I really need help on this.. Compare y=x^2 with y = -2 (x+3)^2-1

Mathematics

1 answer:

klio [65]4 years ago

7 0

Answer:

See below

Step-by-step explanation:

While both are quadratic, second-degree equations, there are a number of differences. y = -2(x+3)^2 - 1 has been flipped over the x-axis (thanks to the negative symbol in front of the 2), stretched by a scale factor of 2, and translated 1 unit down (hence the - 1) and 3 units to the left (full equation for a quadratic equation in vertex form is y = a(x-h)^2 + k. Note the minus sign before the h. This means that the equation was horizontally translated in the opposite direction of what the + or - sign suggests). y=x^2, of course, hasn't had any of these transformations happen to it.

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Ten college students were randomly selected. Their grade point averages (GPAs) when they entered the program were between 3.5 a

solong [7]

Answer:

The 95% confidence interval for the true slope is (0.03985, 0.14206).

Step-by-step explanation:

For the regression equation:

$\hat y=\alpha +\hat \beta x$

The (1 - α)% confidence interval for the regression coefficient or slope $(\hat \beta )$ is:

$Ci=\hat \beta \pm t_{\alpha/2, (n-2)}\times SE(\hat \beta )$

The regression equation for current GPA (Y) of students based on their GPA's when entering the program (X) is:

$\hat Y=3.584756+0.090953 X$

The summary of the regression analysis is:

Predictor Coefficient SE t-stat p-value

Constant 3.584756 0.078183 45.85075 5.66 x 10⁻¹¹

Entering GPA 0.090953 0.022162 4.103932 0.003419

The regression coefficient and standard error are:

$\hat \beta = 0.090953\\SE (\hat \beta)=0.022162$

The critical value of t for 95% confidence level and 8 degrees of freedom is:

$t_{\alpha/2, n-2}=t_{0.05/2, 10-2}=t_{0.025, 8}=2.306$

Compute the 95% confidence interval for $(\hat \beta )$ as follows:

$CI=\hat \beta \pm t_{\alpha/2, (n-2)}\times SE(\hat \beta )\\=0.090953\pm 2.306\times 0.022162\\=0.090953\pm 0.051105572\\=(0.039847428, 0.142058572)\\\approx (0.03985, 0.14206)$