What are the zeros of g(x) = x3 + 6x2 − 9x − 54?

Step-by-step explanation:

kaitlyn score is   6 points above the mean

                                 z-score =  6 / 12   = .5

kiersten score is 34.8 above the mean   z-score = 34.8/29 = 1.2

rebecca score is .54 above the mean   z -score = .54/ .3 = 1.8

rebecca scored the highest percentile (highes z-score)  of the three....the best

Answer:

[tex]x = 4\sqrt{3}[/tex]

Step-by-step explanation:

The triangle is right-angled

In a right-angled triangle, the following equality holds

[tex]\tan x = \dfrac{\text{Side Opposite x}}{\text{Side adjacent to x}}\\\\\tan 30 = \dfrac{4}{x}\\\\[/tex]

[tex]\tan 30 = \dfrac{1}{\sqrt{3}}\\\\\dfrac{1}{\sqrt{3}} = \dfrac{4}{x}\\\\[/tex]

Cross multiply:
[tex]1 \times x = 4 \times \sqrt{3}\\\\x = 4\sqrt{3}[/tex]

It is important to be aware that the distribution of sample means x may not match the distribution of the original x values exactly, due to sampling variability.

if the original distribution of x values is known to be normalize size needs to be large, then the sample size does not need to be restricted in order to claim that the distribution of sample means x taken from random samples of a given size is normal. The sample size should still be at least 30, as this is necessary to produce a reliable result.

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The likelihood of pulling a spade followed by a face card from a deck is 3/52, or roughly 0.0588. (rounded to four decimal places).

The probability of drawing a spade from a deck of 52 cards is 13/52, or 1/4, since there are 13 spades in the deck.

Once a spade has been drawn and not returned to the deck, there are now 51 cards left in the deck, of which 12 are face cards (the Jack, Queen, and King of each suit). Therefore, the probability of drawing a face card given that a spade has already been drawn is 12/51.

To find the probability of both events happening together (drawing a spade and then a face card), we can multiply the probabilities of each event:

P(drawing a spade, then drawing a face card) = P(drawing a spade) x P(drawing a face card given that a spade has been drawn)

= (1/4) x (12/51)

= 3/52

Therefore, the probability of drawing a spade, then drawing a face card out of a deck is [tex]3/52[/tex]  , or approximately [tex]0.0588[/tex] (rounded to four decimal places).

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As per the volume, the dimension of an Extra Large Box is 3.78 feet.

Let's call the length of one side of the cube "s". Since the volume of the cube is given as 512 cubic feet, we can set up an equation to relate the volume to the length of one side:

Volume of cube = s³ = 512 cubic feet

To solve for "s", we can take the cube root of both sides of the equation:

s = ∛512

We can simplify this expression by finding the prime factorization of 512:

512 = 2⁹

Therefore, we can rewrite the expression for "s" as:

s = ∛2⁹

Using the properties of exponents, we know that the cube root of 2^9 is the same as 2 raised to the power of (1/3) times 9:

s = [tex]2^{1/3} \times 9^{1/3}[/tex]

We can simplify this expression further by recognizing that 9 is a perfect cube, and its cube root is 3:

s = [tex]2^{1/3} \times 3[/tex]

Therefore, the length of one side of the cube-shaped box is:

s = [tex]2^{1/3} \times 3[/tex] feet

Since all sides of the cube are equal in length, the dimensions of the box are:

Length = Width = Height = [tex]2^{1/3} \times 3[/tex] feet = 3.78 feet.

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Complete Question:

Head Stevedore loads Extra Large Boxes, also in the shape of perfect cubes. The volume of each box is 512 cubic feet. What are the dimension of an Extra Large Box?

Let m and n be two arbitrary integers. We want to prove that the relation R is an equivalence relation, i.e. it is reflexive, symmetric, and transitive.

Reflexive:  We must show that mRm for all m ∈ Z.

Since the difference of m and m is 0, which is an even number, we have mRm.

Therefore, the relation R is reflexive.

Symmetric: We must show that if mRn, then nRm.

Let mRn, i.e., the difference of m and n is an even number.

Then the difference of n and m is also an even number.

Therefore, nRm, and the relation R is symmetric.

Transitive: We must show that if mRn and nRp, then mRp.

Let mRn and nRp, i.e., the difference of m and n is an even number and the difference of n and p is also an even number.

The sum of the difference of m and n and the difference of n and p is the difference of m and p, which is an even number.

Therefore, mRp, and the relation R is transitive.


Since the relation R is reflexive, symmetric, and transitive, it is an equivalence relation.

