HELP ME IDK WHAT TO DO!!!

an artificial lake is in the shape of a rectangle and has an area of 9/20 square mile. the width of the lake is 1/5 the length o

Gnom [1K]

X=length
y=width
Area (rectangle)=length x width
we suggest this system of equations:
y=x/5
xy=9/20
solve this system of equations by substitution method:
x(x/5)=9/20
x²/5=9/20
Least common multiple=20
4x²=9
x²=9/4
x=⁺₋√(9/4)
we have two solutions:
x₁=-3/2 it does not validate
x₂=3/2 ⇒y=x/5=(3/2)/5=3/10

The dimensions of the lake are:
lengh=3/2 miles
width=3/10 miles

to check:
Area=3/2 miles x 3/10 miles=9/20 miles².
width is 1/5 the length of the lake ⇒ 3/10 miles=(3/2) /5 miles

7 0

4 years ago

an instructor grades on a curve by assuming that grades on a test are normally distributed. If the average grade is 80 and the s

sweet [91]

Step-by-step explanation:

According to normal distribution curves following a bell curve formula, it is followed that

34% of people will lie between Mean point and one positive standard deviation

34% of people will lie between Mean point and one negative standard deviation

14% of people will lie between Mean point and two positive standard deviation

14% of people will lie between Mean point and two negative standard deviation

2% of people will lie between Mean point and three positive standard deviation

2% of people will lie between Mean point and three negative standard deviation

Now following this principle, we are given that mean score is 80 and standard deviation is 8, hence the interval for grades would be;

Grade A ; Score greater than 96

Grade B : Score greater than 88 and less than or equal to 96

Grade C : Score greater than 80 and less than or equal to 88

Grade D : Score greater than 72 and less than or equal to 80

Grade E : Score greater than 64 and less than or equal to 72

Grade F : Score less than 64

3 0

3 years ago

A college job placement office collected data about students’ GPAs and the salaries they earned in their first jobs after gradua

frez [133]

Answer:

X is the GPA

Y is the Salary

Standard deviation of X is 0.4

Standard deviation of Y is 8500

E(X)=2.9

E(Y)=47200

We are given that The correlation between the two variables was r = 0.72

a) $y = a+bx$

$b = \frac{\sum(x_i-\bar{x})(y_i-\bar{y})}{\sum(x_i-\bar{x})^2} = \frac{r \times \sqrt{var(X) \times Var(Y)}}{Var(X)} = \frac{0.72 \times \sqrt{0.4^2 \times 8500^2}}{0.4^2} = 15300$

$a=y-bx = 47200-(15300 \times 29) = 2830$

So, slope = 15300

Intercept = 2830

So, equation : $y = 2830+15300x$

b) Your brother just graduated from that college with a GPA of 3.30. He tells you that based on this model the residual for his pay is -$1880. What salary is he earning?

$y = 2830+15300 \times 3.3 = 53320$