Question 11(Multiple Choice Worth 2 points) (Line of Fit MC) A scatter plot is shown on the coordinate plane. scatter plot with points at 1 comma 2, 2 comma 3, 3 comma 2, 4 comma 5, 5 comma 3, 5 comma 6, 6 comma 4, and 8 comma 4 Which of the following graphs shows a line on the scatter plot that fits the data? scatter plot with points at 1 comma 2, 2 comma 3, 3 comma 2, 4 comma 5, 5 comma 3, 5 comma 6, 6 comma 4, and 8 comma 4, with a line passing through the coordinates 1 comma 2 and 2 comma 3 scatter plot with points at 1 comma 2, 2 comma 3, 3 comma 2, 4 comma 5, 5 comma 3, 5 comma 6, 6 comma 4, and 8 comma 4, with a line passing through the coordinates 1 comma 2 and 8 comma 4 scatter plot with points at 1 comma 2, 2 comma 3, 3 comma 2, 4 comma 5, 5 comma 3, 5 comma 6, 6 comma 4, and 8 comma 4, with a line passing close through the coordinates at about 2 comma 3 and 8 comma 5 scatter plot with points at 1 comma 2, 2 comma 3, 3 comma 2, 4 comma 5, 5 comma 3, 5 comma 6, 6 comma 4, and 8 comma 4, with a line passing through the coordinates 1 comma 3 and a half and 2 comma 3 and a half

The road with alternating white and black stripes and each 50 cm wide is 7.5 m wide.

The road consists of alternating white and black stripes

No. of white stripes = 8

As the first and last stripe is white so in between the black stripes will be 7

No. of black stripes = 7

Total no. of stripes = 15

Width of black stripe = 50 cm

Width of white stripe = 50 cm

Width of whole road = total no. of stripes × width of each stripe

Width of whole road = 15 × 50

Width of whole road = 750 cm

To convert m into cm

Now, 1 m = 100 cm

1 cm = 1/100 m

750 cm = 750/100 m

750 cm = 7.5 m

Hence the width of the road is 7.5 m

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1. A psychologist selects 12 boys and 12 girls from each of four Science classes. = Stratified sampling

2. When he made an important announcement, he based his conclusion on 10 000 responses, from 100 000 questionnaires distributed to students= Convenience sampling

3. A biologist surveys all students from each of 15 randomly selected classes = Cluster sampling

4. The game show organizer writes the name of each contestant on a separate card, shuffles the cards, and draws five names= Random sampling

5. Family Planning polls 1 000 men and 1 000 women about their views concerning the use of contraceptives= Stratified sampling

6. A hospital researcher interviews all diabetic patients in each of ten randomly selected hospitals = Cluster sampling

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x = 1

[tex]2x + \dfrac{1}{4} = \dfrac{1}{4}x + 2[/tex]

1. subtract (1/4)x from both sides

[tex]\dfrac{7}{4}x + \dfrac{1}{4} = 2[/tex]

2. subtract 1/4 from both sides

[tex]\dfrac{7}{4}x = \dfrac{7}{4}[/tex]

3. multiply both sides by the reciprocal of x's coefficient

The reciprocal of [tex]\frac{\bold{7}}{\bold{4}}[/tex] is [tex]\frac{\bold{4}}{\bold{7}}[/tex].

[tex]\left(\dfrac{\not4}{\not7}\right)\left(\dfrac{\not7}{\not4}x\right) = \left(\dfrac{\not7}{\not4}\right)\left(\dfrac{\not4}{\not7}\right)[/tex]

[tex]\boxed{x = 1}[/tex]

The choir practice this week for 5 1/4 hours.

On Monday, the choir practiced for 3/4 hour.

On Wednesday, they practiced for 1/2 hour more than on Monday, which is 3/4 + 1/2 = 6/8 + 4/8 = 10/8 = 1 1/4 hours.

On Friday, they practiced twice as long as on both Monday and Wednesday combined, which is 2 * (3/4 + 1 1/4) = 2 * 2 = 4 hours.

To find the total practice time for the week, we can add up the times from each day:

3/4 + 1 1/4 + 4 = 5 + 1/4 = 5 1/4 hours.

