What are some advantages and disadvantages of using this repayment plan?

Answer:

Its a triangular prism

The surface area is ( perimeter x height of prism) + (2 x area of base triangle)

Exponents that are positive go to the right. Exponents with negative signs shift to the decimal to the left. A result is stated in scientific notation when it is a power of ten multiplied by a number that is greater than or equal to one but less than ten.

One approach is to fully write out each number after it has been converted from scientific notation, then calculate the total of the two values and translate the result back into scientific notation.

Therefore, The 2, 4, 5, and final 0 in the number 0.2540 are all significant because they come after the decimal point and follow numbers. Scientific notation does not consider exponential digits to be significant; 1.12106 has three significant digits, 1, 1, and 2.

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Answer:

the probability of selecting a blue sock from the laundry basket is 2/7 or approximately 0.2857.

Step-by-step explanation:

The probability of selecting a blue sock can be found by dividing the number of blue socks by the total number of socks in the basket:

Probability of selecting a blue sock = Number of blue socks / Total number of socks

Probability of selecting a blue sock = 4 / 14

Simplifying the fraction by dividing both the numerator and denominator by 2 gives:

Probability of selecting a blue sock = 2 / 7

Answer:

i show the answer in picture

Answer: C

Step-by-step explanation:

If q is an odd number and the median of q consecutive integers is 120, then the largest of these integers is option (A) (q-1) / 2 + 120

The number q is an odd number

The median of q consecutive integers = 120

Consider the q = 3

Then three consecutive integers will be 119, 120, 121

The largest number = 121

Substitute the value of q in each options

Option A

(q-1) / 2 + 120

Substitute the value of q

(3-1)/2 + 120

Subtract the terms

=2/2 + 120

Divide the terms

= 1 + 120

= 121

Therefore, largest of these integers is (q-1) / 2 + 120

I have answered the question in general, as the given question is incomplete

The complete question is

if q is an odd number and the median of q consecutive integers is 120, what is the largest of these integers?

a) (q-1) / 2 + 120

b) q/2 + 119

c) q/2 + 120

d) (q+119)/2

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The discriminant of equation qx² + rx + s = 0 is  r² - 4qs. Option c is correct.

The discriminant of the quadratic equation qx² + rx + s = 0 is given by the expression b² - 4ac, where a, b, and c are the coefficients of the quadratic equation.

In this case, the coefficients are:

a = q

b = r

c = s

Therefore, the discriminant is:

b² - 4ac = r² - 4qs

Hence, the  discriminant of qx² + rx + s = 0 is  r² - 4qs.

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Answer: (2,9)

Step-by-step explanation:

The point lies in the region that is shaded by both inequalities.

Answer: 2.5, 5.3

Step-by-step explanation:

\(-16t^2 +126t=217\\ \\ 16t^2 -126t+217=0\\\\t=\frac{-(-126) \pm \sqrt{(-126)^2 -4(16)(217)}}{2(16)}\\\\t \approx 2.5, 5.3\)

Answer:

8 meters of ribbon

Step-by-step explanation:

because a square has four sides so if you do 2×4 =8

Answer:

I hope it helps it was the best I can do.

Step-by-step explanation:

-x12 + x7 = 0

Factorization:

−x7(x − 1)(x4 + x3 + x2 + x + 1) = 0

Solutions based on Jenkins–Traub algorithm:

x1 = … = x7 = 0

x8 = 1

x9 = −0.809017 − 0.587785i

x10 = −0.809017 + 0.587785i

x11 = 0.309017 − 0.951057i

x12 = 0.309017 + 0.951057i

Step-by-step explanation:

10 pieces, each 2/5 m long are

10 × 2/5 = 20/5 = 4 m

that leaves 8 - 4 = 4 m of ribbon.

now, from these 4 m, pieces of 3/4 m are cut.

how many ?

as many as times 3/4 fit into 4 :

4 / 3/4 = 4/1 / 3/4 = 4×4 / (1×3) = 16/3 = 5 1/3.

that means, she got max. 5 pieces of 3/4 m. and she has 1/3 m ribbon left.

