find a equation of the line through the point(-6,-5) and (-1,1))
a. y=6/5x + 11/5
b. y=6/5x - 11/6
c. y=6/5x
or
d. y=6/5x + 11/6

Step-by-step explanation:

The side opposite to the 4 degree angle is 18 m

The side adjacent the 4 degree angle is x

tan(4) = opposite / adjacent

adjacent = opposite / (tan(4))

adjacent = 18 / tan(4)

adjacent = 257.41

Forecast for August with an alpha of 0.7 and forecast of June 164 are to be calculated using exponential smoothing.

Exponential smoothing is a forecasting method that uses a weighted average of past time-series values to forecast future values. In a time-series, data values are obtained over time, such as over months or years, and then plotted in sequence.

The weights decrease exponentially as the observations get older. The EWMA formula used in exponential smoothing is as follows:

St=α × Yt-1+(1-α) × St-1

where,St is the smoothed statistic for the current period tYt is the original observation for the current period tSt-1 is the smoothed statistic for the previous period t-1α is the smoothing parameter 0≤α≤1, and usually close to 0.1.The smoothed statistic is also called the exponentially weighted moving average (EWMA).Calculating the forecast for August with alpha=0.7 and forecast of June 164 as follows:

St=α × Yt-1+(1-α) × St-1St =0.7 × 164+(1-0.7) × 164St=164

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The lower quartile of the given set of numbers 4, 6, 7, 8, 10, 12, 20 is 6. So, correct option is B.

To find the lower quartile of a set of numbers, we first need to arrange them in ascending order. The given numbers are: 4, 6, 7, 8, 10, 12, 20.

Next, we divide the data set into four equal parts. The lower quartile is the median of the lower half of the data set. In this case, the lower half is 4, 6, and 7.

To determine the median of the lower half, we need to find the middle value. Since we have an odd number of data points in the lower half, the middle value is the single value between the two extremes. In this case, the median is 6.

This corresponds to option (b) in the given choices: 6.

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To Prove :

[tex] \sf \dfrac{\cos9\degree + \sin9\degree }{\cos9\degree- \sin9\degree} = \tan54\degree \\ \\ [/tex]

[tex] \sf LHS = \dfrac{\cos9\degree + \sin9\degree }{\cos9\degree- \sin9\degree} \\ \\ [/tex]

[tex] \sf RHS = \tan54\degree \\ \\ [/tex]

Solving for LHS :

Divide numerator and denominator by cos9°

[tex]\longrightarrow \: \: \sf\dfrac { \dfrac{ \cos9\degree}{ \cos9\degree} + { \dfrac{\sin9}{ \cos9} }}{ \dfrac{ \cos9\degree }{ \cos9\degree} - { \dfrac{ \sin9\degree}{ \cos9\degree}}} \\ \\ [/tex]

[tex] \longrightarrow \: \: \sf\dfrac{1 + { \dfrac {\sin9\degree}{ \cos9\degree}} }{1 - { \dfrac{ \sin9\degree}{ \cos9\degree} }} \\ \\ [/tex]

[tex]\longrightarrow \: \: \sf \dfrac{1 + \tan9\degree}{1 - \tan9\degree} \\ \\ [/tex]

[tex]\longrightarrow \: \: \sf{ \dfrac{ \tan45 \degree + \tan9\degree}{1 - \tan45 \degree \tan9 \degree}} \: \: \: \: \: \: \: \sf \red{\bigg( \tan(A + B) = \dfrac{ \tan A + \tan B}{1 - \tan A \tan B} \bigg)} \\ \\ [/tex]

[tex]\longrightarrow \: \: \sf \tan(45\degree + 9\degree) \\ \\ [/tex]

[tex]\longrightarrow \: \: \sf \tan54 \degree[/tex]

LHS = RHS

Answer:

£320 000

Step-by-step explanation:

Let x = original price of the boat.

When you reduce a price by 15%, what percent remains in the sale price?

100% - 15% = 85%

After reducing the price of the boat by 15%, the sale price is 85% of the original price.

85% of x = £272 000

0.85x = 272 000

Divide both sides by 0.85

x = 320 000

Answer: £320 000

The two numbers that multiply to give 24 and add to give -11 are -3 and -8.