Conclusion: The relation R is an equivalence relation.

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The equation that correctly represents this situation is c(t) = 45 + 45(t-2). This equation states that the total number of credits the student will have after t semesters is equal to 45 (the number of credits they had before beginning college) plus 45 times the number of semesters after two (t-2).

To explain this equation in more detail, we need to break it down. First, the student had some credits earned while in high school, so the equation starts off with c(t) = 45, which is the number of credits the student had before beginning college.

Next, 45(t-2) represents the number of credits earned in the additional semesters since college began. The t-2 part of the equation means that the total number of credits earned in the additional semesters starts at zero for t = 2. Then, for each additional semester, 45 credits are added. So, for example, when t = 5, 45 credits are added to the initial 45 credits the student had before beginning college, resulting in 90 credits.Therefore, the equation  c(t) = 45 + 45(t-2)  correctly represents this situation.

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The given parametric equations are X(t) = -3/2 + 2t and y(t) = vt² + 16, the rate of change of y with respect to "t" when t = 3 is 6v

We have the parametric equations that are X(t) = -3/2 + 2t and y(t) = vt² + 16.

At time t, the rate of change of y with respect to t is given by the derivative of y with respect to t, that is dy/dt.

So, y(t) = vt² + 16

Differentiating with respect to t, we get

⇒ dy/dt = 2vt.

Now, t = 3 gives us,

y(3) = v(3)² + 16 ⇒ 9v + 16.

Therefore, the rate of change of y with respect to t at t = 3 is

dy/dt ⇒ 2vt ⇒ 2v(3) ⇒ 6v.

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The crοss-sectiοn in the image  shοwn appears tο be a trapezοid.

A quadrilateral with at least οne pair οf parallel sides. In this crοss-sectiοn, the tοp and bοttοm sides are parallel, while the οther twο sides are nοt parallel is  trapezοid. There are times when people define trapezoids as having at least one pair of opposite sides that are similar, and other times they say there is just one pair.

The parallelogram satisfies the "at least one" interpretation of the definition because it has two pairs of opposite sides that are similar, which qualifies it as both a trapezoid and a parallelogram. The parallelism does not support the "one and only one" interpretation of the description. How students respond to this depends on how they describe themselves.

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Angle L can be calculated as follows:

angle L = angle - 180 LMO stands for angle. MNO \sangle L = 180 - 150 - 30 degree angle L = 0

As a result, angle L is 0 degrees.

1. We know that sides AB and CD of trapezoid ABCD are parallel. We can use the tangent function to find the length of side AD because angle B is a right angle and angle ABD is 45 degrees:

AD/AB = tan(45)

AD=AB * tan(45) AD=AB

As a result, AD = 10.

2. We know that the sides PQ and RS of the trapezoid PQRS are parallel. We can use the sine function to find the length of side PS because angle Q is a right angle and angle PSQ is 60 degrees:

PS/QS sin(60) =

PS = sin * QS (60)

5 * sqrt = PS (3)

As a result, PS = 5*sqrt (3).

3. We know that the sides UV and WX of a trapezoid UVWX are parallel. We can use the cosine function to find the length of side WU because angle V is a right angle and angle WVU is 30 degrees:

WU/UV cos(30) =

UV * cos WU (30)

5 * sqrt(3) / 2 = WU

As a result, WU = (5/2)*sqrt (3).

4. We know that the sides LM and NO of the trapezoid LMNO are parallel. We can use the sine function to find the length of side MO because angle L is a right angle and angle MNO is 30 degrees:

MO/NO sin(30) =

MO = 4 / 2 MO = NO * sin(30)

As a result, MO = 2.

Because angles MNO and LMO add up to 180 degrees, we can calculate angle LMO as follows:

LMO angle = 180 - angle LMO MNO angle = 150 degrees

Finally, because angle N is a right angle, we can calculate angle L as follows:

angle L = angle - 180 LMO stands for angle. MNO\sangle L = 180 - 150 - 30 degree angle L = 0

As a result, angle L is 0 degrees.

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a) Sample size if the lower control limit is to be nonzero: 50
b) Sample size if the probability of detecting a shift to 0.04 is to be 0.50: 100

a) How large should the sample size be if the lower control limit is to be nonzero?

n = (2σ / d)²We know that:

Center line (CL) = 0.01

Sigma (σ) = LCL = 0.005

d = Centerline - LCL = 0.01 - 0.005 = 0.005

Substituting the values in the formula, we get

n = (2 * 0.005 / 0.01)²= 50 Hence, if the lower control limit is to be nonzero, the sample size should be 50.

b) How large should the sample size be if we wish the probability of detecting a shift to 0.04 to be 0.50?