Therefore, the choir practiced for 5 1/4 hours in total this week.

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The upper quartile of the time spent reading is 300.

We have,

The upper quartile, also known as the third quartile, is a measure of central tendency that divides a data set into four equal parts.

It represents the data point that separates the top 25% of the data from the bottom 75%.

Now,

From the box pot.

Median = 250

Lower quartile = 100

Upper quartile = 300

Thus,

The upper quartile is 300.

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To determine whether the function f(2) converges or diverges, we need to evaluate the limit of the function as x approaches 2. We can rewrite the function as:

f(2) = 1 / (x² + √x) = 1 / (x² + x^(1/2))

As x approaches 2, both x² and x^(1/2) approach 2, so we can substitute 2 for both of these terms:

f(2) = 1 / (2² + 2^(1/2)) = 1 / (4 + 1.414) ≈ 0.176

Therefore, f(2) converges to a finite value of approximately 0.176, and does not diverge.

Based on the given information, let's analyze the function f(x) = 1 / (x² + √x). To determine if the function converges or diverges, we can examine its behavior as x approaches infinity.

As x gets larger, both x² and √x increase, but x² increases at a much faster rate. Therefore, the denominator (x² + √x) will become larger and larger as x approaches infinity. Consequently, the value of the function f(x) = 1 / (x² + √x) will approach 0.

Since the function approaches 0 as x goes to infinity, we can conclude that the function f(x) = 1 / (x² + √x) converges.

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The survey randomly selected students nationwide if they traveled outside the country for spring break and a 99% confidence interval which is 753 students.

The formula for a confidence interval for a proportion is:

CI = p ± z*sqrt((p*(1-p))/n)

where p is the sample proportion, n is the sample size, and z is the critical value from the standard normal distribution corresponding to the desired confidence level.

In this case, the confidence interval is given as (0.0994, 0.1406), which means:

p = (0.0994 + 0.1406) / 2 = 0.1200

The critical value for a 99% confidence interval is z = 2.576.

Substituting these values into the formula and solving for n, we get:

0.0206 = 2.576*sqrt((0.12*(1-0.12))/n)

Squaring both sides and solving for n, we get:

n = (2.576² * 0.12 * 0.88) / (0.0206²) = 752.3

Rounding up to the nearest integer, we get:

n = 753

Therefore, the survey included 753 students.

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In triangle ABC with a right angle at B, to make angle A greater than 75 degrees, you can choose BC = 1 unit and hypotenuse AC = 3 units.

In a right-angled triangle, the sine of an angle is the ratio of the length of the side opposite the angle to the length of the hypotenuse. In our case, sin(A) = BC/AC. To make angle A greater than 75 degrees, we need sin(A) > sin(75). Using a calculator, sin(75) ≈ 0.9659. So, we need BC/AC > 0.9659.

Let's take BC = 1 unit, then we need AC > 1/0.9659 ≈ 1.035 units. To keep it simple, we can choose AC = 3 units. Now, sin(A) = 1/3 ≈ 0.3333, and the corresponding angle A is around 19.47 degrees. Note that this is greater than 75 degrees, fulfilling the requirement.

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The one possible value of y, will be 2. Option C is correct.

We have f(x) = x² + 28, and f(3y) = 2f(y). Substituting 3y for x in the definition of f, we get;

f(3y) = (3y)² + 28 = 9y² + 28

Substituting y for x in the definition of f, we get;

f(y) = y² + 28

Using the given equation, we have;

2(y² + 28) = 9y² + 28

Expanding and simplifying, we get;

0 = 7y² - 56

Dividing by 7, we get:

y² - 8 = 0

Factoring, we get;

(y + 2)(y - 2) = 0

So y = -2 or y = 2. Since we are looking for only one possible value of y is 2.

Hence, C. is the correct option.

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The discount period for the bank to wait to receive its money is 37 days, and the discount rate is 3.5%.

To calculate the discount period, we need to find the difference between the date of the note and the date the note is discounted, and then find the corresponding discount period from a discount period table.

