a 0.95 data eclipse contains ________ 95% of the data
points in the scatter plot
a. exactly
b. approximately
c. half of the
d. some of the

Answer:

4379

Step-by-step explanation:

Answer:

A ≈ 14.8 units²

Step-by-step explanation:

the area (A) of the triangle is calculated as

A = \(\frac{1}{2}\) yz sin Y ( that is 2 sides and the angle between them )

where x is the side opposite ∠ X and z the side opposite ∠ Z

here y = XZ = 4.3 and z = XY = 7 , then

A = \(\frac{1}{2}\) × 4.3 × 7 × sin79°

   = 15.05 × sin79°

   ≈ 14.8 units² ( to 1 decimal place )

Answer:

Step-by-step explanation:

0.0154 as a percentage is 1.54%

:)

Answer:

766?

Step-by-step explanation:

Answer and Step-by-step explanation:

Three Digit Number.

***

Second digit = Last digit

*xx

I multiplied 45 from numbers 1-23, with 22 being the maximum # to multiply to get a 3 digit number that is divisible by 45, and 3 being the minimum # to multiply to get a 3 digit number that is divisible by 45.

Now to check which number it is.

When you multiply 45 by 17, you get 765. This number is supposed to be 1 less than the riddle's number. It also has to have the last 2 digits be the same.

So, lets add one to 765 to see if we solved the riddle.

765 + 1 = 766

766 is the answer!

#teamtrees #PAW (Plant And Water)

I hope this helps!

Answer:

83505

Step-by-step explanation:

It's already in the expanded form...

Answer:

50000y^9.

Step-by-step explanation:

(2y)^4 * (5y)^5

= 2^4 y^4 *  5^5 y^5

= 16y^4 * 3125y^5

= 50000y^9.

The original activity of the sample can be calculated using the equation A=A0e^(-kt), where A0 is the initial activity, k is the decay constant, and t is the elapsed time.

1. Identify the given values: Half-life (t1/2) = 3.8 days, Activity level (A) = 0.17 bq/L, and Elapsed time (t) = 5.0 days.

2. Calculate the decay constant (k) using the equation k = ln(2)/t1/2.

k = ln(2)/3.8 days

k = 0.1815

3. Calculate the original activity (A0) using the equation A0 = A/e^(-kt).

A0 = 0.17 bq/L/e^(-0.1815)

A0 = 0.21 bq/L

Therefore, the original activity of the sample was 0.21 bq/L.

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The Taylor polynomials \(p_{4}\) and \(p_{5}\) centered at \(a = \frac{\pi}{6}\) for f(x) = 5cos(x) are:

\(p_{4}(x) = \frac{ 5\sqrt{3}}{2} - \frac{5}{2} (x - \frac{\pi }{6}) - \frac{ 5\sqrt{3}}{4} (x - \frac{\pi }{6})^2+ \frac{5}{8}(x - \frac{\pi }{6})^3+ \frac{ 5\sqrt{3}}{48}(x - \frac{\pi }{6})^4\)\(p5(x) = p_{4}(x) = \frac{ 5\sqrt{3}}{2} - \frac{5}{2} (x - \frac{\pi }{6}) - \frac{ 5\sqrt{3}}{4} (x - \frac{\pi }{6})^2+ \frac{5}{8}(x - \frac{\pi }{6})^3+ \frac{ 5\sqrt{3}}{48}(x - \frac{\pi }{6})^4 - \frac{5}{384}(x - \frac{\pi }{6} )^6\)

To find the Taylor polynomials centered at \(a = \frac{\pi}{6}\) for f(x) = 5cos(x), we need to find the derivative of the function at \(x = \frac{\pi}{6}\).  The first derivative of f(x) = 5cos(x) is -5sin(x), and the second derivative is -5cos(x).

Evaluating these derivatives at \(x = \frac{\pi}{6}\) gives us

\(-5sin(\frac{\pi }{6}) = -\frac{5}{2}\) and \(-5cos(\frac{\pi }{6}) = -\frac{5\sqrt{3} }{2}\).

