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elena-s [515]
3 years ago
15

Find the x intercepts of the parabola with vertex (-1,-16) and y intercept (0,-15)

Mathematics
1 answer:
Arlecino [84]3 years ago
8 0
<span> first, write the equation of the parabola in the required form: </span>
<span>(y - k) = a·(x - h)² </span>

<span>Here, (h, k) is given as (-1, -16). </span>
<span>So you have: </span>
<span>(y + 16) = a · (x + 1)² </span>

<span>Unfortunately, a is not given. However, you do know one additional point on the parabola: (0, -15): </span>

<span>-15 + 16 = a· (0 + 1)² </span>
<span>.·. a = 1 </span>

<span>.·. the equation of the parabola in vertex form is </span>
<span>y + 16 = (x + 1)² </span>

<span>The x-intercepts are the values of x that make y = 0. So, let y = 0: </span>

<span>0 + 16 = (x + 1)² </span>
<span>16 = (x + 1)² </span>

<span>We are trying to solve for x, so take the square root of both sides - but be CAREFUL! </span>

<span>± 4 = x + 1 ...... remember both the positive and negative roots of 16...... </span>

<span>Solving for x: </span>
<span>x = -1 + 4, x = -1 - 4 </span>
<span>x = 3, x = -5. </span>

<span>Or, if you prefer, (3, 0), (-5, 0). </span>
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Answer:

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The average score for the for the 22 participants of the program, \overline x_1 = 6.59

The standard deviation for the 22 participants of the program, s₁ = 4.10

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The confidence interval for unknown standard deviation, σ, is given by the following expression;

\left (\bar{x}_{1}- \bar{x}_{2}  \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}

α = 1 - 0.99 = 0.01

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The degrees of freedom, df = 22 - 1 = 21

t_{\alpha /2} = t_{0.005, \, 21} = 1.721

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The 99% confidence interval for the difference in average is therefore given as follows;

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2. The possible issues in the calculations are;

a. The confidence level used in the confidence interval can influence the result of the confidence interval observed

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