please help me please if some one answers my question next i will post 100 points free if some one answers this question please

Mathematics

2 answers:

Fiesta28 [93]2 years ago

7 0

Answer:

$(0,60) = = > the \: vertical \: displacement \: of \: the \: automobile (y-intercept) \\ (6,0) = = > the \: horizontal \: displacement \: of \: the \: automobile (x-intercept) \\ equation \\ y = mx + c \\note = > m = \frac{(0 - 60)}{(6 - 0)} \\ m = - 10 \\ consider \: point \: (6,0) \\ from \: y = mx + c \\ substitute \: for \: y \: and \: x \\ 0 = (m)(6) + c \\ 0 = ( - 10 \times 6) + c \\ 0 = - 60 + c \\ c = 60 \\ substitute \: for \: c \: in \: y = mx + c \\ y = - 10x +60 \\ 10y = - x + 6$

kari74 [83]2 years ago

4 0

Answer:

Step-by-step explanation:

a) the points (0,60) and (6.0) are the intercept points for the graph of the car slowing down ( accelerating in the opposite direction than it's traveling)

we already know the slope of the line b/c that the rate at which it's slowing .. -10 =m which is the slope of the line .. down 10

use the point-slope formula y-y1 = m(x-x1) and plug in either given point for (x1,y1)

b) y -60 = -10(x-0)

y= -10x+60

thats the slope-intercept form of the equation

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Answer:

99% CI:

$-4.637\leq\mu_v-\mu_o\leq 3.737$

Step-by-step explanation:

We have to calculate a 99%CI for the difference of means for the vegan and the omnivore.

First, we have to estimate the standard deviation

$s_{Md}=\sqrt{\frac{2MSE}{n_h}}$

The MSE can be calculated as

$MSE=\frac{(SSE_1+SSE_2)}{df} =\frac{(n_1s_1)^2+(n_2*s_2)^2}{n_1+n_2-2}\\\\MSE=\frac{(85*1.05)^2+(96*1.20)^2}{85+96-2}=\frac{7965.56+13272.04}{179} =118.65$

The weighted sample size nh can be calculated as

$n_h=\frac{2}{1/n_1+1/n_2}=\frac{2}{1/85+1/96}=\frac{2}{0.0222} =90.16$

The standard deviation then becomes

$s_{Md}=\sqrt{\frac{2MSE}{n_h}}=\sqrt{\frac{2*118.65}{90.16}}=\sqrt{2.632} =1.623$

The z-value for a 99% confidence interval is z=2.58.

The difference between means is

$\Delta M=M_v-M_o=5.10-5.55=-0.45$

Then the confidence interval can be constructed as

$\Delta M-z*s_{Md}\leq\mu_v-\mu_o\leq \Delta M+z*s_{Md}\\\\ -0.45-2.58*1.623\leq\mu_v-\mu_o\leq -0.450+2.58*1.623\\\\-0.450-4.187\leq\mu_v-\mu_o\leq-0.450+4.187\\\\-4.637\leq\mu_v-\mu_o\leq 3.737$