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The probability that the end-of-month volume will remain greater than 18 million m³ is approximately 0.1922.

a)The mean water volume in the lake at the end of the month can be calculated using the formula given below:

Mean water volume = Starting water volume + Total rainfall + Total flow - Total evaporation - Water drawn from the lake

Given:

Starting water volume = 20 million m³

Total rainfall = random variable with mean = 1 million m³ and standard deviation = 0.5 million m³

Total flow = random variable with mean = 8 million m³ and standard deviation = 2 million m³

Total evaporation = 3 million m³

Water drawn from the lake = 10 million m³

Now, let's calculate the mean water volume at the end of the month.

Mean water volume = Starting water volume + Total rainfall + Total flow - Total evaporation - Water drawn from the lake= 20 + 1 + 8 - 3 - 10= 16 million m³

Therefore, the mean water volume at the end of the month is 16 million m³.

The standard deviation of the water volume in the lake at the end of the month can be calculated using the formula given below:

σ = √{σr² + σf² + σe²}

σr = standard deviation of rainfall = 0.5 million m³

σf = standard deviation of flow = 2 million m³

σe = standard deviation of evaporation = 1 million m³σ = √{σr² + σf² + σe²}σ = √{0.5² + 2² + 1²}= √{5.25}≈ 2.29 million m³

Therefore, the standard deviation of the water volume in the lake at the end of the month is approximately 2.29 million m³.b)Given that all the random variables in the problem are normally distributed, we can find the probability that the end-of-month volume will remain greater than 18 million m³ using the z-score formula.

z = (x - μ) / σ

Where,

z = z-scorex = 18 μ = 16σ = 2.29

Now, let's calculate the z-score.

z = (x - μ) / σ= (18 - 16) / 2.29= 0.87

Using the z-table, we can find that the probability of z being less than 0.87 is 0.8078.

Therefore, the probability of the end-of-month volume being greater than 18 million m³ is:

1 - 0.8078 = 0.1922 (rounded to 4 decimal places)

Hence, the probability that the end-of-month volume will remain greater than 18 million m³ is approximately 0.1922.

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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:

10.500

Step-by-step explanation:

step by step explenation

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:

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

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Answer:

I believe it would be 4.563 x 10^8

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.

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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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:

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:

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.

Answer:

Option 3 Or number 5

Step-by-step explanation:

Answer:

8

Step-by-step explanation:

Eight people are at least 169.5 cm tall because the frequency of people that height is five. The frequency of people taller than 169.5 (because it said 'at least') is three. 5 + 3 = 8

Answer: -3

Step-by-step explanation: DeltaMath

Answer:

3.14

Step-by-step explanation:

1x1xpi

1x3.14=3.14

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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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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Answer:

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

Answer: 4529

Step-by-step explanation:

Hope it helps Have a good Day

Just divide it

Answer:

14

Step-by-step explanation:

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Answer:

2.75 x (c) = 500

Step-by-step explanation:

i think this is it

The probability for more than 22% of the given data say their class sizes are larger than 30 is equal to 0.0864, or 8.64%.

To find the probability that more than 22% of the randomly selected teachers say their class sizes are larger than 30,

Use the binomial distribution.

Let us denote the probability of a teacher saying their class size is larger than 30 as p.

19% of the teachers responded with a class size larger than 30, we can estimate p as 0.19.

Now, calculate the probability using the binomial distribution.

find the probability of having more than 22% of the 760 teachers .

which is equivalent to more than 0.22 × 760 = 167 teachers saying their class sizes are larger than 30.

P(X > 167) = 1 - P(X ≤167)

Using the binomial distribution formula,

P(X ≤167) = [tex]\sum_{i=0}^{167}[/tex] [C(760, i) × [tex]p^i[/tex] × [tex](1-p)^{(760-i)[/tex]]

where C(n, r) represents the combination 'n choose r' the number of ways to choose r items from a set of n.

Using a statistical calculator, the probability P(X ≤ 167) is determined to be approximately 0.9136.

This implies,

The probability of having more than 22% of the randomly selected teachers say their class sizes are larger than 30 is,

P(X > 167)

= 1 - P(X ≤ 167)

≈ 1 - 0.9136

≈ 0.0864

Therefore, the probability for the given condition is approximately 0.0864, or 8.64%.

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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:

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 _

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:

Step-by-step explanation:

Answer:

The equation would be y = 75 + 20x

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

Answer:

The guy above me is completely wrong hahaha.

The correct answer should have a 63%.

It's probably D. The terminology is a little confusing.

The equation of the output looks like:

-7.611 + .186(x) = y

x --> Arm span

y --> Foot span

The linear relationship is formed with the armspan.

Answer: It’s D

Step-by-step explanation:

I took the test

Answer:

1. -34

Step-by-step explanation: