(15 points)Need help on math homework... here take sum point for helping!

Gwar [14]

8. The mean of the grades (74) is substantially below the median (88), so the distribution is not symmetrical. Of course, half the grades are below the median, just as half are above the median. However, the mean being 88-74 = 14 points below the median means that the lower half of the grades will have an average at least twice that much, or 28 points, below the median. The distribution of grades must extend quite a bit further to the left of the median than it does to the right. Hence ...
B. The distribution is skewed left.

9. It seems likely the distribution has a number of low grades pulling the average down. There certainly exists the possibility that at least one of them qualifies as an outlier by being more than 1.5 times the IQR below the first quartile. That rule, however, only applies when the distribution is relatively symmetrical, which this one is not. There does not appear to be any recommended way to describe an outlier when the distribution is skewed and has a long tail.*
TRUE: The distribution likely has an outlier.

_____
* Math instruction these days rarely recognizes such subtleties. Since it is probable that at least one value is well below the bottom quartile, I've shown my guess at the expected answer as TRUE. In a real set of grades, I expect the tail of the distribution to have enough low grades that they cannot be considered to be outliers.

8 0

3 years ago

What is the distance between the points M-1,5)<br>and Z(6,2)?

nirvana33 [79]

Answer:7.61577

Step-by-step explanation:

8 0

3 years ago

Which system of inequalities represents region Z?

ANTONII [103]

Answer:

The system of inequalities is

$y\geq 3x-2$

$y$

Step-by-step explanation:

step 1

Find the equation of the solid blue line

Let

$A(0,-2), B(1,1)$

Find the slope of AB

$m=(1+2)/(1-0)=3$

The equation of the line into slope intercept form is equal to

$y=mx+b$

we have

$m=3$

$b=-2$ ----> the point A is the y-intercept

substitute

$y=3x-2$ -----> equation of the solid blue line

The solution of the inequality is the shaded area above the solid line

therefore

The first inequality is

$y\geq 3x-2$

step 2

Find the equation of the dashed red line

Let

$C(0,2), D(5,1)$

Find the slope of CD

$m=(1-2)/(5-0)=-1/5$

The equation of the line into slope intercept form is equal to

$y=mx+b$

we have

$m=-1/5$

$b=2$ ----> the point C is the y-intercept

substitute

$y=-(1/5)x+2$ -----> equation of the dashed red line

The solution of the inequality is the shaded area below the dashed line

therefore

The second inequality is

$y$

The system of inequalities is

$y\geq 3x-2$

$y$

8 0

3 years ago

Helpppppp meeeee pleaseee

dedylja [7]

Answer:

the total cost would be $65.35

6 0

3 years ago

If a substance decays at a rate of 25% every 10 years, how long will it take 512 grams of the substance to decay to 121.5 grams

Juliette [100K]

Answer:

It will take 50 years to decay from 512 grams to 121.5 grams.

Step-by-step explanation:

The decay formula :

$N=N_0e^{-\lambda t}$

where

N= amount of substance after t time

N₀= initial of substance

t= time.

A substance decays at a rate 25% every 10 years.

So, remaining amount of the substance is = (100%-25%)= 75%

$\frac{N}{N_0}=\frac{75\%}{100\%}=\frac{75}{100}=\frac34$ , t= 10

$N=N_0e^{-\lambda t}$

$\Rightarrow \frac {N}{N_0}=e^{-\lambda t}$

$\Rightarrow \frac34 =e^{-\lambda .10}$

Taking ln both sides

$\Rightarrow ln|\frac34| =ln|e^{-\lambda .10}|$

$\Rightarrow ln|\frac34|=-10\lambda$

$\Rightarrow \lambda=\frac{ ln|\frac34|}{-10}$

Now , N₀= 512 grams, N= 121.5 grams, t=?

$N=N_0e^{-\lambda t}$

$\therefore 121.5=512e^{-\frac{ln|\frac34|}{-10}.t}$

$\Rightarrow 121.5=512e^{\frac{ln|\frac34|}{10}.t}$

$\Rightarrow \frac{121.5}{512}=e^{\frac{ln|\frac34|}{10}.t}$

Taking ln both sides

$\Rightarrow ln|\frac{121.5}{512}|=ln|e^{\frac{ln|\frac34|}{10}.t}|$

$\Rightarrow ln|\frac{121.5}{512}|={\frac{ln|\frac34|}{10}.t}$

$\Rightarrow t=\frac{ln|\frac{121.5}{512}|}{\frac{ln|\frac34|}{10}}$

$\Rightarrow t=\frac{10.ln|\frac{121.5}{512}|}{{ln|\frac34|}}$

⇒t=50 years

It will take 50 years to decay from 512 grams to 121.5 grams.

8 0

4 years ago