The probability of detecting a shift to 0.04 is denoted by β and is calculated using the following formula:

β = Φ [(-Zα/2 + Zβ) / √ (p₀q₀/n)], Where, Φ is the standard normal distribution function, Zα/2 is the critical value for the normal distribution at the (α/2)th percentile, Zβ is the critical value for the normal distribution at the βth percentile, p₀ is the assumed proportion of nonconforming items, q₀ is 1 – p₀, and n is the sample size.

In order to determine the sample size, we must first select a value for β. If we select a value for β of 0.50, then β = 0.50. This implies that we have a 50% chance of detecting a shift if one occurs. Since the exact value for p₀ is unknown, we assume that p₀ = 0.01, which is equal to the center line.

n = (Zα/2 + Zβ)² p₀q₀ / β², Substituting the values in the formula, we get,

n = (Zα/2 + Zβ)² p₀q₀ / β²= (1.96 + 0.674)² (0.01) (0.99) / 0.50²= 99.7 ≈ 100

Hence, if we wish the probability of detecting a shift to 0.04 to be 0.50, the sample size should be 100.

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As per the equations, the two dogs are either 11 years old and 4 years old, or 4 years old and 11 years old.

Factoring is the process of breaking down a mathematical expression or number into its component parts, which can then be multiplied together to give the original expression or number. In algebra, factoring is often used to simplify or solve equations.

An equation is a statement that shows the equality of two expressions, typically separated by an equals sign (=). The expressions on both sides of the equals sign are called the "left-hand side" and "right-hand side" of the equation.

In the given question,

Let's call the age of the first dog "x" and the age of the second dog "y". We know that their ages add up to 15, so we can write:

x + y = 15

We also know that their ages multiply to 44, so we can write:

x * y = 44

Now we have two equations with two unknowns, which we can solve simultaneously to find the values of x and y.

One way to do this is to use substitution. From the first equation, we can solve for one variable in terms of the other:

y = 15 - x

We can substitute this expression for y into the second equation:

x × (15 - x) = 44

Expanding the left side, we get:

15x - x^2 = 44

Rearranging and simplifying, we get a quadratic equation:

x^2 - 15x + 44 = 0

We can factor this equation as:

(x - 11)(x - 4) = 0

Using the zero product property, we know that this equation is true when either (x - 11) = 0 or (x - 4) = 0.

Therefore, the possible values of x are x = 11 and x = 4.If x = 11, then y = 15 - x = 4, which means that the ages of the two dogs are 11 and 4.

If x = 4, then y = 15 - x = 11, which means that the ages of the two dogs are 4 and 11.

Therefore, the two dogs are either 11 years old and 4 years old, or 4 years old and 11 years old.

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The number of bacteria present with the given growth rate at t=5 is [tex]N(5) = 7,500 * e^2[/tex] and option x is the correct answer.

An exponential growth pattern is one in which the rate of increase is proportionate to the value of the quantity being measured at any given time. This indicates that the amount by which the quantity increases in each period is a constant proportion of the quantity's present value. Many branches of mathematics and science, such as physics, biology, and finance, utilise exponential growth. Modeling population expansion, the spread of infectious illnesses, the decay of radioactive materials, and the behavior of financial assets are all popular applications.

Given that, the number of bacteria present was 7,500.

The exponential growth is given by the formula:

[tex]N(t) = N(0) * e^{(kt)}[/tex]

Substituting the values N(0) = 7,500 and the growth rate is k = 2/5 we have:

[tex]N(5) = 7,500 * e^{(2/5 * 5)}\\N(5) = 7,500 * e^2[/tex]

Hence, the number of bacteria present at t=5 is [tex]N(5) = 7,500 * e^2[/tex] and option x is the correct answer.

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The cost of sales for liquor, expressed as a percentage is 19.39%.

To compute for the cost of sales for liquor, expressed as a percentage given that you have spent $2,500 on liquor for your bar and your bar sales have been $12,890, you can use the formula:

Cost of sales = (Cost of Goods Sold / Total Sales) × 100

where, Cost of Goods Sold (COGS) = Beginning Inventory + Purchases - Ending Inventory (or the total cost of goods sold during the period)

Total Sales = Gross sales - Sales discounts - Sales returns and allowances

Let's compute for the COGS first:

COGS = Beginning Inventory + Purchases - Ending Inventory

= $0 + $2,500 - $0

= $2,500

Total Sales = Gross sales - Sales discounts - Sales returns and allowances

= $12,890 - $0 - $0

= $12,890

Cost of sales is obtained as follows:

Cost of sales = (Cost of Goods Sold / Total Sales) × 100

= ($2,500 / $12,890) × 100

= 19.39%.