To find these numbers, you can use a system of equations. Let x and y be the two numbers. Then we have:

We can use the second equation to solve for one of the variables in terms of the other. For example, we can solve for x:

x + y = -11

x = -11 - y

We can then substitute this expression for x into the first equation:

xy = 24

(-11 - y)y = 24

Expanding and rearranging, we get:

y^2 + 11y + 24 = 0

This is a quadratic equation that we can solve using factoring or the quadratic formula. Factoring, we get:

(y + 3)(y + 8) = 0

So either y + 3 = 0 or y + 8 = 0. This means that y can be -3 or -8. Substituting each of these values into x = -11 - y, we get:

If y is -3, then x is -11 - (-3) = -8

If y is -8, then x is -11 - (-8) = -3.

Translated question "Numbers that multiply to give me 24 and add to give me -11"

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The next two terms in this arithmetic progression is 7.60 and 7.61.

An arithmetic prοgressiοn (AP) is a sequence where the differences between every twο cοnsecutive terms are the same. Fοr example, the sequence 2, 6, 10, 14, … is an arithmetic prοgressiοn (AP) because it fοllοws a pattern where each number is οbtained by adding 4 tο the previοus term. A real-life example οf an AP is the sequence fοrmed by the annual incοme οf an emplοyee whοse incοme increases by a fixed amοunt οf $5000 every year.

We know the Arithmetic progression formula:

[tex]\rm a_{n}=a_{1}+(n-1)d[/tex]

[tex]\rm a_n[/tex] = the nᵗʰ term in the sequence

[tex]\rm a_1[/tex] = the first term in the sequence

d = the common difference between terms

Here 4 terms are given

5th and 6th terms are to be found,

Thus,

a₁ = 9.56

d = (9.56 -9.57) = 0.01

n = 5

Then

5th term is :

aₙ = a₁ + (n - 1)d

aₙ = 9.56 + (5 - 1)0.01

aₙ = 9.56 + (4)0.01

aₙ = 9.56 + 0.04

aₙ = 7.60

6th term is :

aₙ = a₁ + (n - 1)d

aₙ = 9.56 + (6 - 1)0.01

aₙ = 9.56 + (6)0.01

aₙ = 9.56 + 0.05

aₙ = 7.61

Thus, The next two terms in this arithmetic progression is 7.60 and 7.61.

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The slope of the line parallel to the function: y = 50 - 0.5x include the following: C. -0.5.

In Mathematics and Geometry, parallel lines simply refers to two (2) lines that are always the same or equal distance apart and never meet.

Generally speaking, two (2) lines are considered as parallel under the following standard conditions:

Slope, m₁ = Slope, m₂.

From the function, we have:

y = 50 - 0.5x

Slope, m = -0.5

In this context, we can reasonably infer and logically deduce that the slope of the two lines would be equal and the same:

Slope of line 1 = Slope of line 2

-0.5 = -0.5

In conclusion, we can logically deduce that the slope of the given line is equal to -0.5.

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Answer: yes, yes, no

Explanation: In order for a figure to be a triangle, the two smallest sides must add up to be larger than the biggest side. The reason why the last one isn't a triangle is because 6+8 EQUALS 14 and is not greater than 14.

Since the sequence is always decreasing and cannot go below the lower bound, it becomes increasingly closer to the lower bound, and thus converges to a limit.

A sequence is said to be decreasing if each term is smaller than the previous one, and bounded below if there exists a lower bound such that no term in the sequence is smaller than this bound. In this case, the sequence is decreasing and bounded below by 0.

When a sequence is both decreasing and bounded below, it must have a limit because the terms cannot continue to decrease indefinitely, as they cannot go below the lower bound (in this case, 0). This means that as we progress through the terms of the sequence, the difference between consecutive terms will decrease and eventually approach a constant value.

The limit of a sequence is a value that the terms of the sequence approach as they get closer and closer to the end. In this situation, the limit is the smallest possible value that the sequence can approach, without going below the lower bound.

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The outcomes of the statistical analysis can give medical professionals and patients important knowledge regarding the possible advantages of the new procedure.

The study of random events or phenomena is the focus of the mathematical field of probability. It is a way to gauge how likely something is to happen. It is represented by a number between 0 and 1, where 0 denotes the event's impossibility and 1 denotes its certainty. There are numerous ways to calculate probabilities, including utilizing theoretical or mathematical models, empirical data, and experimental data.

given

A statistical analysis would be useful to evaluate the effectiveness of the novel technique to the standard procedure based on the study's design.

The recovery periods of the two patient groups might be compared using statistical techniques like hypothesis testing and confidence intervals to assess the efficacy of the new therapy.

Researchers can establish whether the new approach is statistically significantly superior than the usual procedure in terms of minimizing recovery time by comparing the recovery times of the two groups.

