A water sample shows 0. 066 grams of some trace element for every cubic centimeter of water. Shandra uses a container in the shape of a right cylinder with a radius of 5. 7 cm and a height of 13. 9 cm to collect a second sample, filling the container all the way. Assuming the sample contains the same proportion of the trace element, approximately how much trace element has Shandra collected? Round your answer to the nearest tenth

The OLS regression results for the equation including gift, mails/year, giftlast and propresp are reported as including a sample size and r-squared value. The r-squared value is higher than that from the simple regression that omits giftlast and propresp.

The estimation of the equation gift 5 b0 1 b1mailsyear 1 b2giftlast 1 b3propresp 1 u by ols is done to assess the effect of various factors on the gift amount. The OLS regression results for the equation include the sample size and r-squared value. The sample size indicates the number of observations used in the estimation and the r-squared value indicates the goodness-of-fit of the model. A higher r-squared value indicates a better fit. In this case, the r-squared value of the equation with all four factors is higher than that of the simple regression model that omits giftlast and propresp. This implies that the addition of the two additional factors has improved the prediction accuracy of the model. The results of the regression should be interpreted with caution, as the addition of more variables may lead to overfitting of the data.

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Answer:

The conclusion is false.

Step-by-step explanation:

[tex]\frac{57}{3} = 19\\\\\frac{9}{3} = 3[/tex]

Although 3 and 9 do share some multiples, they must be divisible by 3. For example. 36/3 = 12, and 12 is divisible by 3. However, 57/3 = 19, which is not divisible by 3.

Answer: El mínimo común múltiplo de 18, 30 y 42 es 630.

Step-by-step explanation:

j

Answer:

630. yes yes yyesss

Answer:

The highest number of trials would most likely lead to an experimental probability of randomly selecting a sugar cookie that is closest to the theoretical probability of randomly selecting a sugar cookie. Therefore, option C, which suggests conducting one experiment with 500 trials, would be the most appropriate method of testing.

With 500 trials, there would be a large enough sample size to reduce the impact of random variations and increase the accuracy of the results. This would allow us to obtain a more reliable estimate of the experimental probability, which would be closer to the theoretical probability of selecting a sugar cookie.

Options A and D involve a small number of trials, which may not provide sufficient data for accurate estimations. Option B involves a large number of experiments with a small number of trials, which may not capture the overall probability of selecting a sugar cookie across all experiments. Therefore, option C is the best choice for obtaining a reliable estimate of the experimental probability.

(Please could you kindly mark my answer as brainliest you could also follow me so that you could easily reach out to me for any other questions)

Answer: I belive the answer is 0lxoly+39.56

Step-by-step explanation:

SIKE

The rental cοst tο drive 80 km is $35.

We knοw that the rental cοst is directly prοpοrtiοnal tο the number οf kilοmeters driven. Let's call the rental cοst C and the number οf kilοmeters driven D. Then we can write:

C = kD

where k is the cοnstant οf prοpοrtiοnality. We need tο find the value οf k in οrder tο answer the questiοn.

We knοw that the rental cοst tο drive 200 km is $87.50, sο we can plug in these values and sοlve fοr k:

87.50 = k × 200

k = 87.50/200

k = 0.4375

Nοw that we knοw the value οf k, we can use the same equatiοn tο find the rental cοst tο drive 80 km:

C = kD

C = 0.4375 × 80

C = 35

Therefοre, the rental cοst tο drive 80 km is $35.

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Therefore , the solution of the given problem of percentage comes out to be F. 0.159.

A% is a figure or integer that in statistics is shown as a percentage of 100. Additionally, the words "pct.," "pct.," and "pc" are rarely used. But even so, the "%" indicator is commonly used to denote it. There are no dimensions; the percentage total is flat. Percentages are really just numbers because the numerator is always 100. To indicate that a number is a percentage, either the percent mark (%) or the phrase "percent" must appear before the number.

Here,

Divide by 100 to transform a percentage to a decimal. Therefore, we must locate a decimal between 0.11 and 0.17 in order to determine the figure that corresponds to a percentage between 11% and 17%.

The choices are as follows:

=> F. 0.159

=> G. 0.013

=> H. 11/1,000

=> J. 6/100

The decimal 0.159 falls within this range (0.11 to 0.17), so the solution is F. 0.159.

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The required probability of the temperature of a person which has normal distribution is given by 0.9500.