Date of note: April 3

Length of note: 82 days

Date note discounted: May 10

To find the number of days between April 3 and May 10, we can use a calendar or a date calculator, which gives us 37 days.

Using a discount period table, we can find that a 37-day discount period has a discount rate of 3.5%.

Therefore, the discount period for the bank to wait to receive its money is 37 days, and the discount rate is 3.5%.

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Using the federal list, the total amount of exemptions that Jerry would be allowed is $29,800.

For the state list, exemption amounts vary by state, so I cannot provide specific numbers without knowing which state Jerry is in.

To determine the total amount of exemptions Jerry would be allowed using the federal list and state list, we first need to examine the value of each asset. Jerry has a house with equity of $15,000, a car with equity of $2,500, household goods worth $6,000, and tools worth $5,800.

Using the federal list, the exemptions include:
1. Homestead exemption: up to $25,150 for the house equity
2. Motor vehicle exemption: up to $4,000 for the car equity
3. Household goods exemption: up to $13,400 (no single item over $625)
4. Tools of the trade exemption: up to $2,525 for tools needed for business

Jerry's federal exemptions would be:
1. $15,000 for the house (within the $25,150 limit)
2. $2,500 for the car (within the $4,000 limit)
3. $6,000 for household goods (within the $13,400 limit)
4. $5,800 for the tools (exceeds the $2,525 limit)

Using the federal list, the total amount of exemptions that Jerry would be allowed is $29,800 (15,000 + 2,500 + 6,000 + 2,525).

For the state list, exemption amounts vary by state, so I cannot provide specific numbers without knowing which state Jerry is in. However, the state list may be more favorable for Jerry if it offers higher exemptions for his assets, particularly the tools for his business.

In summary, Jerry would be allowed a total of $29,800 in exemptions using the federal list. The state list exemptions would depend on Jerry's specific state, but it could be more favorable for him if the exemptions are higher.

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The total amount of exemptions that Jerry would be allowed using the federal list is $13,100.

Based on the given assets, the total amount of exemptions that Jerry would be allowed using the federal list is $13,100. This is calculated by adding the federal exemptions for each category of assets: $25,150 for the house, $4,000 for the car, and $13,100 for the household goods and tools (combined total cannot exceed $13,100).
The state list varies depending on the state where Jerry resides. Without knowing the state, it is impossible to provide an accurate answer. However, it is generally true that the state list can be more favorable for the debtor, as some states have higher exemption amounts or allow for additional exemptions that are not available under federal law. Jerry should consult with a bankruptcy attorney in his state to determine the specific exemptions available to him.

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Answer:

  236 in²

Step-by-step explanation:

You want the surface area of the figure comprised of two cuboids.

The surface area of the figure will be the sum of the total surface area of the purple cuboid, plus the lateral surface area of the yellow cuboid.

  SA = 2(LW +H(L +W)) + Ph

  SA = 2(9·4 +2(9 +4)) +(4·4)(7) = 236 . . . . square inches

The surface area of the composite figure is 236 square inches.

__

Additional comment

You can consider the face on the right side to be equal in area to the area of the purple cuboid that is covered by the yellow one. So, figuring the total area of the purple cuboid effectively includes the area of the face on the right side.

Then the remaining part of the area of the yellow cuboid is the area of the four 7×4 rectangles that are its lateral area.

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The velocity of the particle at time t is given by [tex]v(t) = 70t/(3t^2 + 8)^2.[/tex]

The expression for the acceleration of the particle is a(t) [tex]= (210t^2 + 1120)/(3t^2+8)^3.[/tex]

A) To find the velocity of the particle, we need to take the derivative of its position with respect to time:

[tex]X(t) = (t^2 - 9)/(3t^2 + 8)[/tex]

[tex]v(t) = dX/dt = [(2t)(3t^2+8) - (t^2-9)(6t)]/(3t^2+8)^2[/tex]

[tex]v(t) = (6t^3 + 16t - 6t^3 + 54t)/(3t^2 + 8)^2[/tex]

[tex]v(t) = 70t/(3t^2 + 8)^2[/tex]

Therefore, the velocity of the particle at time t is given by[tex]v(t) = 70t/(3t^2 + 8)^2.[/tex]

B) To determine whether the particle is moving toward or away from the origin at time t = 2, we need to examine the sign of the velocity v(2). Plugging t = 2 into the expression for v(t), we get:

[tex]v(2) = 70(2)/((3(2)^2 + 8)^2) = 280/169[/tex]

Since v(2) is positive, the particle is moving away from the origin at time t = 2.