Therefore, the cost of sales for liquor, expressed as a percentage is 19.39%.

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Work Shown:

x = previous total weight

2% of x = 0.02x = amount of weight lost

x - 0.02x = 0.98x = current total weight

0.98x = 1519

x = 1519/0.98

x = 1550 pounds

Check:

2% of 1550 = 0.02*1550 = 31

The group collectively lost 31 pounds.

1550-31 = 1519

The answer is confirmed.

Note that 98% of 1550 = 0.98*1550 = 1519

Answer:

Step-by-step explanation:

2(2x-5) = 18 (FOIL)

4x-10=18

4x = 28

x= 7

Answer:

x = -2

Step-by-step explanation:

2(2x - 5) = -18

Divide both sides by 2.

2x - 5 = -9

Add 5 to both sides.

2x = -4

Divide both sides by 2.

x = -2

Given:

81x = 243x

= 243 / 81x

= 3

x = 3

As a result, Lara made a $6,262.71 initial deposit into her savings account.

A credit card user who pays 12% interest (or any other loan type that charges compound interest) will double their debt in six years. The rule can also be applied to determine how long it takes for inflation to cause money's value to decrease by half.

To calculate the initial investment, we can apply the continuous compounding formula:

A = Pe(rt)

Where:

A = the current balance ($7,000.00)

P = the initial deposit (unknown)

r = the annual interest rate (11% or 0.11 as a decimal)

t = the time in years (1 year)

Plugging in these values, we get:

$7,000.00 = Pe(0.11 * 1)

A shorter version of the exponential expression:

$7,000.00 = Pe0.11

$7,000.00 = P * 1.1166 (rounded to 4 decimal places)

Dividing both sides by 1.1166:

P = $6,262.71 (rounded to the nearest cent)

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The probability that exactly 10 out of 23 cars will not have engine failure is 0.007638.

Step-by-step explanation: First, calculate the probability of an engine failing in five years with no routine maintenance, which is 27.5%. This can be written as a decimal, 0.275.Next, calculate the probability of an engine not failing in five years with routine maintenance. This probability is 100%-27.5% = 72.5%, written as a decimal 0.725.

Now, using the Binomial Distribution formula (nCr), calculate the probability of exactly 10 engines not failing out of 23 cars, where n = 23, r = 10 and p = 0.725. The equation would be [tex](23C10)*(0.725^{10})*(0.275^{13}) = 0.0076379904[/tex]

Finally, round the result to 6 decimal places, giving an answer of 0.007638.

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Answer:

A questionnaire can collect quantitative, qualitive or both types of data.

Step-by-step explanation:

Answer:

Categorical data

Step-by-step explanation:

Data that relates to certain categories e.g males, females or any types of car

There are 59 groups of 3 girls, or 177 girls in total, taking classes at the school.

A ratio is a means to indicate the relative sizes of two or more items for the purpose of comparison. It can be shown with a colon or as a fraction. Mathematicians employ ratios for a variety of purposes, including comparing numbers, scaling up or down, and resolving proportions. Moreover, ratios can be employed in other mathematical processes, simplified, and transformed to percentages or decimals.

Given that for every three girls there are 4 boys in class.

Thus, the proportion can be given as:

4x = 236

x = 59

Now, the proportion of girls are 3x.

3(59) = 177 girls.

Hence, there are 59 groups of 3 girls, or 177 girls in total, taking classes at the school.

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Answer:

Step-by-step explanation:

To graph the line with slope 3 passing through the point (1,-5), we can use the point-slope form of the equation of a line, which is:

y - y1 = m(x - x1)

where m is the slope, and (x1, y1) is a point on the line.

Substituting the given values, we get:

y - (-5) = 3(x - 1)

Simplifying the equation, we get:

y + 5 = 3x - 3

Subtracting 5 from both sides, we get:

y = 3x - 8

Now we have the equation of the line in slope-intercept form, which is y = mx + b, where m is the slope and b is the y-intercept. Comparing this equation to the one we just found, we can see that the slope is 3 and the y-intercept is -8.

To graph the line, we can plot the y-intercept at (0,-8) and then use the slope to find other points on the line. Since the slope is 3, we can go up 3 units and to the right 1 unit to find another point on the line. We can repeat this process to find more points on the line, and then connect them to graph the line.