Temperature of a person follows a normal distribution.

Use the Empirical Rule to estimate the probability.

Temperature of a randomly selected person will be within 2.42 standard deviations of its mean.

Empirical Rule states that for a normal distribution is,

About 68% of the data falls within one standard deviation of the mean.

About 95% of the data falls within two standard deviations of the mean.

About 99.7% of the data falls within three standard deviations of the mean.

Probability of temperature of a randomly selected person within 2.42 standard deviations of its mean = 95%

95%

= 95/100

= 0.95

=0.9500 ( round this to four decimal places )

Therefore, the probability that the temperature of a randomly selected person is 0.9500.

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The length of PQ is 16 units. ST is the midsegment of PQR, UV is the midsegment of STR, and WX is the midsegment of UVR. It is reasonable to assume that the three midsegments have equal lengths since they are part of the same triangle.

In a triangle, the midsegment is a line segment that connects the midpoints of two sides of the triangle and is parallel to the third side. In the given triangle PQR, the length of PQ is 16 units. ST is the midsegment of PQR, UV is the midsegment of STR, and WX is the midsegment of UVR. Since all the midsegments are part of the same triangle, it is reasonable to assume that they have equal lengths. The triangle midsegment theorem, which asserts that the length of a triangle's midsegment is equal to half of the length of the third side of the triangle, may be used to demonstrate this.. In this case, the length of ST, UV, and WX would all be equal to 8 units, since the third side of each triangle is 16 units. This indicates that it is reasonable to assume that the midsegments of the triangle have equal lengths.

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The given points of the graph represent a nonlinear relationship.

A linear graph is a type of graph where the points plotted create a straight line when connected. The slope of a linear graph indicates how steep the line is, while the y-intercept indicates the point where the line crosses the y-axis.

If the given points (1,1), (2,2), (3,6), (4,24) form a linear or nonlinear relationship with and without plotting the graph, we can calculate the slope between each pair of points.

The slope of a line between two points [tex](x_{1} , y_{1})[/tex] and [tex](x_{2} , y_{2})[/tex] is given by the formula:

slope = [tex]\frac{y_{2}-y_{1} }{x_{2}-x_{1}}[/tex]

Let's calculate the slope between the first two points (1,1) and (2,2):

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

Now, let's calculate the slope between the second and third points (2,2) and (3,6):

slope = (6 - 2) / (3 - 2) = 4/1 = 4

Next, let's calculate the slope between the third and fourth points (3,6) and (4,24):

slope = (24 - 6) / (4 - 3) = 18/1 = 18

As we can see, the slopes between the pairs of points are not equal. In fact, they are increasing at a faster and faster rate. Therefore, the relationship between the x and y values is nonlinear.

To determine the specific type of nonlinear relationship, we can try to fit a curve to the data points. In this case, we can see that the points follow an exponential pattern, where the y-values increase rapidly as x increases. The equation that fits this pattern is y = [tex]2^x[/tex].

Therefore, the given points (1,1), (2,2), (3,6), (4,24) represent a nonlinear relationship that follows the exponential function y = [tex]2^x[/tex].

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There is a 9.1% chance that a chicken of this particular breed will lay 248 eggs in a year.

A normal distribution is a continuous probability distribution described by its mean (μ) and standard deviation (σ). The formula for a normal distribution is:

[tex]P(x) = (1/σ√2π) e^(-(x-μ)^2/2σ^2)[/tex]

In this case, the mean (μ) is 225 and the standard deviation (σ) is 10.

To calculate the probability of a specific number of eggs being laid, we plug the mean and standard deviation into the normal distribution formula. For example, to calculate the probability of a chicken laying 248 eggs:

P(248) = 0.091

This means that there is a 9.1% chance that a chicken of this breed will lay 248 eggs in a year.

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Bartlett Company is in a healthy financial position, with strong liquidity, consistent profitability, and diverse revenue streams.

The Strengths are defined as,

The company has a healthy current ratio, which indicates that it can meet its short-term obligations efficiently.

The company has been consistently generating profits over the years, and its net income has been increasing.

Bartlett Company generates revenue from multiple sources, which reduces the risk of dependence on a single source.

The Weaknesses are defined as

An increasing debt level can put pressure on the company's financial position in the long run.

Bartlett Company's return on equity (ROE) is relatively lower compared to the industry average, indicating that the company is not generating as much profit from its equity as its peers.

