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3 years ago

7 people can pave a driveway in 9 hours. how long will it take for 11 people to pave a driveway?

Mathematics

1 answer:

disa [49]3 years ago

8 0

Answer:

Step-by-step explanation:

If 7 people can pave the drive in 9 hours, we can use that information in equation form to find out how long it would take 1 person to pave the drive alone.

$\frac{7}{x}=\frac{1}{9}$ and cross multiply

x = 63. That means that it would take 1 person, working alone, 63 hours to get the job done. If that be the case, and assuming that the 2 added people work at the same rate as do the first 9, we simply divide that time alone into 11 parts.

It would take 11 people $5.\bar{27}$ hours to pave that driveway.

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#10 i The table shows the admission costs (in dollars) and the average number of daily visitors at an amusement park each the pa

lions [1.4K]

The line of best fit is a straight line that can be used to predict the

average daily attendance for a given admission cost.

Correct responses:

The equation of best fit is; $\underline{ \hat Y = 1,042 - 4.9 \cdot X_i}$

The correlation coefficient is; r ≈ -0.969

<h3>Methods by which the line of best fit is found</h3>

The given data is presented in the following tabular format;

$\begin{tabular}{|c|c|c|c|c|c|c|c|c|}Cost, (dollars), x&20&21&22&24&25&27&28&30\\Daily attendance, y&940&935&940&925&920&905&910&890\end{array}\right]$

The equation of the line of best fit is given by the regression line

equation as follows;

$\hat Y = \mathbf{b_0 + b_1 \cdot X_i}$

Where;

$\hat Y$ = Predicted value of the ith observation

b₀ = Estimated regression equation intercept

b₁ = The estimate of the slope regression equation

$X_i$ = The ith observed value

$b_1 = \mathbf{\dfrac{\sum (X - \overline X) \cdot (Y - \overline Y) }{\sum \left(X - \overline X \right)^2}}$

$\overline X$ = 24.625

$\overline Y$ = 960.625

$\mathbf{\sum(X - \overline X) \cdot (Y - \overline Y)} = -433.125$

$\mathbf{\sum(X - \overline X)^2} = 87.875$

Therefore;

$b_1 = \mathbf{\dfrac{-433.125}{87.875}} \approx -4.9289$

Therefore;

The slope given to the nearest tenth is b₁ ≈ -4.9

$b_0 = \mathbf{\dfrac{\left(\sum Y \right) \cdot \left(\sum X^2 \right) - \left(\sum X \right) \cdot \left(\sum X \cdot Y\right)} {n \cdot \left(\sum X^2\right) - \left(\sum X \right)^2}}$

By using MS Excel, we have;

n = 8

∑X = 197

∑Y = 7365

∑X² = 4939

∑Y² = 6782675

∑X·Y = 180930

(∑X)² = 38809

Therefore;

$b_0 = \dfrac{7365 \times 4939-197 \times 180930}{8 \times 4939 - 38809} \approx \mathbf{1041.9986}$

The y-intercept given to the nearest tenth is b₀ ≈ 1,042

The equation of the line of best fit is therefore;

$\underline{\hat Y = 1042 - 4.9 \cdot X_i}$

The correlation coefficient is given by the formula;

$\displaystyle r = \mathbf{\dfrac{\sum \left(X_i - \overline X) \cdot \left(Y - \overline Y \right)}{ \sqrt{\sum \left(X_i - \overline X \right)^2 \cdot \sum \left(Y_i - \overline Y \right)^2} }}$

Where;

$\sqrt{\sum \left(X - \overline X \right)^2 \times \sum \left(Y - \overline Y \right)^2} = \mathbf{446.8121}$

$\sum \left(X_i - \overline X \right) \times \left(Y - \overline Y\right) = \mathbf{-433.125}$

Which gives;

$r = \dfrac{-433.125}{446.8121} \approx \mathbf{-0.969367213}$

The correlation coefficient given to the nearest thousandth is therefore;

Correlation coefficient, r ≈ -0.969

Learn more about regression analysis here:

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7 0

2 years ago

Determine whether the graph represents a proportional relationship.

olya-2409 [2.1K]

No, it is not a proportional relationship because the graph does not go through the origin. A proportional relationship is supposed to be a straight line, starting from the origin. It starts on the number 4.

8 0

3 years ago

From a large number of actuarial exam scores, a random sample of scores is selected, and it is found that of these are passing s

Mnenie [13.5K]

Supposing 60 out of 100 scores are passing scores, the 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).

The lower limit is 0.5.
The upper limit is 0.7.

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $\alpha$ , we have the following confidence interval of proportions.