Revenues= 27000 expenses= 18000 net income?

A wire 6 meters long is cut into two pieces. One piece is bent into a square for a frame for a stained glass ornament, while the

zhenek [66]

Answer:

Used wire in circle x = 2.64 m

Used in square L - x = 3.36 m

Total wire used 6 m

Step-by-step explanation:

We have a wire of 6 meters long.

We will cut it a distance x from one end, to get two pieces

x and 6 - x

We are going to use the piece x to get the circle then

So Perimetr of a circle is 2π*r (r is the radius of the circle) then:

x = 2*π*r ⇒ r = x/2*π

And area would be A(c) = π* (x/2*π)² ⇒ A(c) = x²/4π

From 6 - x we will get a square, and as the perimeter is 4 times the side

we have

( 6 - x )/ 4 is the side of the square

And the area is A(s) = [( 6 - x ) /4]²

Total area as function of x is

A(t) = A(c) + A(s)

A(x) = x²/4π + [ ( 6 - x ) / 4 ]²

A(x) = x²/4π + (36 + x² - 12x) /16

A(x) = 1 / 16π [ 4x² + 36π + πx² - 12π x ]

Taking drivatives on both sides of the equation we get:

A´(x) = 1/ 16π [8x +2πx - 12π]

A´(x) = 0 ⇒ 1/ 16π [8x +2πx - 12π] = 0

[8x +2πx - 12π] = 0

8x + 6.28x - 37.68 = 0

14.28x - 37.68 = 0 ⇒ x = 37.68 /14.28

x = 2.64 m length of wire used in the circle

Then the length L for the side of the square is

(6 - x )/4 ⇒ ( 6 - 2.64 )/ 4 ⇒ 3.36 / 4

L = 0.84 m total length of wire used in the square is

3.36 m

And total length of wire used is 6 m

The function is a quadratic function and "a" coefficient is positive then is open upward parabola there is not a maximun

5 0

3 years ago

rewona [7]

X = 5/3 and m∠SPY = 8 1/3°.

Since PQ bisects the angle, the two angles formed by the bisector are equal:

11/2x - 5 = 4x - 5/2

We will first multiply everything by 2 to eliminate the fractions:
(11/2x)*2 - 5*2 = 4x*2 - (5/2)*2
11x - 10 = 8x - 5

Subtract 8x from each side:
11x - 10 - 8x = 8x - 5 - 8x
3x - 10 = -5

Add 10 to both sides:
3x - 10 + 10 = -5 + 10
3x = 5

Divide both sides by 3:
3x/3 = 5/3
x = 5/3

Now we will plug this in for x in both smaller angles and add them together to find the measure of ∠SPY:

11/2(5/3) - 5 + 4(5/3) - 5/2
55/6 - 5 + 20/3 - 5/2

We will rewrite everything using a denominator of 6:
55/6 - 30/6 + 40/6 - 15/6 = (55-30+40-15)/6 = 50/6 = 8 2/6 = 8 1/3°

3 0

3 years ago

Some scientists believe alcoholism is linked to social isolation. One measure of social isolation is marital status. A study of

frez [133]

Answer:

1) H0: There is independence between the marital status and the diagnostic of alcoholic

H1: There is association between the marital status and the diagnostic of alcoholic

2) The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

3) $\chi^2 = \frac{(21-33.143)^2}{33.143}+\frac{(37-41.429)^2}{41.429}+\frac{(58-41.429)^2}{41.429}+\frac{(59-46.857)^2}{46.857}+\frac{(63-58.571)^2}{58.571}+\frac{(42-58.571)^2}{58.571} =19.72$

4) $df=(rows-1)(cols-1)=(3-1)(2-1)=2$

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >19.72)=5.22x10^{-5}$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.72,2,TRUE)"

Since the p value is lower than the significance level so then we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

Diag. Alcoholic Undiagnosed Alcoholic Not alcoholic Total

Married 21 37 58 116

Not Married 59 63 42 164

Total 80 100 100 280

Part 1

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is independence between the marital status and the diagnostic of alcoholic

H1: There is association between the marital status and the diagnostic of alcoholic

The level os significance assumed for this case is $\alpha=0.05$

Part 2

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

Part 3

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total col * total row}{grand total}$

And the calculations are given by:

$E_{1} =\frac{80*116}{280}=33.143$

$E_{2} =\frac{100*116}{280}=41.429$

$E_{3} =\frac{100*116}{280}=41.429$

$E_{4} =\frac{80*164}{280}=46.857$

$E_{5} =\frac{100*164}{280}=58.571$

$E_{6} =\frac{100*164}{280}=58.571$

And the expected values are given by:

Diag. Alcoholic Undiagnosed Alcoholic Not alcoholic Total

Married 33.143 41.429 41.429 116

Not Married 46.857 58.571 58.571 164

Total 80 100 100 280

And now we can calculate the statistic:

$\chi^2 = \frac{(21-33.143)^2}{33.143}+\frac{(37-41.429)^2}{41.429}+\frac{(58-41.429)^2}{41.429}+\frac{(59-46.857)^2}{46.857}+\frac{(63-58.571)^2}{58.571}+\frac{(42-58.571)^2}{58.571} =19.72$

Part 4

Now we can calculate the degrees of freedom for the statistic given by:

$df=(rows-1)(cols-1)=(3-1)(2-1)=2$

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >19.72)=5.22x10^{-5}$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.72,2,TRUE)"

Since the p value is lower than the significance level so then we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.

7 0

3 years ago

Simplify 4 + (30 + 2) Justify your steps

Sveta_85 [38]

Answer:

The answer is 36

Step-by-step explanation:

1) Add 30 + 2 and get 32

2) Then you just add 32+4 and get 36

6 0

2 years ago

What is a variable? HELP HELP

bonufazy [111]

♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫

➷ A variable is an unknown number that is represented by a letter, for example, x.

It is usually found in an algebraic expression.

✽

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

TROLLER

5 0

3 years ago