The company has a low dividend payout ratio, which means that it is not distributing much of its profits to shareholders in the form of dividends.

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Answer:

y = -2x² + 7x - 11

Step-by-step explanation:

To find the equation of a parabola that passes through the given points, we need to first determine the form of the equation.

A general form of a quadratic equation is y = ax^2 + bx + c, where a, b, and c are constants. To determine the values of a, b, and c, we substitute the coordinates of each point into the equation and solve the resulting system of equations.

Substituting the first point (-8, -309), we get:

-309 = a(-8)^2 + b(-8) + c

Substituting the second point (-5, -132), we get:

-132 = a(-5)^2 + b(-5) + c

Substituting the third point (1, 6), we get:

6 = a(1)^2 + b(1) + c

Substituting the fourth point (2, 1), we get:

1 = a(2)^2 + b(2) + c

Substituting the fifth point (6, -99), we get:

-99 = a(6)^2 + b(6) + c

Expanding the terms, we get a system of five equations:

64a - 8b + c = -309

25a - 5b + c = -132

a + b + c = 6

4a + 2b + c = 1

36a + 6b + c = -99

Solving the system of equations, we get:

a = -2

b = 7

c = -11

Therefore, the equation of the parabola that passes through the given points is:

y = -2x^2 + 7x - 11

So the answer is y = -2x² + 7x - 11. (^2 is just another version of ²)

Hope this helps!! I'm sorry if it's wrong. If you need more help, ask me! :]

Answer:

[tex]\large\boxed{\mathtt{7.\bar{4}}}[/tex]

Step-by-step explanation:

[tex]\textsf{We are asked to identify the rational number out of given options.}[/tex]

[tex]\textsf{First, let's review what a rational number is.}[/tex]

[tex]\textsf{A Rational Number is a number that terminates.}[/tex]

[tex]\textsf{A Rational Number is also any number that can be represented as a fraction.}[/tex]

[tex]\textsf{Let's identify if our given options are Rational Numbers.}[/tex]

[tex]\mathtt{\sqrt{60} = 7.74.... (Irrational)}[/tex]

[tex]\mathtt{7.\bar{4} = 7 \frac{4}{9} \ (Rational) \ \checkmark}[/tex]

[tex]\mathtt{\sqrt{29} = 5.38.... \ (Irrational)}[/tex]

[tex]\mathtt{\pi = 3.14159.... \ (Irrational)}[/tex]

[tex]\sqrt{60} , \sqrt{29} , \textsf{and} \ \pi \textsf{ are irrational due to the decimals never terminating.}[/tex]

[tex]\textsf{Hence, our final answer is}[/tex] [tex]\mathtt{7.\bar{4}.}[/tex]

(A) The basis in which the space work is done is around 2. The powers of 10's logarithmic values can be used to determine this. (b) They could have two fingers for each hand, giving infants a total of five fingers.

a. The space creatures are most likely employed in a base other than base 10. The values of log 10, log 100, and log 1000 do not match the expected values for base 10, which should be 1, 2, and 3, respectively. Instead, the values given indicate that the base is slightly less than ten, because log 10 is less than one, log 100 is less than two, and log 1000 is less than three.

To be more specific, using the calculator's approximation, we can use the change of base formula to find the value of log base b for each of these numbers:

log base b of 10 = log 10 / log b ≈ 0.926628408

log base b of 100 = log 100 / log b ≈ 1.853256816

log base b of 1000 = log 1000 / log b ≈ 2.779885224

By dividing the known values of log 10, log 100, and log 1000, we can find log b:

0.473606797 log b = log 1000 / log 100

As a result, the space creatures are most likely based on 9.526.

b. It is impossible to determine how many fingers the space creatures have based solely on the calculator information and logarithm values. The space creatures' base could be related to the number of fingers they have, but we don't have enough information to tell.

The inverses of exponential functions, or those that "undo," the exponential functions, are the logarithmic functions.

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Answer:To solve this problem, we need to standardize the values of 55 and 63 to z-scores and then use the standard normal distribution table to find the area between those two z-scores.

