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3 years ago

How would I go about solving this?

Mathematics

1 answer:

garik1379 [7]3 years ago

8 0

If you draw more of those triangles, there will be 6 that can fit, so find area of 1 triangle and multiply it by 6. Write that number down and then do Pi r squared to find the area of the circle, then do circle area minus triangles area when you get that, divide it by 6. That is the area of the white region so then do Pi R squared again and then subtract the white area from that

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krok68 [10]

Answer:

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Step-by-step explanation:

Think of it like this-

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2 years ago

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2 years ago

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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 ≈<u> -0.969</u>

<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<em> i</em>th observation

b₀ = Estimated regression equation intercept

b₁ = The estimate of the slope regression equation

$X_i$ = The <em>i</em>th 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;

<u>Correlation coefficient, r ≈ -0.969</u>

Learn more about regression analysis here:

brainly.com/question/14279500

7 0

3 years ago

3. The curve C with equation y=f(x) is such that, dy/dx = 3x^2 + 4x +k

Andreas93 [3]

a. Given that y = f(x) and f(0) = -2, by the fundamental theorem of calculus we have

$\displaystyle \frac{dy}{dx} = 3x^2 + 4x + k \implies y = f(0) + \int_0^x (3t^2+4t+k) \, dt$

Evaluate the integral to solve for y :

$\displaystyle y = -2 + \int_0^x (3t^2+4t+k) \, dt$

$\displaystyle y = -2 + (t^3+2t^2+kt)\bigg|_0^x$

$\displaystyle y = x^3+2x^2+kx - 2$

Use the other known value, f(2) = 18, to solve for k :

$18 = 2^3 + 2\times2^2+2k - 2 \implies \boxed{k = 2}$

Then the curve C has equation

$\boxed{y = x^3 + 2x^2 + 2x - 2}$

b. Any tangent to the curve C at a point (a, f(a)) has slope equal to the derivative of y at that point:

$\dfrac{dy}{dx}\bigg|_{x=a} = 3a^2 + 4a + 2$

The slope of the given tangent line $y=x-2$ is 1. Solve for a :

$3a^2 + 4a + 2 = 1 \implies 3a^2 + 4a + 1 = (3a+1)(a+1)=0 \implies a = -\dfrac13 \text{ or }a = -1$

so we know there exists a tangent to C with slope 1. When x = -1/3, we have y = f(-1/3) = -67/27; when x = -1, we have y = f(-1) = -3. This means the tangent line must meet C at either (-1/3, -67/27) or (-1, -3).

Decide which of these points is correct:

$x - 2 = x^3 + 2x^2 + 2x - 2 \implies x^3 + 2x^2 + x = x(x+1)^2=0 \implies x=0 \text{ or } x = -1$

So, the point of contact between the tangent line and C is (-1, -3).

7 0

2 years ago