The outcomes of the statistical analysis can give medical professionals and patients important knowledge regarding the possible advantages of the new procedure.

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The probability of drawing at least one spade when drawing two cards with replacement from a standard deck of cards is approximately 0.26%.

When drawing two cards at random with replacement from a standard 52-card deck, there are 4 possible outcomes for the first card being a spade and 4 possible outcomes for the second card being a spade.

However, there is also 1 possible outcome where both cards are spades, which would be counted twice if we add the outcomes for the first and second cards separately. Therefore, the total number of outcomes where at least one card is a spade is 4 + 4 - 1 = 7.

The total number of possible outcomes when drawing two cards with replacement is 52 x 52 = 2,704.

Therefore, the probability that at least one card is a spade when drawing two cards with replacement is:

P(at least one spade) = 7 / 2,704 ≈ 0.0026

So, the probability that at least one card is a spade is approximately 0.0026 or 0.26%.

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Thus, the discounted price on the retail price of each piece of pottery is found to be: Discounted price = $11.025.

The interest rate that is adjusted to the projected cash flows of either an investment to determine its present value is referred to as the discount rate. The expected rate of return on investment that businesses or investors anticipate. The viability of an investment can be determined by computing the net present value via discounting.

Cost price =  $4.50

Percentage increase = 250%.

Marked price = 4.50 + 2.5*4.50

Marked price = $15.75

Discount = 30%

Discounted price = 15.75 - 0.30*15.75

Discounted price = $11.025

Thus, the discounted price on the retail price of each piece of pottery is found to be: Discounted price = $11.025.

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The perimeter of the figure, given the rectangles and the total area, can be found to be 48 inches

The total area is 75 in² which means that the areas of all 5 rectangles add up to 75 in². We can use this to solve for x to be:

5 × ( x × ( x - 2 )) = 75 in²

5 × ( x² - 2x ) = 75 in²

x² - 2x = 15 in²

(x - 5)(x + 3) = 0

x - 5 = 0 or x + 3 = 0

x = 5 or x = -3

Dimensions of a rectangle cannot be negative so x is equal to 5.

The perimeter is then:

= ( x - 2 ) + x + ( x - 2 ) + x + (x - 2 ) + x + ( x - 2 ) + x + (x - 2 ) + x + ( x -2 ) + x

= ( 5 - 2 ) + 5 + ( 5 - 2 ) + 5 + (5 - 2 ) + 5 + ( 5 - 2 ) + 5 + (5 - 2 ) + 5 + ( 5  -2 ) + 5

= 48 inches

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Answer:

53°

Step-by-step explanation:

it’s vertical, verticals are always identical, I think  

The present value of the trust which grows at the rate of interest 8% for 25 years is equal to $ 36504.477

The future value of the trust is equal to  $250,000

Interest rate is equal to 8% per year

To calculate present value of the trust use the formula,

Present Value = Future Value / (1 + r)^n

r is the annual interest rate

n is the number of years.

Let us consider PV represents the present value of the trust.

Present age = 25 years

Maturity age of the trust = 50 years

Number of years used amount to mature is equal to,

= 50 - 25

= 25 years

Substitute the values into the present value formula we have,

PV = 250,000 / (1 + 0.08)^25

Simplify it we get,

⇒ PV = 250,000 / 6.848475

⇒ PV = 36504.477

Therefore, the present value of the trust mature for 25 years is equal to approximately $36504.477.

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If delta and united flights are not equally on time, then the chance that we see a sample statistic of not equally or any statistic (larger or smaller) is larger, 0.05.

This is because the probability of a sample statistic being either larger or smaller (relative to the population) is always greater than the probability of it being equal. The p-value of 0.05 indicates that there is a 5% probability that the difference in on-time performance between Delta and United flights is due to chance.

The p-value of 0.05 indicates that there is a 5% chance that the observed sample statistic (of being larger or smaller than expected) is due to chance alone.

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The complete question is:

Complete the interpretation for the p-value:

If delta and united flights are (equally, not equally) on time, then the chance that we see a sample statistic of ______ or any statistic (larger, smaller) is _______ .

Explanation:

Most people with a mental illness in the United States receive care at some point in their lives, but most receive insufficient or inappropriate care. In any given year, fewer than one-third of people with a diagnosable mental illness obtain treatment.

However, according to a study, most people in the United States with a mental disorder in any given 12-month period receive treatment during the same time frame.