The z-score for a value of 55 is:

z = (55 - 62) / 4 = -1.75

The z-score for a value of 63 is:

z = (63 - 62) / 4 = 0.25

Using a standard normal distribution table, we can find that the area to the left of z = -1.75 is 0.0401 and the area to the left of z = 0.25 is 0.5987. Therefore, the area between z = -1.75 and z = 0.25 is:

0.5987 - 0.0401 = 0.5586

To convert this area back to a percentage, we multiply by 100:

0.5586 * 100 ≈ 55.9%

Therefore, the answer is A. 55.9%.

Step-by-step explanation:

Therefore, the largest balance we should carry on this card to maintain an acceptable ratio is $2,117.75, which is option D.

By typically dividing two figures, ratios contrast them. If you were comparing one data point (A) to another data point, your formula would be A/B. (B). You are multiplying information A by information B, as indicated by this.

To maintain an acceptable credit utilization ratio, it is generally recommended to keep the balance on a credit card below 30% of the credit limit. In this case, the credit limit is $4,930, so we should aim to keep the balance below 30% of $4,930, which is:

0.3 * $4,930 = $1,479

Therefore, the largest balance we should carry on this card is $1,479.

None of the answer choices match this amount exactly, so we can calculate the ratio of each option to the credit limit to see which one is closest to 30%.

A. $2,807.25 / $4,930 ≈ 0.570 --> Not acceptable

B. $2,462.50 / $4,930 ≈ 0.500 --> Not acceptable

C. $2,119.90 / $4,930 ≈ 0.430 --> Not acceptable

D. $2,117.75 / $4,930 ≈ 0.429 --> Acceptable

Therefore, the largest balance we should carry on this card to maintain an acceptable ratio is $2,117.75, which is option D.

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The line is 13 units to the right (or shifted by -13 units).

A mathematical statement known as an equation consists of two algebraic expressions separated by equal signs (=) on either side.

It demonstrates the equality of the relationship between the printed statements on the left and right.

Left side equals right side in all formulas.

To find the values of unknowable variables, which stand in for unknowable quantities, you can solve equations.

A statement is not an equation if it lacks the equals sign.

When two expressions have the same value, a mathematical statement known as an equation will include the symbol "equal to" between them.

According to our question-

The ratio of y to x, which indicates how many units change in y when a specific number of units change in x, is used to express a line's slope.

A has a slope of 2 (x factor) = 2/1.

and as a result, y moves 2 units while x moves 1 unit.

B has a slope of 1 = 1/1.

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Determine the steady-state current in the RLC circuit to be 0.68 A.

The equation for Lq′′ + Rq′ + 1/C q = E is used to find the steady-state current in each RLC circuit. The circuit parameters are L = 0.1 H, C = 0.01 F, R = 3 Ω, and E(t) = 5cos10t - 5sin10t.Lq′′ + Rq′ + 1/C q = E can be written as Ld²q/dt² + Rdq/dt + 1/C q = E(t)The current flowing through the circuit can be determined using this formula. To calculate the current, we can use E(t) = 5cos10t - 5sin10t.The current can be found using the equation q(t) = Q cos(wt - φ), where Q = I/Z, Z = √(R² + (wL - 1/wC)²), w = 2πf, and φ = tan⁻¹((wL - 1/wC)/R). Z, w, and φ can be calculated from the parameters of the circuit.Using the above formulas, the following is obtained.I = 5/√(3² + (10π x 0.1 - 1/(10π x 0.01))²) = 0.68 ADetermine the steady-state current in the RLC circuit to be 0.68 A.

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The variance of the discrete random variable X is equal to 70t.

The variance of discrete random variable X is given by Var(X) = E[X2] – (E[X])2. In this case, the probability function is given as P(X = k) = t (6 – k) for k = 1, 2, 3, 4, 5. Therefore, the expected value of X is the sum of the products of each possible value of X and its corresponding probability. This can be calculated as:

E[X] = 1 x P(X = 1) + 2 x P(X = 2) + 3 x P(X = 3) + 4 x P(X = 4) + 5 x P(X = 5)

E[X] = t (6 - 1) + 2t (6 - 2) + 3t (6 - 3) + 4t (6 - 4) + 5t (6 - 5)

E[X] = t (5) + 2t (4) + 3t (3) + 4t (2) + 5t (1)

E[X] = 5t + 8t + 9t + 8t + 5t

E[X] = 35t

The expected value of X2 is the sum of the products of each possible value of X and its corresponding probability, multiplied by X2. This can be calculated as:

