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1 year ago

True or false: the regression sum of squares (ssr) can never be greater than the total sum of squares (sst).

Mathematics

1 answer:

Valentin [98]1 year ago

8 0

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A piece of rubber tubing maintains a cylindrical shape as it is stretched. At the instant that the inner radius of the tube is 2

professor190 [17]

Answer:

the rate of change in volume is dV/dt = 4π mm³/s = 12.56 mm³/s

Step-by-step explanation:

since the volume V of a cylinder is related with the height H and the radius R through:

V = πR²*H

then the change in time is given by the derivative with respect to time t

dV/dt = (∂V/∂R)*(dR/dt) + (∂V/∂H)*(dH/dt)

the change in volume with radius at constant height is

(∂V/∂R) = 2*πR*H

the change in volume with height at constant radius is

(∂V/∂H) = πR²

then

dV/dt = 2π*R*H *(dR/dt) + πR²*(dH/dt)

replacing values

dV/dt = 2π* 2 mm * 20 mm * (-0.1 mm/s) + π (2 mm) ²* 3 mm/s = 4π mm³/s

dV/dt = 4π mm³/s = 12.56 mm³/s

7 0

1 year ago

If p → q is true, then ~ p → ~ q is __________ true. always sometimes never

Yuliya22 [10]

Answer:

The correct answer is Never

6 0

1 year ago

After dinner 3/4 of a pie remains. if tasha eats 1/6 of the remaining pie

astraxan [27]

Answer: 1/8

Step-by-step explanation: 3/4 x 1/6 = 24/3 (simplify to 1/8)

4 0

2 years ago

Sue’s Corner Market has a markup of 60% on bottled water. If the market sells a bottle of water for $2, find the original amount

lakkis [162]

Markup is the amount added to the cost price of goods to cover overhead and profit.

Sue’s Corner Market has a markup of 60% on bottled water.

Let us say original price was $x.

Now price after markup is $2.

So we can make an equation like:

original price + markup price = price after markup

x + 60% of x =2

$x + 0.6x =2$

$1.6x =2$

dividing both sides by 1.6

x= 1.25

So original price was 1.25 dollars.

5 0

1 year ago

Read 2 more answers

A random sample of 35 undergraduate students who completed two years of college were asked to take a basic mathematics test. The

lisov135 [29]

Answer:

$(75.1 -72.1) -1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= -1.96$

$(75.1 -72.1) +1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= 7.96$

And the confidence interval would be given by:

$-1.96 \leq \mu_1 -\mu_2 \leq 7.96$

And since the confidence interval contains the value 0 we have enough evidence to conclude that we don't have significant differences between the two means at 10% of significance.

Step-by-step explanation:

For this case we have the following info given :

$\bar X_1= 75.1$ represent the sample mean for the scores of the undergraduate students

$s_1 = 12.8$ represent the standard deviation for the undergraduate students

$n_1 =35$ the sample size for the undergraduate

$\bar X_2= 72.1$ represent the sample mean for the scores of the high school students

$s_2 = 14.6$ represent the standard deviation for the high school students

$n_2 =50$ the sample size for the high school

The confidence interval for the true difference of means is given by:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

The degrees of freedom are given by:

$df=n_1 +n_2 -2= 35+50-2=83$

The confidence level is 90% and the significance level is $\alpha=0.1$ and $\alpha/2 =0.05$ then the critical value would be:

$t_{\alpha/2}= 1.99$

And replacing the info we got:

$(75.1 -72.1) -1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= -1.96$

$(75.1 -72.1) +1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= 7.96$

And the confidence interval would be given by:

$-1.96 \leq \mu_1 -\mu_2 \leq 7.96$

And since the confidence interval contains the value 0 we have enough evidence to conclude that we don't have significant differences between the two means at 10% of significance.

8 0

1 year ago