According to the National Survey on Drug Use and Health (NSDUH), about one in five adults (18.5%) had a mental illness in 2019. It is said that roughly 22.3 million people aged 18 and above received care for a mental health condition in the preceding year.

Treatment may involve medication, therapy, support groups, or a combination of these methods. The majority of people with mental illness may significantly benefit from these treatments.

As per a study, most people in the United States with a mental disorder in any given 12-month period receive treatment during the same time frame. This statement is true.

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The analysis of the data to obtain the best fit line equations that model the data, indicates;

5. a. Quadratic

b. y = -0.1179·x² + 2.1124·x + 4.215

c. Please find the completed chart showing the predicted values and the  in the residuals in the following section.

6. a. A linear model may not be appropriate for the data in the residual plot

b. 4

c. 5

4. a. The exponential model of the function is; y = 100·e^(-0.124·t)

b. The weight after 8 weeks is about 37.1 grams

c. The sample will be 4 grams in about 26 weeks

5. a. The equation is; y = 22.703·x + 2.5733

b. The predicted y-value at x = 10 is; y = 229.60

c. The first time is about x ≈ 3.54

The best fit line is a line that is drawn through a set of data points in such a way that it minimizes the sum of squared errors of the data.

5. The best fit model can be obtained by plotting a scatter plot of the graph, which indicates that the best fit line resembles the shape of a parabola.

Therefore;

a. The best fit equation is; Quadratic

b. The best fit equation, obtained using technology is; y = -0.1179·x² + 2.1124·x + 4.215

The square of the correlation coefficient is; R² = 0.9584

The chart can be filled with the best fit equation as follows;

[tex]\begin{tabular}{ | c | c | c | c | c | }\cline{1-4}Distance (foot) & Height (foot) & Predicted Value &Residual \\ \cline{1-4}0 & 4 & 4.215 & -0.215 \\\cline{1-4}2 & 8.4 & 8.16588 & 0.23412 \\\cline{1-4}6 & 12.1 & 13.23804 & -1.13804 \\\cline{1-4}9 & 14.2 & 14.56626 & -0.36626\\\cline{1-4}12 & 13.2 & 13.77228 & -0.57228 \\\cline{1-4}13 & 10.5 & 13.03602 & -2.53602 \\\cline{1-4}15 & 9.8 & 10.8561 & -1.0561 \\\cline{1-4}\end{tabular}[/tex]

The predicted value are obtained by plugging in the x-value into the best fit equation and the residuals is the difference between the actual value and the predicted value.

6. a. A residual plot can be used to as assessment with regards to meeting the assumptions of a linear regression model.

A residual plot that is randomly scattered about zero indicates that a linear model is appropriate for the data.

The data points in the residual plot are not expressed as being randomly scattered around zero.

The pattern that exists in the residual plot indicates that the values are positive for x-values that are either low or high, and the middle x-values have a negative residuals. Therefr;

b. The number of positive residuals = 4

c. The number of negative residuals = 5

4. The general form of the exponential model of a function, y = A × e^(-k·t)  can be used to find the function for the data as follows;

A = The initial amount = The value at 0 = 100

y = The amount of radioactive material at a given time t

e = Euler's number = 2.71828

The datapoints in the table indicates;

88.3 = 100 × e^(-k × 1)

k = -㏑(0.883) ≈ 0.124

The exponential model is therefore; y = 100 × e^(-0.124·t)

b. The weight of the sample after 8 weeks can be obtained by plugging in t = 8 in the exponential function for the weight of the radioactive substance as follows;

y = 100 × e^(-0.124 × 8) ≈ 37.1

The weight of the sample after 8 weeks is about 37.1 grams

c. When the sample is 4 grams we get;

y = 4

4 = 100 × e^(-0.124 × t)

e^(-0.124 × t)  = 4/100 = 1/25

-0.124 × t = ln(1/25) = -ln(25)

t = ln(25)/0.124 ≈ 26

t ≈ 26 weeks

Therefore; The weight will be 4 grams after approximately 26 weeks

5. A linear regression can be performed using MS Excel to obtain the equation that models the data as follows;

y = 22.703·x + 2.5733

The square of the regression coefficient is; R² = 0.9995

a. The most appropriate equation to model the data in the table is; y = 22.703·x + 2.5733

b. The y-value when x = 10 can be predicted by plugging in x = 10 into the  model equation for the data as follows;

y = 22.703 * 10 + 2.5733 ≈ 229.60

Therefore;

c. The first time the y-value is 83, can be found by setting y = 83 in the equation and solve for x as follows;

83 = 22.703·x + 2.5733

22.703·x = (83 - 2.5733)

x = (83 - 2.5733)/22.703 ≈ 3.54

Therefore, the first time the y-value is 83 is when x is approximately 3.54

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Therefore, based on this collection of data during a significance level of 0.05, there is insufficient proof to draw the conclusion that the average annual precipitation in the Amazonian rain forest is less than 200 cm.