Answer: Let us call the centers of the four circles C1, C3, C5, and C7, respectively, where the subscript refers to the radius of the circle. Without loss of generality, we can assume that the tangent point A lies to the right of all the centers, as shown in the diagram below:

                 C7

          o-----------o

       C5 /             \ C3

         /               \

        o-----------------o

                C1

                |

                |

                | l

                |

                A

Let us first find the coordinates of the centers C1, C3, C5, and C7. Since all the circles are tangent to line l at point A, the centers must lie on the perpendicular bisector of the line segment joining A to the centers. Let us denote the distance from A to the center Cn by dn. Then, the coordinates of Cn are given by (an, dn), where an is the x-coordinate of point A.

Using the Pythagorean theorem, we can write the following equations relating the distances dn:

d1 = sqrt((d3 - 2)^2 - 1)

d3 = sqrt((d5 - 4)^2 - 9)

d5 = sqrt((d7 - 6)^2 - 25)

We can solve these equations to obtain:

d1 = sqrt(16 - (d7 - 6)^2)

d3 = sqrt(4 - (d7 - 6)^2)

d5 = sqrt(1 - (d7 - 6)^2)

Now, let us consider the region S that lies inside exactly one of the four circles. This region is bounded by the circle of radius 1 centered at C1, the circle of radius 3 centered at C3, the circle of radius 5 centered at C5, and the circle of radius 7 centered at C7. Since the circles are all tangent to line l at point A, the boundary of region S must pass through point A.

The maximum possible area of region S occurs when the boundary passes through the centers of the two largest circles, C5 and C7. To see why, imagine sliding the circle of radius 1 along line l until it is tangent to the circle of radius 3 at point B. This increases the area of region S, since it adds more points to the interior of the circle of radius 1 without removing any points from the interior of the other circles. Similarly, sliding the circle of radius 5 along line l until it is tangent to the circle of radius 7 at point C also increases the area of region S. Therefore, the boundary of region S must pass through points B and C.

Using the coordinates we obtained earlier, we can find the x-coordinates of points B and C as follows:

x_B = a - 2 - sqrt(9 - (d7 - 6)^2)

x_C = a + 6 + sqrt(9 - (d7 - 6)^2)

To maximize the area of region S, we want to maximize the distance BC. Using the distance formula, we have:

BC^2 = (x_C - x_B)^2 + (d5 - d3)^2

Substituting the expressions we derived earlier for d3 and d5, we get:

BC^2 = 32 - 2(d7 - 6)sqrt(9 - (d7 - 6)^2)

To maximize BC^2, we need to maximize the expression inside the square root. Let y = d7 - 6. Then, we want to maximize:

f(y) = 9y^2 - y^4

Taking the derivative of f(y) with respect to y and setting it equal to zero, we get:

f'(y) = 18y - 4y^3 = 0

This equation has three solutions: y = 0, y = sqrt(6)/2, and y = -sqrt(6)/2. The only solution that gives a maximum value of BC^2 is y = sqrt(6)/2, which corresponds to d7 = 6 + sqrt(6)/2.

Substituting this value of d7 into our expressions for d1, d3, and d5, we obtain:

d1 = sqrt(16 - (sqrt(6)/2)^2) = sqrt(55/2)

d3 = sqrt(4 - (sqrt(6)/2)^2) = sqrt(19/2)

d5 = sqrt(1 - (sqrt(6)/2)^2) = sqrt(5/2)

Using these values, we can compute the coordinates of points B and C as follows:

x_B = a - 2 - sqrt(9 - (sqrt(6)/2)^2) = a - 2 - sqrt(55)/2

x_C = a + 6 + sqrt(9 - (sqrt(6)/2)^2) = a + 6 + sqrt(55)/2

The distance between points B and C is then:

BC = |x_C - x_B| = 8 + sqrt(55)

Finally, the area of region S is given by:

Area(S) = Area(circle of radius 5 centered at C5) - Area(circle of radius 7 centered at C7)

= pi(5^2) - pi(7^2)

= 25pi - 49pi

= -24pi

Since the area of region S cannot be negative, the maximum possible area is zero. This means that there is no point that lies inside exactly one of the four circles. In other words, any point that lies inside one of the circles must also lie inside at least one of the other circles.

Step-by-step explanation:

The probability that X is greater than 3Y is 1/2, and the probability that X + Y is greater than 3 is 1/8.

