Solve this problem. -16r=272

Find the equation of a line that passes through the points (4,0) and (2,4).

Ksju [112]

Answer:

answer is y=-2x+8 is the eq of given line

3 0

3 years ago

Given m||n, find the value of x.<br> +<br> (6x+10)<br> (10x+10)°

ludmilkaskok [199]

Answer:

(6x+10)°=(6x+10)°[vertically opposite angle]

now,

6x+10+10x+10=180°[being cointerior angle]

16x=180-20

x=160%16

x=10°

5 0

3 years ago

Need help with this question

Lilit [14]

Answer: 30

Step-by-step explanation: To solve this problem, we're going to have to use the Pythagorean theorem...

a² + b² = c²

24² + 18² = c²

576 + 324 = c²

c² = 900

$\sqrt{900}$

30

30 = c

I hope this helps!

6 0

3 years ago

A random sample of 20 recent weddings in a country yielded a mean wedding cost of $ 26,388.67. Assume that recent wedding costs

Makovka662 [10]

Answer:

a) 95% confidence interval for the meancost, μ, of all recent weddings in this country = (22,550.95, 30,226.40)

.The 95% confidence interval is from $22,550.95 to $30,226.40.

b) For the interpretation of the result, option D is correct.

We can be 95% confident that the mean cost, μ, of all recent weddings in this country is somewhere within the confidence interval.

c) Option B is correct.

The population mean may or may not lie in this interval, but we can be 95% confident that it does.

Step-by-step explanation:

Sample size = 20

Sample Mean = $26,388.67

Sample Standard deviation = $8200

Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample mean) ± (Margin of error)

Sample Mean = 26,388.67

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error of the mean)

Critical value will be obtained using the t-distribution. This is because there is no information provided for the population mean and standard deviation.

To find the critical value from the t-tables, we first find the degree of freedom and the significance level.

Degree of freedom = df = n - 1 = 20 - 1 = 19.

Significance level for 95% confidence interval

(100% - 95%)/2 = 2.5% = 0.025

t (0.025, 19) = 2.086 (from the t-tables)

Standard error of the mean = σₓ = (σ/√n)

σ = standard deviation of the sample = 8200

n = sample size = 20

σₓ = (8200/√20) = 1833.6

99% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 26,388.67 ± (2.093 × 1833.6)

CI = 26,388.67 ± 3,837.7248

99% CI = (22,550.9452, 30,226.3948)

99% Confidence interval = (22,550.95, 30,226.40)

a) 95% confidence interval for the meancost, μ, of all recent weddings in this country = (22,550.95, 30,226.40)

.The 95% confidence interval is from $22,550.95 to $30,226.40.

b) The interpretation of the confidence interval obtained, just as explained above is that we can be 95% confident that the mean cost, μ,of all recent weddings in this country is somewhere within the confidence interval

c) A further explanation would be that the population mean may or may not lie in this interval, but we can be 95% confident that it does.

Hope this Helps!!!

4 0

3 years ago

Arm Span(x) Height(y)

natita [175]

Answer:

Here's what I get.

Step-by-step explanation:

1. Representation of data

I used Excel to create a scatterplot of the data, draw the line of best fit, and print the regression equation.

2. Line of best fit

(a) Variables

I chose arm span as the dependent variable (y-axis) and height as the independent variable (x-axis).

It seems to me that arm span depends on your height rather than the other way around.

(b) Regression equation

The calculation is easy but tedious, so I asked Excel to do it.

For the equation y = ax + b, the formulas are

$a = \dfrac{\sum y \sum x^{2} - \sum x \sumxy}{n\sum x^{2}- \left (\sum x\right )^{2}}\\\\b = \dfrac{n\sumx y - \sum x \sumxy}{n\sum x^{2}- \left (\sum x\right )^{2}}$

This gave the regression equation:

y = 1.0595x - 4.1524

(c) Interpretation

The line shows how arm span depends on height.

The slope of the line says that arm span increases about 6 % faster than height.

The y-intercept is -4. If your height is zero, your arm length is -4 in (both are impossible).

(d) Residuals

$\begin{array}{cccr}&\textbf{Arm Span} & \textbf{Arm Span}&\\\textbf{Height/in} &\textbf{Actual} & \textbf{Predicted}&\textbf{Residual}\\25 & 19 & 22.3 & -3.3\\40 & 41 & 38.2 & 2.8\\55 & 51 & 54.1 & -3.1\\65 & 67 & 62.6 & 4.4\\ \end{array}$

The residuals appear to be evenly distributed above and below the predicted values.

A graph of all the residuals confirms this observation.

The equation usually predicts arm span to within 4 in.

(e) Predictions

(i) Height of person with 66 in arm span

$\begin{array}{rcl}y& = & 1.0595x - 4.1524\\66 & = & 1.0595x - 4.1524\\70.1524 & = & 1.0595x\\x & = & \dfrac{70.1524}{1.0595}\\\\& = & \textbf{66 in}\\\end{array}\\\text{A person with an arm span of 66 in should have a height of about $\large \boxed{\textbf{66 in}}$}$

(ii) Arm span of 74 in tall person

$\begin{array}{rcl}y& = & 1.0595x - 4.1524\\& = & 1.0595\times74 - 4.1524\\& = & 78.4030 - 4.1524\\& = & \textbf{74 in}\\\end{array}\\\text{ A person who is 74 in tall should have an arm span of $\large \boxed{\textbf{74 in}}$}$

6 0

3 years ago