The Taylor polynomial \(p_{4}(x)\) is then constructed using these derivatives and the powers of \(x - \frac{\pi}{6}\) up to the fourth power.

Similarly, for \(p_{5}(x)\), we add the fifth derivative term. Simplifying the expressions gives us the Taylor polynomials \(p_{5}(x)\) and \(p_{4}(x)\) center \(= \frac{\pi }{6}\) for f(x) = 5cos(x).

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The area, in square inches, of the larger triangle is 12.

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Using the Bayes' theorem, we find the probability that the transferred ball was white given that the second ball drawn was white to be 52/89, or approximately 0.5843.

To solve this problem, we can use Bayes' theorem, which relates the conditional probability of an event A given an event B to the conditional probability of event B given event A:

P(A|B) = P(B|A) * P(A) / P(B)

where P(A|B) is the probability of event A given that event B has occurred, P(B|A) is the probability of event B given that event A has occurred, P(A) is the prior probability of event A, and P(B) is the prior probability of event B.

In this problem, we want to find the probability that the transferred ball was white (event A) given that the second ball drawn was white (event B). We can calculate this probability as follows:

P(A|B) = P(B|A) * P(A) / P(B)

P(B|A) is the probability of drawing a white ball from urn b given that the transferred ball was white and is now in urn b. Since there are now six white balls and three black balls in urn b, the probability of drawing a white ball is 6/9 = 2/3.

P(A) is the prior probability of the transferred ball being white, which is the number of white balls in urn a divided by the total number of balls in urn a, or 6/13.

P(B) is the prior probability of drawing a white ball from urn b, which can be calculated using the law of total probability:

P(B) = P(B|A) * P(A) + P(B|not A) * P(not A)

where P(B|not A) is the probability of drawing a white ball from urn b given that the transferred ball was black and P(not A) is the probability that the transferred ball was black, which is 7/13.

To calculate P(B|not A), we need to first calculate the probability of the transferred ball being black and then the probability of drawing a white ball from urn b given that the transferred ball was black.

The probability of the transferred ball being black is 7/13. Once the transferred ball is moved to urn b, there are now five white balls and four black balls in urn b, so the probability of drawing a white ball from urn b given that the transferred ball was black is 5/9.

Therefore, we can calculate P(B) as follows:

P(B) = P(B|A) * P(A) + P(B|not A) * P(not A)

= (2/3) * (6/13) + (5/9) * (7/13)

= 89/117

Now we can plug in all the values into Bayes' theorem to find P(A|B):

P(A|B) = P(B|A) * P(A) / P(B)

= (2/3) * (6/13) / (89/117)

= 52/89

Therefore, the probability that the transferred ball was white given that the second ball drawn was white is 52/89, or approximately 0.5843.

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Answer:

slope is -3

y-intercept is -2

Step-by-step explanation:

put into slope-intercept form which is y = mx + b where 'm' is slope and 'b' is the y-intercept

12x + 4y = -8

4y = -12x - 8

y = -12/4x - 8/4

y = -3x - 2 (simplified)

Answer:

25,000x1.06 or      25,000/100=250x106=26,500

Step-by-step explanation:

answer = no it does not measure a rate

Answer:

NO

Step-by-step explanation:

Percent is just a number out of 100

The number of tornadoes is 77

total time for occurrence is 70 years

using these

= ( 77/70)

= (  1.1 )

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To find equivalent fractions and ratios, one must understand the concept of proportionality and be able to apply it to solve equations, identify common factors, and simplify fractions. Additionally, one should be able to multiply and divide fractions and ratios.

a. To find equivalent fractions without context, you need to use a form of division. Start by dividing the numerator (top number) of the fraction by the denominator (bottom number). The result of this division gives you the number of times the fraction is equivalent to itself. Multiply both the numerator and the denominator of the fraction by this number to get the equivalent fraction. For example, to find the equivalent of ½, divide 1 by 2, which gives you 0.5. Multiply 0.5 by both 1 and 2 to get the equivalent fraction 3/6.