To find P(X > 3Y), we need to integrate the joint probability density function over the region where X is greater than 3Y. We set up the integral as follows:

P(X > 3Y) = ∫∫[2(x^2)y/81] dy dx

The integration limits are determined by the condition X > 3Y. From 0 ≤ x ≤ 3, we have 0 ≤ 3Y ≤ x, which gives us 0 ≤ Y ≤ x/3. So, the integral becomes:

P(X > 3Y) = ∫[0 to 3] ∫[0 to x/3] [2(x^2)y/81] dy dx

Simplifying the integral, we get:

P(X > 3Y) = ∫[0 to 3] [(x^2)/27] dx

Evaluating the integral, we find P(X > 3Y) = 1/2.

To find P(X + Y > 3), we integrate the joint probability density function over the region where X + Y is greater than 3. We set up the integral as follows:

P(X + Y > 3) = ∫∫[2(x^2)y/81] dx dy

The integration limits are determined by the condition X + Y > 3. From 0 ≤ y ≤ 3, we have 3 - y ≤ X ≤ 3. So, the integral becomes:

P(X + Y > 3) = ∫[0 to 3] ∫[3-y to 3] [2(x^2)y/81] dx dy

Simplifying the integral, we get:

P(X + Y > 3) = ∫[0 to 3] [2y/27] dy

Evaluating the integral, we find P(X + Y > 3) = 1/8.

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If Homer can travel the same distance on the moving walkway as he can on the floor in the same amount of time, then his walking speed and the speed of the moving walkway must be the same. Therefore, Homer's walking speed is 4 mph.

Let's say that the distance Homer needs to travel to get to his gate is D. If he walks on the floor next to the moving walkway, he will travel this distance at his walking speed, which we'll call W. If he walks on the moving walkway, he will travel this distance at the speed of the walkway plus his walking speed, or W + 4 mph.

Since the time it takes for Homer to travel the distance D is the same in both cases, we can set the two equations equal to each other:

D/W = D/(W + 4)

Cross-multiplying gives us:

D(W + 4) = DW

Distributing the D gives us:

DW + 4D = DW

Subtracting DW from both sides gives us:

4D = 0

Dividing both sides by 4 gives us:

D = 0

Since the distance D cannot be zero, this means that the only way for the two equations to be equal is if W and the speed of the moving walkway are the same. Therefore, Homer's walking speed is 4 mph.

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Answer:

9r

Step-by-step explanation:

A total of 55.86 minutes have gone by since Greg started his run on the treadmill.

If Greg has completed 57% of his run, it means he has 43% of his run remaining.

To find out how many minutes have gone by, we can use proportions.

Let's say x is the total number of minutes Greg needs to complete his run:

x = 57% * 98 minutes

Evaluate

x =  minutes

Therefore, approximately 55.86 minutes have gone by since Greg started his run on the treadmill.

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Answer:

The value of 'x' = 3

Step-by-step explanation:

Explanation:-

Given that

              4x - 2(x - 5) = - 7 + 5x + 8

              4x - 2(x) +10 = - 7 + 5x + 8

                   2 x + 10 = 1 + 5 x

Subtracting '2x' on both sides , we get

             2 x - 2 x + 10 = 1 + 5 x - 2 x

                  10 = 1 + 3 x

                 10 -1 = 3x

                   9 = 3 x

Dividing '3' on both sides, we get

                 \(\frac{9}{3} = \frac{3x}{3}\)

               3 = x

∴ The value of 'x' = 3

The probability that the red die shows a three and the blue die shows a number greater than three is 1/12

From the question, we have the following parameters that can be used in our computation:

Red die

Blue die

The sample space of a die is

{1, 2, 3, 4, 5, 6}

using the above as a guide, we have the following:

P(3) = 1/6

P(Greater than 3) = 1/2

So, we have

P = 1/2 * 1/6

Evaluate

P = 1/12

Hence, the probability of a three and and a number greater than three is 1/12

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Answer:

38.28 .........................

Answer:

57.13..................................................

Answer:

In both cases, we have similar figures.

This means that the shape of the figures is the same, but the size is different:

PQST is similar to STNR and to NRPQ

VUYZ is similar to YZWX and to VUWX

this means that, for example, in problem 18, the ratio between ST and NR must be the same as the ratio between PQ and ST. This happens because the measure increases by the same scale factor.

