def:This is what we call the money you get back when you pay for something.

term:

Answer: Materials sink in water depending on how  heavy they are. It is about how heavy something is compared to the same amount (volume) of water. This ratio of an object’s mass to its volume is known as density. Density is what really determines whether something will sink or float.

Answer:

3.14

Step-by-step explanation:

1x1xpi

1x3.14=3.14

Answer:

The equation would be y = 75 + 20x

Answer:

28 in² = 28

Step-by-step explanation:

[tex]a = \frac{a + b}{2}h[/tex]

[tex]a = 6[/tex]

[tex]b = 8[/tex]

[tex]h = 4[/tex]

[tex]a = \frac{6 + 8}{2}4[/tex]

[tex] \frac{14}{2} 4[/tex]

[tex]7 \times 4[/tex]

[tex] = 28[/tex]

Answer = 28 _

Answer:

H

Step-by-step explanation:

If they are similar that means that they are racially the same so just divide 60 by 84 to find the rate and then multiply the rate by 210 to get your answer

Answer:

14

Step-by-step explanation:

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Answer: 4529

Step-by-step explanation:

Hope it helps Have a good Day

Just divide it

Answer:

C

Step-by-step explanation:

When 3 is squared, it becomes 9. That means that both A and B are incorrect.

The powers on u and v are doubled not added to become u^6 and v^8 so that makes C the only possible answer.

The numerical probability of Flora picking out a blue sock is 1 out of 6, or approximately 0.1667, or 16.67%.

To calculate the numerical probability of Flora picking out a blue sock, we need to consider the total number of socks and the number of blue socks in the drawer.

Given:

Total number of socks = 12

Number of red socks = 9

Number of blue socks = 2

Number of green socks = 1

The probability of Flora picking a blue sock can be calculated as the ratio of the number of blue socks to the total number of socks:

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

Probability of picking a blue sock = 2 / 12

Simplifying the fraction, we get:

Probability of picking a blue sock = 1 / 6

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Answer: -3

Step-by-step explanation: DeltaMath

The relative error is found by approximating y(1) relative to the exact value 8e²-16.

The relative error of approximation for the given method can be calculated using the formula:

Relative error = |(approximated value - exact value) / exact value| * 100%

In this case, we have the function y(t) = 8e²-8 (t+1) and need to approximate the value of y(1) using three different methods. To calculate the relative error for each method, we substitute t = 1 into the function and compare the approximated value with the exact value.

In the first paragraph,

To calculate the relative error of approximation for the given methods in estimating y(t), we substitute t = 1 into the function y(t) = 8e²-8 (t+1). By comparing the approximated values with the exact value, we can determine the relative error for each method.

In the second paragraph:

Let's evaluate the function at t = 1:

y(1) = 8e²-8 (1+1) = 8e²-8(2) = 8e²-16

Now, let's consider the three methods for approximating y(1) and calculate their respective relative errors:

Method 1: Approximated value = 8e²-8 (1+1) = 8e²-16

Relative error = |(8e²-16 - 8e²-16) / 8e²-16| * 100% = 0%

Method 2: Approximated value = 8e²-8 (1+1) - 8 = 8e²-24

Relative error = |(8e²-24 - 8e²-16) / 8e²-16| * 100% = |8e²-24 - 8e²-16| / |8e²-16| * 100%

Method 3: Approximated value = 8e²-8 (1+1) + 8 = 8e²-8

Relative error = |(8e²-8 - 8e²-16) / 8e²-16| * 100% = |8e²-8 - 8e²-16| / |8e²-16| * 100%

By calculating the relative error using the above formulas, we can determine the accuracy of each method in approximating y(1) relative to the exact value 8e²-16.

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Answer:

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[tex]10 \times 10 \\ = 10^{2} \\ = 100[/tex]

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Answer:

I can't read the weight of the package due to the image quality, could you type it out please.

There are 91 ways to distribute the 12 identical computers among the elementary, middle, high school, and community college, ensuring that all 12 computers are given out.

To find the number of ways to distribute 12 identical computers among four schools (elementary, middle, high school, and community college) while ensuring that all 12 computers are given out, we can use the concept of stars and bars or the balls and urns method.

Let's represent the distribution using stars and bars. We have 12 identical stars (representing the computers) and 3 identical bars (representing the separators between the four schools). The bars divide the stars into four groups, each representing the number of computers given to each school.

We need to determine the number of ways to arrange the 12 stars and 3 bars. This can be calculated using the formula:

Number of ways = (n + k - 1) choose (k - 1) where n is the number of stars (12) and k is the number of bars (3).

Using this formula, the number of ways to distribute the 12 computers among the four schools is:

Number of ways = (12 + 3 - 1) choose (3 - 1)

            = 14Choosee 2

            = 91

Therefore, there are 91 ways to distribute the 12 identical computers among the elementary, middle, high school, and community colleges, ensuring that all 12 computers are given out.

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Answer:

1. -34

Step-by-step explanation:

The 99% confidence interval for the difference in approximately (-0.409123, -0.095277).

To obtain the confidence interval for the difference in the proportions, we use the formula:

Confidence Interval = (p₁ - p₂) ± Z × √((p₁ × (1 - p₁) / n₁) + (p₂ × (1 - p₂) / n₂))

Where:

Given the parameters:

p₁ = 73 / 201 = 0.3632

p₂ = 56 / 91 = 0.6154

n₁ = 201

n₂ = 91

Z = 2.576

Plugging in the values:

Confidence Interval = (0.3632 - 0.6154) ± 2.576 × √((0.3632 × (1 - 0.3632) / 201) + (0.6154 × (1 - 0.6154) / 91))

Confidence Interval = -0.2522 ± 2.576 × √((0.3632 × 0.6368 / 201) + (0.6154 × 0.3846 / 91))

Confidence Interval = -0.2522 ± 2.576 × √(0.003712)

Confidence Interval = -0.2522 ± 2.576 × 0.060851

Confidence Interval = -0.2522 ± 0.156923

Confidence Interval = (-0.409123, -0.095277)

Therefore, the 99% confidence interval is approximately (-0.409123, -0.095277).