E[X2] = 1 x P(X = 1) x (1)2 + 2 x P(X = 2) x (2)2 + 3 x P(X = 3) x (3)2 + 4 x P(X = 4) x (4)2 + 5 x P(X = 5) x (5)2

E[X2] = t (6 - 1) x (1)2 + 2t (6 - 2) x (2)2 + 3t (6 - 3) x (3)2 + 4t (6 - 4) x (4)2 + 5t (6 - 5) x (5)2

E[X2] = t (5) x (1)2 + 2t (4) x (2)2 + 3t (3) x (3)2 + 4t (2) x (4)2 + 5t (1) x (5)2

E[X2] = 5t + 16t + 27t + 32t + 25t

E[X2] = 105t

Therefore, the variance of X is given by Var(X) = E[X2] – (E[X])2 = 105t – (35t)2 = 70t.

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Answer:

[tex]30x +68[/tex]

Step-by-step explanation:

[tex]6(9+5x)+14[/tex]

[tex]\text{Distribute the 6}[/tex]

[tex]30x +68[/tex]

The proportion of green apples can be estimated in the same way: expressions  Green apple proportion = (number of green apples) / (Total number of fruits)

In mathematics, you can multiply, divide, add, or subtract. An expression is formed as follows: Expression, number, and math operator Numbers, parameters, and functions make up a mathematical expression. It is possible to contrast phrases and expressions. Every mathematical statement that contains variables, numbers, and a mathematical action between them is referred to as an expression. For example, the phrase 4m + 5 is made up of the phrases 4m and 5, as well as the variable m from the provided equation, all separated by the mathematical sign +.

a) Because the basket contains four different types of fruits, the proportion of cherries can be computed as follows:

Cherry proportion = (number of cherries) / (Total number of fruits)

We can't compute the proportion of cherries without knowing how many fruits are in the basket.

b) The proportion of green apples can be estimated in the same way:

Green apple proportion = (number of green apples) / (Total number of fruits)

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1. The parametric equations for an object moving along the unit circle are x = cos(t) and y = sin(t), where t is the parameter. This is because a point on the unit circle is defined by the equation x2 + y2 = 1.

2. The following is an example of a table of values for parametric equations x = 2cost and y = 2sint, with the parameter varying over the domain 0 ≤ t ≤ 2π:

t x y

0 2 0

π/2 0 2

π -2 0

3π/2 0 -2

2π 2 0

3. The following is an example of a table of values for parametric equations x = 4cost and y = 2sint, with the parameter varying over the domain 0 ≤ t ≤ 2π:

t x y

0 4 0

π/2 0 2

π -4 0

3π/2 0 -2

2π 4 0

4. From the results of the above two examples, it can be inferred that if x = a cos t and y = b sin t, then when a = b, the graph will trace a circle with a radius of a/2. When a ≠ b, the graph will trace an ellipse with major and minor axis lengths of a and b, respectively.

A parametric equation is a mathematical equation that uses one or more parameters, or variables, to express an equation. It is used to describe a curve or a surface in two or three dimensions. The parameters are usually denoted by variables, such as t for time, x for the position, or y for the height. Parametric equations allow for the modeling of functions that cannot be expressed in one variable and can be used to solve a variety of problems in mathematics and other sciences.

The parametric equations for an object moving along the unit circle are x = cos(t) and y = sin(t), where t is the parameter. This is because a point on the unit circle is defined by the equation x2 + y2 = 1.It can be inferred that if x = a cos t and y = b sin t, then when a = b, the graph will trace a circle with a radius of a/2. When a ≠ b, the graph will trace an ellipse with major and minor axis lengths of a and b, respectively.

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a. The number of students that failed in all subject is 4

b. The percentage of students who failed one of economic and mathematics is 5.2%

c. The probability that a student at random failed is 1/10

A set is defined as the collect of item with similarities.

n( of student that passed math only) = 6

n( of students that passed account only) = 9

n( of students that passed economics only) = 16-(4+3+2)

= 16-9 = 7

n ( of student that passed math and account only) = 5

n( of student that passed account and economics only) = 2

n ( of student that passed math and economics only) = 18-(5+3+y) = 6

= 18-8-y = 6

= 10-y = 6

y = 4

n( of students that passed all the three subjects) =19-( 5+2+x) = 9

= 19-7-x = 9

= 12 -x = 9

-x = -3

x = 3

a. Therefore no of student that failed all subject =

6+9+7+5+2+4+3+z = 40

36+z = 40

z = 40-36 = 4

b. Student that failed at least math and economics = 4+9 = 13

= 13/40 × 100 = 5.2%

c. probability that a student at random failed = 4/40 = 1/10

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The line passes through the two points (-2,-9) and (3,-16).

We can write an equation in point-slope form using the following formula for the line passing through the point (-2,-9) and having a slope of m = -7/5:

y - y1 = m(x - x1)

where m is the slope of the line and (x1,y1) is the line's point.

Inputting the values provided yields:

y - (-9) = (-7/5)(x - (-2))

Simplifying:

y + 9 = (-7/5)(x + 2)

To get rid of the fraction, multiply both sides by 5. This gives us:

5y + 45 = -7(x + 2)

As we enlarge and rearrange, we obtain:

7x + 5y + 59 = 0

This is the line's standard form equation.

We can locate two points on the line and plot them to create a graph of the line. We can utilise the y-intercept, which is obtained by solving for y while setting x=0 in the equation:

7(0) + 5y + 59 = 0

5y = -59

y = -59/5

The y-intercept is therefore (0,-59/5). Using the slope, we may also locate a different point on the line:

m = -7/5

Beginning at (-2,-9), we can travel 7 units left and 5 units right to reach a different point:

(-2+5, -9-7) = (3, -16) (3, -16)

Hence, the line traverses both of the following two points: (3,-16).

Here is the line's graph:

     |

     |

*    |

 \   |

  \  |

   \ |

    \|

-------*-----------------

     |

    3|

     |

    2|

     |

    1|

     |

     |

     |

     |

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The answer is:

[tex]\sf{y+9=-\dfrac{7}{5}(x+2)}[/tex]

Work/explanation:

We need to determine the line's equation.  I will begin by writing the equation in point slope form:

[tex]\mapsto\phantom{333}\sf{y-y_1=m(x-x_1)}[/tex]

where m = slope and (x₁,y₁) is the point

Plug in the data:

[tex]\sf{y-(-9)=-\dfrac{7}{5}(x-(-2)}[/tex]

Simplify

[tex]\sf{y+9=-\dfrac{7}{5}(x+2)}[/tex]

The approximate distance between points T and U is 3.16 units, rounded to the nearest hundredth.(option c)

The problem states that we need to find the approximate distance between points T and U, rounded to the nearest hundredth. To solve this problem, we can use the distance formula, which is derived from the Pythagorean theorem. The formula is given as:

distance = √[(x₂ - x₁)² + (y₂ - y₁)²]

Here, (x₁, y₁) and (x₂, y₂) represent the coordinates of the two points, and √ denotes the square root operation.

Now, let's apply this formula to find the distance between points T and U. Suppose that T has the coordinates (-2, 1), and U has the coordinates (-1, 4). Then, we can plug in these values into the formula and simplify:

distance = √[(-1 + 2)² + (4 - 1)²]

= √[(1)² + 3²]

= √(1 + 9)

= √10

≈ 3.16

Therefore, we can see that the correct answer is (c) 3.16 units.

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The volume of the given cube is 8,000 cubic units if the surface area is 2,400.

Let's first find the length of one side of the cube using the given surface area.

The surface area of a cube is given by the formula 6s², where s is the length of one side of the cube.

We are given that the surface area of the cube is 2,400. Therefore, we can set up the following equation:

6s² = 2,400

Dividing both sides by 6, we get:

s² = 400

let's see the square root of both sides, we get:

s = 20

So, the length of one side of the cube is 20.

Now, we can find the volume of the cube using the formula V = s³:

V = 20³ = 8,000

Then, the volume of the cube is 8,000 cubic units.

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A bag of peanuts could be divided among 8 children, 9 children, or 10 children with each getting the same number, and with 2 peanuts left over in each case. The smallest number of peanuts that could be in the bag is 26.

This is because 8 children could each get 3 peanuts and there would be 2 peanuts left over, 9 children could each get 2 peanuts and there would be 2 peanuts left over, and 10 children could each get 2 peanuts and there would be 2 peanuts left over.

Small number means a number that is insufficiently large to be statistically significant, as determined by the department. Counting numerals are included in whole numbers as well. Whole numbers begin at zero. As a result, the least whole number is 0. Even though zero has no value, it is employed as a placeholder.

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