With this in mind, we can solve the problem:

18)

ST = 7.5

NR = 5.5

Then the quotient ST/NR is:

ST/NR = 7.5/5.5

And, as we said above:

PQ/ST = ST/NR

PQ/7.5 = 7.5/5.5

PQ = (7.5/5.5)*7.5 = 10.23

19) Here we should have:

YZ/VU = WX/YZ

Then:

22.9/35 = WX/22.9

(22.9/35)*22.9 = WX = 14.98

The rate at which water is being pumped into the tank is 11,760 cm3/min.

          V = (1/3)πr2h

          Using similar triangles:

          (r / 2) = (2 / 6)

          r = 2/3 * 4r = 8/3 cm

          dV / dt = (πr2 / 3)dh / dt

         dV / dt = pump rate - leak rate

      = pump rate - 10,500 cm3/mindh / dt

      = 20 cm/minr = 8/3 cm

        pump rate - 10,500 = (π(8/3)2 / 3) * 20pump rate - 10,500

     = 2114.67pump rate

     = 11,760 cm3/min

Therefore, the rate at which water is being pumped into the tank is 11,760 cm3/min.

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The probability that a child chosen at random from the class has both a cat and a dog is 1/10.

Suppose b is the total number of kids that have both:

• children having a cat Only must be 14 - b

• children having a dog Only must be 19 - b

We also get:

We also know there are 30 kids, therefore

⇒ (14 - b) + b + (19 - b) = 30

⇒ 33 - b = 30

⇒ b = 3

Consequently, we now understand:

3/30 = 1/10

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Complete question is;

In the triangles attached , DF is congruent to MN,DG is congruent to MP, angle D is congruent to angle P. Can you prove that triangle DFG is congruent to MNP.

Answer:

Proved below

Step-by-step explanation:

From the attached triangles, we can see that;

∠D corresponds to ∠M

∠F corresponds to ∠N

∠G corresponds to ∠P

But we are told that ∠D is congruent to ∠P. Thus, since we have 2 other congruent sides in the triangles, we can conclude that Side-Angle-Side Postulate (SAS) congruency theorem that triangle DFG is congruent to MNP.

Answer: This really depends on how long you have to reach this goal.

Let’s say you want to make this amount in one month.

There is about 4 weeks in a month so you’ll need to divide 2112.24 by 4.

That would be 528.06 per week.

To find an amount per day you’ll need to divide 528.06 by 7.

That would be about 75.5 per day.

Please let me know if this doesn’t help.

We prefer t procedures to z procedures for inference about a population mean because t procedures are more appropriate when the sample size is small or when the population standard deviation is unknown.

T procedures take into account the additional uncertainty introduced by estimating the population standard deviation from the sample. Z procedures, on the other hand, assume that the population standard deviation is known, which is often not the case in practice. Therefore, t procedures provide more accurate and reliable estimates of the population mean when the underlying assumptions are met.

In summary, t procedures are preferred when dealing with small sample sizes or unknown population standard deviations, while z procedures are suitable for large sample sizes with known population standard deviations.

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The limit lim(x→∞) x/√(x^2 + 3) is 1, and there is no need to apply L'Hospital's Rule in this case.

When trying to use L'Hospital's Rule to find the limit lim(x→∞) x/√(x^2 + 3), it is important to note that L'Hospital's Rule can only be applied if the function is continuous and differentiable. In this case, the function is continuous and differentiable, but applying L'Hospital's Rule is not necessary as the limit can be evaluated using another method.
First, let's rewrite the given function by dividing both the numerator and the denominator by x:
lim(x→∞) (x/x) / (√(x^2 + 3)/x) = lim(x→∞) 1 / √(1 + 3/x^2)
As x approaches infinity, the term 3/x^2 approaches 0, so the limit becomes:
lim(x→∞) 1 / √(1 + 0) = 1 / √(1) = 1
Therefore, the limit lim(x→∞) x/√(x^2 + 3) is 1, and there is no need to apply L'Hospital's Rule in this case.

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Answer:

2.06

Step-by-step explanation:

The original decimal is 2.0625.

Since the thousandths place is 2, we can't round up. So we can keep the 6 which is in the hundredths place giving us 2.06