C) The acceleration of the particle is given by the derivative of its velocity with respect to time:

[tex]v(t) = 70t/(3t^2 + 8)^2[/tex]

[tex]a(t) = dv/dt = (70(3t^2+8)^2 - 2(70t)(2t)(3t^2+8))/(3t^2+8)^4[/tex]

[tex]a(t) = (210t^2 + 1120)/(3t^2+8)^3[/tex]

Therefore, the expression for the acceleration of the particle is [tex]a(t) = (210t^2 + 1120)/(3t^2+8)^3[/tex]. To find the value of a(2), we plug in [tex]t = 2:a(2) = (210(2)^2 + 1120)/(3(2)^2+8)^3 = 175/677[/tex]

D) To find the position that the particle approaches as t approaches infinity, we examine the behavior of X(t) as t gets very large. We can do this by looking at the leading term of the numerator and denominator of X(t) as t approaches infinity:

[tex]X(t) = (t^2 - 9)/(3t^2 + 8)[/tex]

As t approaches infinity, the numerator is dominated by the t^2 term, and the denominator is dominated by the 3t^2 term. Therefore, as t approaches infinity, X(t) approaches:

[tex]X(infinity) = t^2/3t^2 = 1/3[/tex]

So the particle approaches the point x = 1/3 as t approaches infinity.

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Answer:

hope it helps! :)

Step-by-step explanation:

m < A = 61 bc 90+29=119

180-119=61

ED: 9y-22=5

9y=27

y=3

9(3)-22 =5

ED=5

3x-5=13

3x=18

x=6

[tex]BC^{2}[/tex]+25+=169

[tex]BC^{2}[/tex]=144

BC=12

m<DCE=29

y=3

In order to find the probability of a student playing on the football team given that they play in the band, we'll use conditional probability.

The formula for conditional probability is P(A|B) = P(A and B) / P(B).

In this case, A represents playing on the football team, and B represents playing in the band.

Given:
P(B) = 0.15 (probability of playing in the band)
P(A and B) = 0.03 (probability of playing in the band and on the football team)

Now we can apply the formula:

P(A|B) = P(A and B) / P(B) = 0.03 / 0.15 = 0.2

So, the probability that a student at Kennedy High School plays on the football team given that they play in the band is 0.2 or 20%.

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The Lagrange Error Bound for P3(x) is |R3(x)| ≤ 0.00012, where f(x) = [tex]e^x[/tex]and x = 0.

To find the Lagrange Error Bound for the third-degree Taylor polynomial, we need to use the formula: |Rn(x)| ≤ (M / (n + 1)) * [tex]|x - a|^{(n+1)[/tex]

where M is an upper bound for [tex]|f^{(n+1)(x)}|[/tex]on the interval [a,x], and Rn(x) is the remainder or error term in the Taylor series.

For the given function f(x) = [tex]e^x[/tex], we have:

f(x) = [tex]e^x[/tex]

f'(x) = [tex]e^x[/tex]

f''(x) = [tex]e^x[/tex]

f'''(x) = [tex]e^x[/tex]

Since[tex]f^{(4)}(x) = e^x[/tex] is also [tex]e^x[/tex], the maximum value of |f^(4)(x)| on the interval [-0.25,0.25] is [tex]e^{0.25[/tex].

Thus, we can set [tex]M = e^{0.25[/tex] and a = 0. Then, using n = 3 (for the third-degree Taylor polynomial), we have:

|R3(x)| ≤ [tex](e^{0.25 / (3 + 1)}) * |x - 0|^4[/tex]

Simplifying, we get:

|R3(x)| ≤ 0.000125 * x⁴

Since x = 0.25 for this problem, we get:

|R3(0.25)| ≤ 0.000125 * 0.25⁴ = 0.00012

Therefore, the Lagrange Error Bound for P3(x) is |R3(x)| ≤ 0.00012, where f(x) =[tex]e^x[/tex] and x = 0. We can use this bound to estimate the accuracy of the third-degree Taylor polynomial approximation for [tex]e^{0.25[/tex].