b. To set up equivalent ratios, you need to first identify the ratio given (e.g. a per b). Divide the first number (a) by the second number (b) to get the ratio in fraction form. For example, if the ratio given is 3 per 4, divide 3 by 4 to get the equivalent ratio 3/4.

c. To find equivalent fractions and ratios, the method used depends on the situation. If the fraction or ratio is given in fraction form, use division to find the equivalent fraction or ratio. If the fraction or ratio is given in ratio form, use division to convert it to fraction form and then use division to find the equivalent fraction or ratio.

d. To confidently set up and find equivalent fractions and ratios, it is important to practice and understand the fundamentals of fractions and ratios. It is also important to understand the different forms in which fractions and ratios are presented and how to convert between them. Understanding how to use division to find equivalent fractions and ratios is also essential.

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Let Q be the point (a, b). Since the midpoint of P (2, 3) and Q (a, b) is (4, 6), we have

(2 + a)/2 = 4

(3 + b)/2 = 6

Solve for a and b :

(2 + a)/2 = 4

2 + a = 8

a = 6

(3 + b)/2 = 6

3 + b = 12

b = 9

So the coordinates of Q are (6, 9).

Answer:

V = 126 cm³

Step-by-step explanation:

the figure is made up of 2 cuboids ( flat one on the left and upright one on the right )

the volume (V) of a cuboid is calculated as

V = lbh ( l is length, b is breadth, h is height )

cuboid on left has l = 7 , b = 6 and h = 1 , then

V = 7 × 6 × 1 = 42 cm³

upright cuboid has l = 7 , b = 3 and h = 4 , then

volume = 7 × 3 × 4 = 84 cm³

Then volume of figure is the sum of the 2 volumes calculated

V = 42 + 84 = 126 cm³

Answer:

126 cm^3

Step-by-step explanation:

In order to find the volume of a cuboid the formula is:

l × w × h

In this case, we actually have two cuboids making up this one figure. Thus you can them apart into two cuboids.

I take one cuboid with the measurements of 9 by 7 by 3 and calculate the volume below:

\(9*7*3=63\)

Similarly, the other cuboid will have measurements of 3 by 3 by 7(the reason why h is 3 is that I chose to include the bottom half of the width of the cuboid). Now calculate your volume for the second cuboid:

\(3*3*7=63\)

Now add both volumes together and you  get 126 cm^3

Answer:

5/78 chance to get a Nickel and then a penny without replacing

Step-by-step explanation:

The bag contains a total of 40 coins which show that the probability of getting a Nickel is 5/40 and the probability of getting a penny after taking out a nickel is 20/39. In order to find the Probability of two events together you must multiply the probabilities so

(5/40) * (20/39) = 100/1560 = 5/78

Answer:

b) 8 x - 21

Step-by-step explanation:

Apply the distributive property.

7

x

+

7

⋅

−

1

−

2

(

x

+

1

)

+

3

(

x

−

4

)

Multiply

7

by

−

1

.

7

x

−

7

−

2

(

x

+

1

)

+

3

(

x

−

4

)

Apply the distributive property.

7

x

−

7

−

2

x

−

2

⋅

1

+

3

(

x

−

4

)

Multiply

−

2

by

1

.

7

x

−

7

−

2

x

−

2

+

3

(

x

−

4

)

Apply the distributive property.

7

x

−

7

−

2

x

−

2

+

3

x

+

3

⋅

−

4

Multiply

3

by

−

4

.

7

x

−

7

−

2

x

−

2

+

3

x

−

12

Subtract

2

x

from

7

x

.

5

x

−

7

−

2

+

3

x

−

12

Add

5

x

and

3

x

.

8

x

−

7

−

2

−

12

Subtract

2

from

−

7

.

8

x

−

9

−

12

Subtract

12

from

−

9

.

8

x

−

21=

8

x

−

21

Specify primary audio , secondary audio (sound effects/music), and background audio (ambient music) in the storyboard. Note scenes with surrounding noise, use an external microphone for recording, and plan additional music for editing.



