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11111nata11111 [884]

3 years ago

8

Beach Cruises prints brochures and fliers to advertise their dolphin watching tours. To print, the brochure requires three pages

while the flier requires two. They are limited to 500 sheets of paper, but need to print at least 80 brochures and 100 fliers. Each brochure cost 10 cents to print and each flier cost 6 cents to print. How many of each should their print to minimize their cost? Include: Constraints (inequalities), shaded graph, feasible region, labeled vertices, Objective Function, Max/Min

1 answer:

Nimfa-mama [501]3 years ago

6 0

Answer:

500 divided by 80 equals 6.25. 6.25 divided by 100 equals 0.0625

Step-by-step explanation:

500 divided by 80 equals 6.25. 6.25 divided by 100 equals 0.0625

You might be interested in

235 people are at RSM summer camp. There are 35 more boys than girls, and 70 fewer adults than girls. How many people of each gr

makkiz [27]

Answer:

The number of girls are 90 girls

The number of boys are 125 boys

The number of adults are 20 adults

There are 90 girls, 125 boys and 20 adults at the camp

Step-by-step explanation:

The total number of people is given as 235 people

The number of boys = 35 + The number of girls

The number of adults = The number of girls - 70

Let the total number of girls = G

Therefore, we have;

The number of boys = 35 + G

The number of adults = G - 70

The total number of persons = The number of boys + The number of adults + The number of girls

Therefore;

235 = The number of boys + The number of adults + The number of girls

Which gives;

235 = 35 + G + G - 70 + G = 3·G - 35

235 = 3·G - 35

3·G - 35 = 235

3·G = 235 + 35 = 270

G = 270/3 = 90

The number of girls = 90

The number of boys = 35 + The number of girls = 35 + 90 = 125

The number of boys = 125

The number of adults = The number of girls - 70 = 90 - 70 = 20

The number of adults = 20.

8 0

3 years ago

Can you answer this?

nignag [31]

The answer is the graph b.

4 0

2 years ago

4^4x-5= 8^3x-4 solve

Alexxandr [17]

$4^{4x-5}=8^{3x-4}\\\\(2^2)^{4x-5}=(2^3)^{3x-4}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\2^{2(4x-5)}=2^{3(3x-4)}\iff2(4x-5)=3(3x-4)\qquad\text{use distributive property}\\\\(2)(4x)+(2)(-5)=(3)(3x)+(3)(-4)\\\\8x-10=9x-12\qquad\text{add 10 to both sides}\\\\8x=9x-2\qquad\text{subtract 9x from both sides}\\\\-x=-2\to \boxed{x=2}$

8 0

3 years ago

A cookie factory monitored the number of broken cookies per pack yesterday.

trapecia [35]

Answer:

Confidence Interval - 2.290 < S < 2.965

Step-by-step explanation:

Complete question

A chocolate chip cookie manufacturing company recorded the number of chocolate chips in a sample of 50 cookies. The mean is 23.33 and the standard deviation is 2.6. Construct a 80% confidence interval estimate of the standard deviation of the numbers of chocolate chips in all such cookies.

Solution

Given

n=50

x=23.33

s=2.6

Alpha = 1-0.80 = 0.20

X^2(a/2,n-1) = X^2(0.10, 49) = 63.17

sqrt(63.17) = 7.948

X^2(1 - a/2,n-1) = X^2(0.90, 49) = 37.69

sqrt(37.69) = 6.139

s*sqrt(n-1) = 18.2

$s\sqrt{\frac{n-1}{X^2 _{(n-1), \frac{\alpha }{2} } }$ $\leq \sigma \leq s\sqrt{\frac{n-1}{X^2 _{(n-1), 1-\frac{\alpha }{2} } }$

confidence interval:

(18.2/7.948) < S < (18.2/6.139)

2.290 < S < 2.965

8 0

2 years ago

HELP ASAP PLZZZZ

Tcecarenko [31]

QUESTION 1

The given system of equations is

$3d - e = 7...eqn(1)$
$d + e = 5...eqn(2)$

To solve by linear combination, we add equation (1) to equation (2) to get,

$3d + d= 7 + 5$

$4d = 12$

We divide through by 4 to obtain,

$d = \frac{12}{4}$

$d = 3$

We put d=3 into equation (2) to get,

$3+ e = 5$

$e = 5 - 3$

$e = 2$

$\boxed {The \: solution \: is \: (3, 2)}$

QUESTION 2

The given system is

4x + y = 5 ...eqn(1)

3x + y = 3 ...eqn(2)

To solve by linear combination, we subtract equation (2) from equation (1) to eliminate y from the equation.

This will give us,

$4x - 3x = 5 - 3$

This implies that,

$x = 2$

Put x=3 into equation (1) to get,

$4(2) + y = 5$

$8+ y = 5$

$y = 5 - 8$

$y = - 3$

The solution is

$(2,-3)$

QUESTION 3

We want to solve the system;

a – 2b = –2 ....eqn(1)

2a + 2b = 14...eqn(2)

by linear combination.

We need to add equation (1) to equation (2) to eliminate b.

This implies that,

$2a + a = 14 + - 2$

Simplify,

$3a = 12$

Divide both sides by 3 to get,

$a = 4$
Put a=4 into equation (2) to obtain,

$2(4) + 2b = 14$

$8 + 2b = 14$
$2b = 14 - 8$

$2b = 6$

$b = 3$

The ordered pair in the form (a, b) is

$(4,3)$

QUESTION 4

The given system of equations is

11x + 4y = 18 ...eqn(1)

3x + 4y = 2 ...eqn(2)

We subtract equation (2) from equation (1) to get,

$11x - 3x = 18 - 2$

$8x = 16$

$x = 2$

Put x=2 into equation (2) to obtain,

$3(2) + 4y = 2$

This implies that,

$6 + 4y = 2$

$4y = 2 - 6$

$4y = - 4$

$y=-1$

The correct answer is (2,-1).

QUESTION 5

The given system is ;

2d + e = 8...eqn1

d – e = 4...eqn2

We add the two equations to eliminate e.

This implies that,

$2d + d = 8 + 4$

$3d = 12$

We divide both sides by 3 to get,

$d = 4$

We put d=4 into equation (2) to get,

$4 - e = 4$

$- e = 4 - 4$

$- e = 0$

$e = 0$

The solution is

$(4,0)$

7 0

3 years ago