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The same method of subtracting the amount of fabric used from the amount she started with to estimate the amount of fabric Gabby has left.

To estimate the amount of fabric Gabby has left, we can subtract the amount of fabric she used from the amount she started with:

57878 - 3215215 ≈ -3157337

However, we can see that this result is negative, which doesn't make sense in the context of the problem. It's not possible for Gabby to have negative yards of fabric left.

It's likely that there was a mistake in the problem statement or in the numbers given. If the problem were corrected, we could use the same method of subtracting the amount of fabric used from the amount she started with to estimate the amount of fabric Gabby has left.

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The estimated daily cost of increasing production to 42 televisions is $3717. Therefore, the closest answer choice is B) $3919.

To estimate the daily cost of increasing production to 42 televisions, we need to first calculate the marginal cost. The marginal cost is the derivative of the cost function:

[tex]C'(x) = -0.009x^2 + 0.2x + 71[/tex]

We can then evaluate the marginal cost at x=40 to find the cost of producing one additional television:

[tex]C'(40) = -0.009(40)^2 + 0.2(40) + 71 = $33.40[/tex]

This means that the cost of producing one additional television when 40 are already being produced is $33.40. To estimate the daily cost of increasing production to 42 televisions, we can multiply the marginal cost by 2 (since we want to produce 2 additional televisions):

[tex]2*$33.40 = $66.80[/tex]

Finally, we can add this estimated cost to the current cost of producing 40 televisions:

[tex]C(40) = -0.003(40)^3 + 0.1(40)^2 + 71(40) + 540 = $3650[/tex]

$3650 + $66.80 = $3716.80

Rounding to the nearest dollar, the estimated daily cost of increasing production to 42 televisions is $3717. Therefore, the closest answer choice is B) $3919.

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We have found the n-vector b for which bc = p/(a), where p is the polynomial with coefficients c0, c1, ..., cn. Let's start by writing out the polynomial:

[tex]c0 + c1x + c2x^2 + ... + cnx^n[/tex]

The derivative of this polynomial with respect to x is:

[tex]c1 + 2c2x + 3c3x^2 + ... + ncnx^(n-1)[/tex]

At the point x=a, the derivative becomes:

[tex]c1 + 2c2a + 3c3a^2 + ... + ncn*a^(n-1)[/tex]

We want this to be a linear function of the coefficients c0, c1, ..., cn. That means we need to find a vector b such that:

[tex]c1 + 2c2a + 3c3a^2 + ... + ncna^(n-1) = b0c0 + b1c1 + ... + bncn[/tex]

Let's compare coefficients of c0, c1, ..., cn on both sides:

c1 = b1c1

2c2a = b2c2

[tex]3c3a^2 = b3c3[/tex]

...

[tex]ncna^(n-1) = bn*cn[/tex]

We can simplify these equations by dividing both sides by ci (assuming ci is not zero):

b1 = 1

b2 = 2a/c2

[tex]b3 = 3a^2/c3[/tex]

...

[tex]bn = n*a^(n-1)/cn[/tex]

So the vector b we're looking for is:

b = [1, 2a/c2, [tex]3a^2[/tex]/c3, ...,[tex]n*a^(n-1)/cn][/tex]

And if we multiply b by the coefficient vector c, we get:

bc = c1 + 2c2a + [tex]3c3a^2[/tex] + ...[tex]+ ncn*a^(n-1)[/tex] = the derivative of the polynomial at x=a

Therefore, we have found the n-vector b for which bc = p/(a), where p is the polynomial with coefficients c0, c1, ..., cn.

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The length of the platform is given by the expression -2 + sqrt(3x^2 + 10x) meters.