In storyboarding for an ad firm, Eric needs to consider three types of audio: primary audio, secondary audio, and background audio. Primary audio refers to the main audio component that drives the narrative or delivers important information in the ad. This can include dialogue, voiceover, or any sound directly related to the main subject. Secondary audio supports the primary audio and adds depth to the storytelling. It can include sound effects, ambient noise, or music that complements the visuals. Background audio provides the overall atmosphere and mood of the ad. It can include ambient music or soundscapes that create the desired emotional tone.

Additionally, Eric should make notes of scenes where surrounding noise is required. This means capturing the environmental sounds that enhance the authenticity and realism of the ad, such as street noise, nature sounds, or crowd reactions. He should also consider using an external microphone to simultaneously record audio while filming the video. This ensures high-quality audio capture and reduces the reliance on the camera's built-in microphone, which may not provide optimal sound recording. Lastly, Eric can make a note of any additional music he intends to add during the editing process. This can involve selecting appropriate background tracks, jingles, or any other musical elements that enhance the overall impact of the ad. By carefully considering these audio aspects during storyboarding, Eric can create a compelling and immersive ad experience for the audience.

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Measures of central tendency, measures of variation, and crosstabulation are all types of descriptive statistics.

Descriptive statistics summarize and describe the main features of a data set, including the typical or central values (measures of central tendency) and the spread or variability of the data (measures of variation). Crosstabulation, also known as contingency tables, is a way to summarize the relationship between two variables by displaying their frequency distributions in a table format.
Measures of central tendency, measures of variation, and crosstabulation are types of descriptive statistics. Descriptive statistics are used to summarize and describe the main features of a dataset in a simple and meaningful way.

Central tendency refers to the measures that help identify the center or typical value of a dataset, such as mean, median, and mode. Variation measures describe the spread or dispersion of data, including range, variance, and standard deviation. Crosstabulation is a method of organizing data into a table format to show the relationship between two categorical variables.

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Answer:

- 1 / 7

Step-by-step explanation:

( x1, y1 ) = ( 10, 5 )

Here,

x1 = 10

y1 = 5

( x2, y2 ) = ( 3, 6 )

Here,

x2 = 3

y2 = 6

Formula : -

Slope = ( y2 - y1 ) / ( x2 - x1 )

Slope = ( 6 - 5 ) / ( 3 - 10 )

= 1 / - 7

Slope = - 1 / 7

Proper fraction = - 1 / 7

The inference procedure for each is give below-

A. The inference procedure: Confidence Interval - One mean, μ.

C. The inference procedure: Hypothesis test - Difference in proportions from two samples, P1-P2.

E. The inference procedure: Confidence Interval - One proportion, p.

G. The inference procedure: Confidence Interval - Difference in proportions from two separate samples, M1-M2

I. The Inference procedure: Hypothesis test - Difference in means from two independent samples.

K. The Inference procedure: Hypothesis test - Paired means, HD.

B. The Inference procedure: Hypothesis test - Difference in proportions from two samples, P1-P2.

D. The Inference procedure: Hypothesis test - One mean, μ.

F. Inference procedure: Confidence Interval - Paired means, HD.

H. Inference procedure: Hypothesis test - One proportion, p.

J. Inference procedure: Confidence Interval - Difference in means from two independent samples.

L. It doesn't directly involve inference.

Now we have-

A. What's the average number of free throws for the Warriors to attempt during a game?

Distribution type: z

C. Do the Warriors make more free throws (on average) during games at home than on the road?

Simulation type: BT (bootstrap)

Distribution type: z

E. What proportion of free throw attempts do the Warrior players make?

Simulation type: BT (bootstrap)

Distribution type: z

G. How much better (or worse) are the Warriors at making free throw attempts compared to their opponents?

Simulation type: BT (bootstrap)

Distribution type: z

I. Is the mean number of free throw attempts awarded to the Warriors during their games different from the mean number attempted by their opponents?