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Answer:

Step-by-step explanation:

Answer:

2.75 x (c) = 500

Step-by-step explanation:

i think this is it

Answer:

The average rate of change for the depth of the river measured as feet per hour is approximately 0.3 feet/hour

Step-by-step explanation:

The depth of the river in feet with time is given by the function with the attached

From the graph, we have;

The depth of the river at hour t = 9 is f(9) = 18 feet

The depth of the river at hour t = 18 is f(18) = 21 feet

The average rate of change, A(x), for the depth of the river measured as feet per hour is given as follows;

[tex]A(X) = \dfrac{f(b) - f(a)}{b - a}[/tex]

Therefore, for the river, we have;

[tex]A(X) = \dfrac{f(18) - f(9)}{18 - 9} = \dfrac{21 - 18}{18 -9} = \dfrac{3}{9} =\dfrac{1}{3}[/tex]

The average rate of change for the depth of the river measured as feet per hour A(X) = 1/3 feet/hour

By rounding the answer to the nearest tenth, we have;

A(X) = 0.3 feet/hour.

Answer:

A,C and D are the correct options.

Answer:

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Step-by-step explanation:

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Answer:

use pythagoras theorem for solving this

value of x is 12 cm

Answer:

the area is 42 square units

Step-by-step explanation:

lets break it down

12 x 1 = 12

12 x 5 = 60

60 / 2 = 30

12 + 30 = 42

Answer:

10.500

Step-by-step explanation:

step by step explenation

Answer:

4

Step-by-step explanation:

From the above question:

1/4 cup of sand = 1 sand art

1 cup of sand = x

Cross Multiply

x × 1/4 cup of sand = 1 cup of sand × 1 sand art

x = 1 cup of sand × 1 sand art/1/4 cup of sand

x = 1 ÷ 1/4

x = 1 × 4

x = 4 pieces sand art

Therefore, Duncan can make 4 pieces of sand art.

In an experiment, there are many variables including independent and dependent variables. Therefore, the correct option is a.

Many including independent and dependent variables.

Variables are any feature, amount, or state that can be quantified or measured in any way. In a study, the term variable refers to any feature that can be changed or manipulated.

Variables in an Experiment. In an experiment, an independent variable is a variable that is changed or manipulated by the experimenter, and a dependent variable is a variable that is measured in response to the independent variable.

An independent variable is a variable that is changed or manipulated in an experiment by the researcher. The independent variable is the one that the researcher controls in order to examine its effect on the dependent variable.

The dependent variable is the variable that is measured in response to changes in the independent variable. This is the variable that the experimenter is interested in studying and is affected by the independent variable.

An experiment is a scientific study in which the researcher manipulates one or more independent variables in order to observe the effect on the dependent variable.

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The tri-linear inequality that estimates how much the drug will lower a typical patient's systolic blood pressure using a 90% confidence level is 34.5 < µ < 36.5.

Given that the sample mean of a blood-pressure drug is 35.5 for a sample size of 767 and standard deviation 15.2, to estimate how much the drug will lower a typical patient's systolic blood pressure, we use the following formula of a confidence interval:

Confidence interval = sample mean ± margin of error,

where the margin of error = z(α/2) * (σ/√n),

σ = 15.2, the standard deviation

n = 767, sample size

α = 0.10, level of significance

z(α/2) = 1.645 (from a standard normal distribution table)

Plugging in the values,

Margin of error = 1.645 * (15.2 / √767)≈ 1.02

Confidence interval = 35.5 ± 1.02≈ 34.5 < µ < 36.5

Therefore, the blood-pressure drug will lower a typical patient's systolic blood pressure within the range of 34.5 and 36.5.

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a. To solve the initial value problem using Laplace transforms, we start by taking the Laplace transform of both sides of the given differential equation. The Laplace transform of y(t) is denoted as Y(s). The Laplace transform of the second derivative y"(t) can be expressed as s²Y(s) - sy(0) - y'(0), where y(0) and y'(0) are the initial conditions. The Laplace transform of 48t is simply 48/s².

Applying the Laplace transform to the given differential equation, we get:

s²Y(s) - sy(0) - y'(0) + 16Y(s) = 48/s²

Substituting the initial conditions y(0) = 5 and y'(0) = 2, we have:

s²Y(s) - s(5) - 2 + 16Y(s) = 48/s²

Simplifying this equation gives the corresponding algebraic equation in terms of Y(s).

b. Now, we solve the equation obtained in part (a) for Y(s). Rearranging the terms, we have:

(s² + 16)Y(s) = 48/s² + s(5) + 2

Combining like terms, we get:

(s² + 16)Y(s) = (48 + 5s² + 2s) / s²

Dividing both sides by (s² + 16), we obtain:

Y(s) = (48 + 5s² + 2s) / (s²(s² + 16))

So, Y(s) is equal to the Laplace transform of y(t).

c. To find y(t), we take the inverse Laplace transform of Y(s) obtained in part (b). We can use partial fraction decomposition and the properties of Laplace transforms to simplify the expression and find the inverse Laplace transform.

Taking the inverse Laplace transform of Y(s), we find:

y(t) = L^(-1){Y(s)} = L^(-1){(48 + 5s² + 2s) / (s²(s² + 16))}

The inverse Laplace transform can be calculated using tables or software, and it yields the solution y(t) to the initial value problem.

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i had this but your numbers aren't the same i thought i could help you..