Answer: 31.8 years

Step-by-step explanation:

The formula to calculate the simple interest is:

I = Prt

where:

I = Interest

P = Principal amount

r = Rate of interest

t = Time (in years)

We want to know how long it will take for the investment to double in value. That means the interest earned should be equal to the principal amount. So, we can write:

2P = P + I

Simplifying:

P = I

Now, we can substitute the values given in the problem:

1400 = 14000.044t

Simplifying:

t = 1400/(1400*0.044) = 31.82 years

Therefore, it will take approximately 31.8 years for the investment to double in value.

Answer:

0

Step-by-step explanation:

0

Thank me...........

The slopes of the two line are equal. Hence, the two lines are parallel.

A line's slope is a gauge of the line's steepness. The ratio of the vertical change (change in y) to the horizontal change (change in x) between any two locations on the line is what is meant by this term. When a line moves from left to right, the slope might be positive, negative, zero, or undefined. When a line moves from left to right, the slope can be negative (when the line is vertical). The slope is determined using the following formula and is represented by the letter m:

m = (y2 - y1) / (x2 - x1)

where (x1, y1) and (x2, y2) are any two points on the line.

Given that, the first line passes through the points (-5, -5) and (-3, -3).

The slope is given by:

slope = (change in y) / (change in x)

slope = (-3 - (-5)) / (-3 - (-5)) = 1

The second line passes through the points (3, 1) and (4, 2).

slope = (2 - 1) / (4 - 3) = 1

The slopes of the two line are equal. Hence, the two lines are parallel.

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(a) To predict how many supermarkets are expected to adopt the new procedure over a long period of time, we can analyze the behavior of the differential equation using a phase portrait.

The equation can be rewritten as dN/N = (1-0.0005N)dt. Integrating both sides, we get ln|N| = t - 0.0005N^2/2 + C, where C is the constant of integration. Solving for N, we have:

N(t) = +/- sqrt((2ln|N| - 2C)/0.001)

We can see that the solutions are of the form of a hyperbola, with N approaching the asymptotes y=0 and y=2000. The equilibrium point is N=0, which is unstable, and the critical point is N=2000, which is stable.

Therefore, over a long period of time, we expect the number of supermarkets using the computerized checkout system to approach 2000.

(b) To solve the initial-value problem, we can use the separation of variables:

dN/N = (1-0.0005N)dt

ln|N| = t - 0.00025N^2 + C

N(0) = 1

Substituting N=1 and t=0, we get C=0. Therefore, the solution is:

ln|N| = t - 0.00025N^2

N = e^(t-0.00025N^2)

Using a graphing utility, we can plot the solution curve for N(t):

The graph confirms that the solution curve approaches 2000 as t increases.

When t=15, we can evaluate N(15) using the solution:

N(15) = e^(15-0.00025N^2)

Rounding to the nearest integer, we get N(15) = 1719.

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Answer:

x = -13/7

y = 54/7

Step-by-step explanation:

Y = 5x + 17                          Y = -2x + 4

5x + 17 = -2x + 4

7x + 17 = 4

7x = -13

x = -13/7

Not put -13/7 in for x and solve for y

y = 5(-13/7) + 17

y = 54/7

So, the answer is x = -13/7 and y = 54/7

Answer:   x = -13 / 7, y = 54/7

Step-by-step explanation:

To eliminate a variable, we can substitute y for 5x + 17

We get 5x + 17 = -2x + 4

7x = -13

x = -13 / 7

Substituting x into the 2nd equation y = 5 * -13 / 7 + 17

y = 119/7 - 65/7

y = 54/7

The survey ratings provide valuable feedback for drivers who are looking for quality tires for their vehicles. These ratings can give insight on steering responsiveness, how quickly a tire wears, and overall customer satisfaction. Drivers can also see how their tire of choice stacks up against other tires, making it easier to make an informed decision when shopping for tires.

The Tire Rack is America’s leading online distributor of tires and wheels. They conduct extensive testing to ensure that customers are provided with the most suitable tires for their vehicles, driving styles, and driving conditions. They also use an independent consumer survey to help drivers gain insight on long-term tire experiences.

The following data show survey ratings on a 1-10 scale, with 10 being the highest rating, of 18 maximum performance summer tires, as of February 3, 2009 (Tire Rack website). The variables are Steering (steering responsiveness), Tread Wear (quickness of wear), and Buy Again (overall tire satisfaction and desire to purchase the same tire again). The Yokohama Aegis LS4 has a Steering rating of 7.9, Tread Wear rating of 7, and Buy Again rating of 7.1. The Dunlop SP20 FE has a Steering rating of 5.7, Tread Wear rating of 5, and Buy Again rating of 3.3.

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