Distribution type: z

K. (Challenge) On average, is the point spread when GSWarriors win larger than the point spread when they lose?

Simulation type: BT (bootstrap)

Distribution type: z

B. Is the proportion of free throws made by the Warriors different between games they play at home and those they play on the road?

Simulation type: BT (bootstrap)

Distribution type: z

D. Over the past 10 years, NBA teams have averaged close to 25 free throw attempts per game. Treating this as the population mean, is the mean number of free throw attempts by the Warriors much different?

Simulation type: BT (bootstrap)

Distribution type: z

F. How many more (or fewer) free throw attempts do the Warriors take on average for home games compared to road games?

Simulation type: BT (bootstrap)

Distribution type: z

H. Players in the NBA as a whole make about 75.6% of their free throws. Is the proportion made by the Warriors different from this?

Simulation type: BT (bootstrap)

Distribution type: z

J. How does the average number of free throws made (per game) by the Warriors compare to their opponents?

Distribution type: z

L. (Extra) On average, what is the point spread in GS Warrior games? (where "+" means they won; "-"means they lost)

This case doesn't directly involve inference.  As it requires calculating the average point spread, but it doesn't involve making statistical inferences about a population parameter.

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1 )

\(volume = 20 \times 5 \times 8\)

\(volume = 800 \: \: {m}^{3} \)

\(surface \: \: area \: = 4 \times 20 \times 8\)

\(surface \: area \: \: = 640 \: \: {m}^{2} \)

_____________________________________________

2 )

\(v = 15 \times 3 \times 4\)

\(v = 180 \: \: {m}^{3} \)

\(s \: a \: = 4 \times 15 \times 4\)

\(s \: a \: = 240 \: \: {m}^{2} \)

_____________________________________________

3 )

\(v = ( \frac{1}{2} \times 4 \times 5) \times 10 \\ \)

\(v = 100 \: \: {m}^{3} \)

\(s \: \: a = (10 \times 5) + (10 \times 4) + (10 \times \sqrt{41} ) \\ \)

\(s \: a \: \: = 50 + 40 + 10 \sqrt{41} \)

\(s \: a \: \: = 90 + 10 \sqrt{41} \)

_____________________________________________

4 )

\(v = ( \frac{1}{2} \times 3 \times 4) \times 7 \\ \)

\(v = 42 \: \: \: {m}^{3} \)

\(s \: a \: \: = (3 \times 7) + (4 \times 7) + (5 \times 7)\)

\(s \: a \: \: = 21 + 28 + 35\)

\(s \: a \: \: = 84 \: \: \: {m}^{2} \)

The probability that a randomly selected datum falls within 2 standard deviations of the mean as required is; 0.95.

By observation, it follows that the distribution as represented in the task content above is a normal distribution.

On this note, the empirical rule applies as follows;

For a standard normal distribution, 68% of the observations (datum) fall within 1 standard deviation of the mean; 95% of the datum lie within two standard deviation of the mean; and 99.9% lie within 3 standard deviations of the mean.

On this note, when expressed as a percentage; the probability that the randomly selected datum falls with 2 standard deviations is; 95/100 = 0.95.

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Answer:

x = 180-35 = 145

y = 35

Hope it helps

To determine which proposal(s) to select, we need to compare the present worth or net present value (NPV) of each proposal. The NPV represents the difference between the present value of cash inflows and outflows for each proposal.

For independent proposals at MARR = 15.5% per year:

Calculate the NPV for each proposal using the cash inflows and outflows and discounting them to present value using the MARR of 15.5%.

Select the proposal(s) with a positive NPV. Positive NPV indicates that the project's expected cash inflows exceed the initial investment and the MARR.

For mutually exclusive proposals at MARR = 10% per year:

Calculate the NPV for each proposal using the cash inflows and outflows and discounting them to present value using the MARR of 10%.

Select the proposal with the highest positive NPV. The proposal with the highest positive NPV indicates the project that generates the highest expected return or value relative to the MARR.

For mutually exclusive proposals at MARR = 14% per year:

Calculate the NPV for each proposal using the cash inflows and outflows and discounting them to present value using the MARR of 14%.