Hypothesis The amount of pleasure increases with daily solar exposure. The presumed reason in this case and the factor that is independent is sun exposure. The expected impact, or dependent variable, is the degree of happiness.

To determine if the data provide sufficient that the annual rainfall in the Amazon rainforest is different than 200 cm, we can perform a hypothesis test with the following hypotheses:

Null hypothesis: The true mean annual rainfall in the Amazon rainforest is equal to 200 cm.

Alternative hypothesis: The true mean annual rainfall in the Amazon rainforest is different than 200 cm.

We can use a two-sided t-test with a significance level of 0.05 to test this hypothesis.

We must first determine the population mean and standard variation.

x = (188 + 232 + 210 + 198 + 202 + 193 + 219 + 202 + 252 + 156 + 184 + 222)/12 = 202.17 cm

Sample standard deviation, s = 26.770 cm

Next, we can calculate the t-statistic:

t = (x - μ) / (s / sqrt(n)) = (202.17 - 200) / (26.770 / sqrt(12)) = 0.491

Using a t-distribution table with 11 degrees of freedom (n-1), at a significance level of 0.05, we find the critical values to be ±2.201. Since the calculated t-statistic of 0.491 is within the range of -2.201 to 2.201, we fail to reject the null hypothesis.

Therefore, there is not sufficient evidence to conclude that the annual rainfall in the Amazon rainforest is different than 200 cm based on this sample data at a significance level of 0.05.

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From the given information provided, the percentage of children who travel by bicycle is 15%.

According to the given information, out of 20 children, 3 of them travel to school by bicycle. To calculate the percentage of children who travel to school by bicycle, we need to divide the number of children who travel by bicycle by the total number of children and multiply by 100.

Percentage of children who travel to school by bicycle = (Number of children who travel by bicycle / Total number of children) x 100%

= (3/20) x 100%

= 15%

Therefore, the percentage of children who travel to school by bicycle is 15%.

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the number of more 4's printed than 8's is 188 - 59 = 129.

We can approach this problem by counting the number of times the digit 4 appears and the number of times the digit 8 appears in the page numbers from 1 to 488.

First, let's count the number of times the digit 4 appears. We can break down the counting into three cases:

The digit 4 appears in the units place (page numbers 4, 14, 24, ..., 484): there are 49 such page numbers (from 4 to 484, incrementing by 10).

The digit 4 appears in the tens place (page numbers 40 to 49, 140 to 149, ..., 440 to 449): there are 50 such page numbers.

The digit 4 appears in the hundreds place (page numbers 400 to 488): there are 89 such page numbers.

Thus, the total number of times the digit 4 appears is 49 + 50 + 89 = 188.

Next, let's count the number of times the digit 8 appears. We can break down the counting into two cases:

The digit 8 appears in the tens place (page numbers 80 to 89, 180 to 189, ..., 480 to 489): there are 50 such page numbers.

The digit 8 appears in the hundreds place (page numbers 800 to 880): there are 9 such page numbers.

Thus, the total number of times the digit 8 appears is 50 + 9 = 59.

Therefore, the number of more 4's printed than 8's is 188 - 59 = 129.

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Answer: I'm pretty bad at math, but I made it more simple for you :)

Step-by-step explanation:

The answer is y = 4 + 5x/4

I hope it helped a bit sorryy

The percent that reported mushroom as their favorite from the pizza shop survey would be 10 %.

The percent that instead reported cheese or pepperoni as their favorite is 62.5 %.

Out of 400 people, the number that would say their favorite was sausage is 50 people .

The percent that prefer mushroom would be:

= Number of those who chose mushroom / Number surveyed

= 8 ( 35 + 8 + 15 + 12 + 10 )

= 8 / 80

= 10 %

The percentage that instead chose cheese or pepperoni:

= ( 15 + 35 ) / ( 35 + 8 + 15 + 12 + 10 )

= 50 / 80

= 62.5 %

The number out of 400 that would be expected to choose sausage :

= 10 / ( 35 + 8 + 15 + 12 + 10 ) x 400

= 50 people

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Player A's weight was 0.208 standard deviations below the mean weight of the football team.