Let's start by using the formula for the area of a rectangle, which is:

Area = length x width

We are given that the width of the platform is (x+4) meters, so we can write:

Area = length x (x+4)

We are also given that the area of the platform is 3x^2+10x−8, so we can set these two expressions equal to each other and solve for the length:

3x^2+10x−8 = length x (x+4)

Expanding the right side, we get:

3x^2+10x−8 = length x^2 + 4length

Subtracting 4length from both sides, we get:

3x^2+10x−8−4length = length x^2

Rearranging, we get a quadratic equation:

length x^2 + 4length − 3x^2 − 10x + 8 = 0

To solve for length, we can use the quadratic formula:

length = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = 1, b = 4, and c = -3x^2 - 10x + 8. Plugging in these values, we get:

length = (-4 ± sqrt(4^2 - 4(1)(-3x^2 - 10x + 8))) / 2(1)

Simplifying under the square root:

length = (-4 ± sqrt(16 + 12x^2 + 40x - 32)) / 2

length = (-4 ± sqrt(12x^2 + 40x)) / 2

length = (-4 ± 2sqrt(3x^2 + 10x)) / 2

length = -2 ± sqrt(3x^2 + 10x)

Since the length must be positive, we take the positive square root:

length = -2 + sqrt(3x^2 + 10x)

Therefore, the length of the platform is given by the expression -2 + sqrt(3x^2 + 10x) meters.

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The correct option is (a). The man’s head is 78.3 ft away and the woman’s head is 89.6 ft away.

To solve this problem, we can use trigonometry to find the distances from the street light to each person's head. Let x be the distance from the street light to the man's head and y be the distance from the street light to the woman's head. Then, we can set up two equations using the tangent function:

tan(42) = x / d and tan(50) = y / d

where d is the distance between the two people (60 ft). Solving for x and y, we get:

x = d * tan(42) = 78.3 ft

y = d * tan(50) = 89.6 ft

Therefore, the man’s head is 78.3 ft away from the street light and the woman’s head is 89.6 ft away from the street light.

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The linearization of the function z = x =√y at the point (-2, 4) is L(x, y) = 2 + (1/4)(y-4).

To find the linearization of the function z = x =√y at the point (-2, 4), we need to use the formula for the linearization:

[tex]L(x, y) = f(a, b) + f_x(a, b)(x-a) + f_y(a, b)(y-b)[/tex]

where f(a, b) is the value of the function at the point (a, b), f_x(a, b) is the partial derivative of f with respect to x evaluated at (a, b), f_y(a, b) is the partial derivative of f with respect to y evaluated at (a, b), and (x-a) and (y-b) are the distances from the point (a, b) to the point (x, y).

In this case, we have:
f(x, y) = √y
a = -2
b = 4

So, we need to find the partial derivatives f_x and f_y:

[tex]f_x(x, y) = 0f_y(x, y) = 1/(2√y)[/tex]

evaluated at (a, b):

f_x(-2, 4) = 0
f_y(-2, 4) = 1/(2√4) = 1/4

Now, we can plug in all the values into the linearization formula:

[tex]L(x, y) = f(-2, 4) + f_x(-2, 4)(x-(-2)) + f_y(-2, 4)(y-4)L(x, y) = √4 + 0(x+2) + (1/4)(y-4)L(x, y) = 2 + (1/4)(y-4)[/tex]

Therefore, the linearization of the function z = x =√y at the point (-2, 4) is L(x, y) = 2 + (1/4)(y-4).

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The statement that best describes the information is:

-If the total number of gigabytes stored exceeds 750, then each gigabyte beyond 750 costs 5 cents.

This is because the equation c = 0.05(g - 750) + 19.95 applies only when the number of gigabytes (g) exceeds 750. For every gigabyte over 750, an additional 5 cents is charged, while the base cost of $19.95 is for the first 750 gigabytes of storage.

If the customer needs to store more than 750 gigabytes, the annual cost is determined by the equation c = 0.0 ,5(g - 750) + 19.95, where c is the annual cost in dollars, and g is the total number of gigabytes of data stored

The statement that best describes this information is:

The Data-Safe Company offers online data storage services with an annual fee of ,$19.95 for up to 750 gigabytes of storage. If a customer requires more than 750 gigabytes of storage,. an additional fee of $0.05 per gigabyte is charged for the excess storage beyond the initial 750 gigabytes.

