60°

n = 8√3

Case 7

x = 3 / sin 45°

x = 3√2

y = 3 / tan 45°

y = 3

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The type of relationship that is indicated by the residual plot is given as follows:

Non-linear relationship.

A function is classified as exponential if when the input variable is changed by one, the output variable is multiplied by a constant.

A function is classified as linear if when the input variable is changed by one, the output variable is increased/decreased by a constant.

For this problem, we have an alternative pattern, where the function neither multiplied nor added/subtracted by a constant value when the input is increased by one, hence we merely classify the relationship as a non-linear relationship.

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The amount that you would save in annual fuel expenses would be $9, 333. 32

The amount that you would have saved over the five years would be $52, 854.63

The amount saved in fuel when using the hybrid is:

= Fuel usage by SUV - Fuel usage by Hybrid

= ( 30, 000 / 9 x 4 ) - ( 30, 000 / 30 x 4 )

= $9, 333.32

If you saved this amount as an annuity, it would come out to:

=  ( 9, 333.32 / 12 ) x  [ ( 1 + 0. 004333 ) ^ ( 12 x 5 ) - 1 ] / 0. 004333

= 777. 78 x 0. 2957 / 0. 004333

= $ 52, 854.63

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Answer:3.77

2.51 are correct

Step-by-step explanation:We can solve this quadratic equation in cos(θ) by using the substitution u = cos(θ):

-7u^2 + 3u + 7 = 0

Multiplying both sides by -1, we get:

7u^2 - 3u - 7 = 0

We can use the quadratic formula to solve for u:

u = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = 7, b = -3, and c = -7. Substituting these values, we get:

u = (3 ± sqrt(9 + 196)) / 14

u = (3 ± 5sqrt(5)) / 14

Therefore, either:

The domain of the function g(x) is given as follows:

B. All values of x such that x ≥ 0.

The domain of a function is the set of all the possible input values that can be assumed by the function.

The function for this problem is defined as follows:

[tex]g(x) = \sqrt{8x}[/tex]

The square root function assumes values that are greater than or equals to zero, hence the domain of the function is obtained as follows:

8x ≥ 0.

x ≥ 0.

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Answer:

Step-by-step explanation:

35.26    rounded to  35.3

8.32   rounded to        8.3

6.78    rounded to       6.8

35.3 + 8.3 = 43.6

43.6 + 6.8 = 50.4

The answer is 50.4.

39.26 rounds to 35.3

8.32 rounds to 8.3

6.78 rounds to 6.8

When you add them together, you get 50.4.

When it comes to interest, most Certificates of Deposit (CDs) will allow periodic interest payouts. So, correct option is B.

A CD is a type of savings account that allows you to earn a fixed interest rate on your deposit for a specific period of time, called the term. Typically, the longer the term of the CD, the higher the interest rate offered.

However, during the term of the CD, the funds are locked in, meaning that you cannot withdraw the principal without incurring a penalty.

While some CDs may allow for unlimited withdrawals of interest, this is not common, and may still come with restrictions or penalties. Loans against principal are also not typically allowed with CDs, as the funds are meant to be held for a set term.

Therefore, the most common option available for CD holders is to receive periodic interest payouts, which can be monthly, quarterly, or annually, depending on the terms of the CD. This allows the CD holder to earn interest on their deposit while still receiving some income during the term of the CD.

So, correct option is B.

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Answer:

We can use the formula for compound interest:

A = P(1 + r/n)^(nt)

where A is the ending amount, P is the principal (starting amount), r is the interest rate (as a decimal), n is the number of times the interest is compounded per year, and t is the time in years.

We know P = $5700, A = $6100, n = 1 (since the interest is compounded annually), and t = 12. We can solve for r:

$6100 = $5700(1 + r/1)^(1*12)

$6100/$5700 = (1 + r)^12

1.0702 = (1 + r)^12

log(1.0702) = log[(1 + r)^12]

0.0291 = 12log(1 + r)

0.00243 = log(1 + r)

10^(0.00243) = 1 + r

r = 0.0059 or 0.59%

Therefore, the interest rate of the savings account is 0.59%.

The quotient of the long division of the polynomial (6x³ - 4x² + 11)/(x+3)  is 6x² - 22x + 66.

The quotient of the polynomial divided by a factor of x + 3 is determined by applying long division method as shown below;

6x³ - 4x² + 11 ÷  x + 3

                   6x² - 22x + 66

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

  x + 3   √ 6x³ - 4x² + 11

               - (6x³ + 18x²)

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

                         -22x² + 11

                    -   (-22x² - 66x)

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

                                  66x + 11

                            -  (66x + 198)

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

                                      -187

Thus, the quotient of the long division of the polynomial is obtained as   6x² - 22x + 66.

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Find the quotient of the polynomial using long division

6x3 − 4x2 + 11/x + 3

The resultant between the two vectors is 21.17 lb

Parallelogram law of vectors says that if two vectors acts on a point, the resultant of this vectors can be represented by the diagonal from the point of intersection.

R² = x²+y2+2xycos (tetha)

where x and y are the vectors

When x = 20.6 and y = 17.4

R² = 20.6²+17.4²+2(20.6)(17.4)cos 112.9

R² = 727.12+716.88cos 112.9

R² = 727.12 -278.95

R² = 448.17

R = √448.17

R = 21.17 lb

Therefore the resultant is 21.17lb

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Answer:

y=mx + b

Step-by-step explanation:

y=3x-4

3 is the slope and -4 is the y intercept

The approximate height of the pole in centimeters is 17,500 centimeters.

We have,

Let's start by calculating the total length of the pole, which is buried plus the part that protrudes from the ground.

We know that the part that protrudes is 25 meters and that it represents one-seventh of the total length.

25 = L/7

where L is the total length of the pole.

We can solve for L by multiplying both sides of the equation by 7:

L = 7 x 25

  = 175 meters

Now we need to convert this length to centimeters.

Since 1 meter is equal to 100 centimeters,

175 meters x 100

= 17,500 centimeters

Therefore,

The approximate height of the pole in centimeters is 17,500 centimeters.

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The complete question.

The Clarisse company, dedicated to the telephone business, is expanding its coverage, due to this, it is placing poles, where the seventh part of a pole is buried, and protrudes 25 meters from the ground. Calculate the approximate height of the pole in centimeters.