To calculate the standard deviation, we first need to calculate the mean weight of the football team

Mean weight = (sum of all weights) / (number of team members)

(174+176+178+184+185+185+185+185+188+190+200+202+205+206+210+211+211+212+212+215+215+220+223+228+230+232+241+241+242+245+247+250+250+259+260+260+265+265+270+272+273+275+276+278+280+280+285+285+286+290+290+295+302) / 52

= 225.5 pounds

Now we can calculate the standard deviation using the following formula:

Standard deviation = sqrt((sum of (x - mean)^2) / N)

Where x is the weight of a team member, N is the total number of team members.

We can simplify this formula by calculating the variance first:

Variance = (sum of (x - mean)^2) / N

So we have

Variance = ((174-225.5)^2 + (176-225.5)^2 + ... + (302-225.5)^2) / 52

= 10764.35

Now we can calculate the standard deviation

Standard deviation = sqrt(Variance)

= sqrt(10764.35)

= 103.76

To find out how many standard deviations above or below the mean Player A's weight was, we can use the following formula

Z-score = (x - mean) / standard deviation

Where x is Player A's weight, mean is the mean weight of the team, and standard deviation is the standard deviation we just calculated.

So we have

Z-score = (204 - 225.5) / 103.76

= -0.208

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The given question is incomplete, the complete question is:

Following are the published weights (in pounds) of all of the team members of Football Team A from a previous year.

174, 176, 178, 184, 185, 185, 185, 185, 188, 190, 200, 202, 205, 206, 210, 211, 211, 212, 212, 215, 215, 220, 223, 228, 230, 232, 241, 241, 242, 245, 247, 250, 250, 259, 260, 260, 265, 265, 270, 272, 273, 275, 276, 278, 280, 280, 285, 285, 286, 290, 290, 295, 302

median = 241

the first quartile = 205.5

the third quartile = 272.5

Assume the population was Football Team A. When Player A played football, he weighed 204pounds. How many standard deviations above or below the mean was he?

We can expect approximately 90 mosquitoes in the trap at the end of the 13-week mosquito season.

Based on the given data, we can observe that the number of mosquitoes is increasing over time, and the increase is not constant. So, it would be unreasonable to continue counting individual mosquitoes after a certain point.

To estimate the number of mosquitoes at the end of the 13-week mosquito season, we can assume that the rate of increase remains constant over time. Using this assumption, we can find the average increase in mosquitoes per week, which is the difference between the number of mosquitoes in the last week and the number in the first week, divided by the number of weeks:

Average increase = (27 - 8) / 3 = 6.33

Then, we can use this average increase to estimate the number of mosquitoes in the 13th week:

Number of mosquitoes in the 13th week = 27 + 6.33 × 10 = 90.3

Since we cannot have a fraction of a mosquito, we can round this estimate to the nearest whole number.

Therefore, we can expect approximately 90 mosquitoes in the trap at the end of the 13-week mosquito season.

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[tex]=\frac{1}{15} +2\\[/tex]

⇒We can only add the numbers if they have the same denominator, to do so we can first convert 2 into a fraction

[tex]=\frac{1}{15} +\frac{2(15)}{1(15)} \\=\frac{1}{15} +\frac{30}{15} \\=\frac{31}{15} \\[/tex]

There are 100,000 possible five-digit student ID numbers that can be created if repeat digits are allowed (10^5).

However, if repeats are not allowed, there are only 9 x 9 x 9 x 9 x 9 = 590,490 possible student ID numbers that can be created (9^5).

The reason for the difference is that when repeats are allowed, you can use the same digit multiple times, for example "11111".

When repeats are not allowed, you can only use each digit once, for example "12345". The possible permutations of the digits 0-9 increase as more digits are added to the ID number, meaning that the more digits there are, the more possible combinations there are.

Therefore, when repeats are not allowed, the number of possible combinations is reduced significantly. In the case of five-digit student ID numbers, the difference is almost twofold.

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Using trigonometric relations we can see that u = 4.28m

First we can find the length of the shared side using the trigonometric relation:

tan(56°) = x/1.65m

1.65m*tan(56°) = x

2.45m = x

Then the length of the bottom side of the right triangle in the right side is iven by:

tan(43°) = 2.45m/y

Solving for y:

y =  2.45m/tan(43°)

y = 2.63m

Then the length of u is:

u = 1.65m + 2.63m = 4.28m

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