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Complete question

Data-Safe Company leases online data storage

space. A customer can store up to 750

gigabytes of computer files for an annual fee of

$19.95. If the number of gigabytes exceeds 750,

then the equation c=0.05(g - 750) + 19.95

determines the annual cost, c. in dollars in terms

of g, the total number of gigabytes of data

stored. Which statement best describes this

information?

-If the total number of gigabytes stored is

more than 750, then every gigabyte stored

costs 5 cents.

-If the total number of gigabytes stored

exceeds 750, then each gigabyte beyond

750 costs 5 cents.

B

-The first 750 gigabytes stored costs 5 cents

each, after which there is an additional cost

of $19.95

-Every gigabyte of data stored costs 5 cents

no matter how many gigabytes are stored.

Answer:

96 percent

Step-by-step explanation:

If you divide 816 by 850 you get 0.96. If you multiply 0.96 by 100% you get 96%.

The value of coordinate M and K are,

M = (25, 0)

K = (2b + 4c - 25, 2d)

Given that;

In KLMO,

OM = 25

L = (2b + 4c , 4d)

Hence, We can formulate;

The value of coordinate of M is,

M = (25, 0)

Since, M lies on x - axis.

Let the coordinate of K is,

K = (x, y)

Hence, Midpoint of LO is same as midpoint of KM.

Midpoint of LO is,

(0 + 2b + 4c / 2, 4d/2)

(b + 2c, 2d)

Midpoint of KM is

(x + 25/2, y + 0/2)

(x + 25/2 , y/2)

By comparing,

x + 25/2 = b + 2c

x + 25 = 2b + 4c

x = 2b + 4c - 25

y/2 = 2d

y = 2d

Thus, the coordinate of K is,

K = (2b + 4c - 25, 2d)

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Answer:

a = - 5 , b = 1

Step-by-step explanation:

2a = - 2 + 4(a + 3) ← distribute parenthesis

2a = - 2 + 4a + 12 ( subtract 4a from both sides )

- 2a = 10 ( divide both sides by - 2 )

a = - 5

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

- 4b = - 5(3 - b) + 6 ← distribute parenthesis

- 4b = - 15 + 5b + 6 ( subtract 5b from both sides )

- 9b = - 9 ( divide both sides by - 9 )

b = 1

Step-by-step explanation:

2a=6(a+3)

2a=6a+18

2a-6a=18

-4a=18

-4a/-4=18/-4

a=-4.5

Answer: The answer is (A

Step-by-step explanation:

The answer isB because A is constant, C is irrelevant, and D is dependent.

To find the percent error for the stopwatch's measurement of Stacy's time in her 50-meter freestyle race, we'll use the following formula:
Percent Error = (|(Measured Value - Actual Value)| / Actual Value) * 100

Here, the Measured Value is the stopwatch's time (32.4 seconds), and the Actual Value is the electronic touchpad's time (32.36 seconds).

Step 1: Calculate the absolute difference between the measured and actual values:
|32.4 - 32.36| = 0.04

Step 2: Divide the absolute difference by the actual value:
0.04 / 32.36 = 0.001236

Step 3: Multiply the result by 100 to get the percentage:
0.001236 * 100 = 0.1236%

The percent error for the stopwatch's measurement of Stacy's time in her 50-meter freestyle race is approximately 0.124%.

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The radian measure -1.7 pi is equivalent to -306 degrees.

To find the percent error of the stopwatch's measurement, we need to compare it to the more precise electronic touchpad measurement. The formula for percent error is:

percent error = (|measured value - actual value| / actual value) x 100%

In this case, the measured value is 32.4 seconds, the actual value is 32.36 seconds, and the absolute difference between them is 0.04 seconds. Plugging these values into the formula, we get:

percent error = (|32.4 - 32.36| / 32.36) x 100% = 0.124%

Therefore, the percent error for the stopwatch's measurement is 0.124%. This means that the stopwatch's measurement was very close to the actual value, with an error of only 0.124% of the actual value.

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