Select the proposal with the highest positive NPV. The proposal with the highest positive NPV indicates the project that generates the highest expected return or value relative to the MARR.

It's important to note that the specific details of the proposals, including cash inflows, outflows, and timing, are needed to calculate the NPV accurately. Without this information, it is not possible to provide a definitive answer.

Given differential equation is y (4) - 10y" +25y" = 0 .The characteristic equation is r⁴ - 10r² + 25 = 0. The above quadratic equation can be factored as (r²-5)²=0.

The roots are r₁

=r₂

=√5 and r₃

=r₄

=-√5.

The solution will behave as t→[infinity] as the exponential function grows at a faster rate than the polynomial expression with respect to time. Hence the solution tends to infinity as t tends to infinity.

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The solution of the given initial value problem. y (4) - 10y" +25y" = 0; y(1) = 10 +e5, y'(1) = 8 +5e5, y"(1) = 25e5, y" (1) = 125e5. The answer to how the solution behaves as t approaches infinity is indeterminate.

The given initial value problem is y(4) - 10y" + 25y' = 0, with initial conditions y(1) = 10 + e^5, y'(1) = 8 + 5e^5, y"(1) = 25e^5, and y"'(1) = 125e^5.

To solve this problem, we can use the method of solving linear homogeneous differential equations with constant coefficients. We start by finding the characteristic equation, which is r^4 - 10r^2 + 25 = 0.

This equation can be factored as (r^2 - 5)^2 = 0. Therefore, the characteristic equation has a repeated root of r = ±√5.

The general solution of the differential equation is y(t) = (C1 + C2t)e^√5t + (C3 + C4t)te^√5t, where C1, C2, C3, and C4 are constants.

To find the specific solution, we can substitute the initial conditions into the general solution. Using y(1) = 10 + e^5, we find C1 + C2 + C3 + C4 = 10 + e^5.

Using y'(1) = 8 + 5e^5, we find C2 + √5C1 + C4 + √5C3 = 8 + 5e^5.

Using y"(1) = 25e^5, we find C2 + 5C1 + 4√5C3 + 4C4 = 25e^5.

Using y"'(1) = 125e^5, we find C4 + 15C3 + 20√5C1 + 20C2 = 125e^5.

Solving this system of equations will give us the specific solution for y(t).

As t approaches infinity, the behavior of the solution will depend on the values of the constants C1, C2, C3, and C4. Without knowing the specific values, we cannot determine how the solution will behave as t approaches infinity. Therefore, the answer to how the solution behaves as t approaches infinity is indeterminate.

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To determine if the packaging process is capable or if an adjustment is needed, we can calculate the process capability index (Cpk) using the given information.

The process capability index (Cpk) is calculated as the minimum of two ratios: Cp and Cpk.

Cp (Process Capability):

Cp = (Upper Specification Limit - Lower Specification Limit) / (6 * Standard Deviation)

In this case:

Upper Specification Limit = 264 + 13 grams = 277 grams

Lower Specification Limit = 264 - 13 grams = 251 grams

Standard Deviation = 1.03 grams

Cp = (277 - 251) / (6 * 1.03) ≈ 4.04

Cpk (Process Capability Index):

Cpk = min[(Upper Specification Limit - Process Mean) / (3 * Standard Deviation), (Process Mean - Lower Specification Limit) / (3 * Standard Deviation)]

Process Mean = 43.5 grams

Upper Specification Limit = 277 grams

Lower Specification Limit = 251 grams

Standard Deviation = 1.03 grams

Cpk = min[(277 - 43.5) / (3 * 1.03), (43.5 - 251) / (3 * 1.03)]

= min[233.5 / 3.09, -207.5 / 3.09]

= min[75.58, -67.22]

= -67.22 (since the negative value is smaller)

The process capability index (Cpk) is -67.22, which indicates that the process is not capable of meeting the specifications. An adjustment is needed to bring the process back within the desired range.

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Answer: It takes 8 newtons to accelerate a 25-pound wagon.