In the Triangle below, find X A. 10.3 B. 9.6 C. 84 D. 9

Factor the expression completely. -50x^2 +45

stich3 [128]

Answer:

Step-by-step explanation:

-5(10x^2 - 9)

8 0

3 years ago

I need help asappppppp!!!!!!!!!

trapecia [35]

An=A(n-1)-5.9

The next two are -18.5 and -24.4

You subtract 5.9 from the previous number to get the new number

7 0

2 years ago

The Springer Dog Food Company makes dry dog food from two ingredients. The two ingredients (A and B) provide different amounts o

Nataly [62]

Answer:

a) Objective function (minimize cost):

$C=0.50A+0.20B$

Restrictions

Proteins per pound: $16A+8B\leq 12$

Vitamins per pound: $4A+8B\leq 6$

Non-negative values: $A,B\geq0$

b) Attached

c) The optimum solution (minimum cost) is 0 pounds of ingredient A and 0.75 pounds of ingredient B. The cost is $0.15 per ration.

d) The optimum solution changes. The cost is now 0 pounds of ingredient A and 0.625 pounds of ingredient B. The cost is $0.125 per ration.

Step-by-step explanation:

a) The LP formulation for this problem is:

Objective function (minimize cost):

$C=0.50A+0.20B$

Restrictions

Proteins per pound: $16A+8B\leq 12$

Vitamins per pound: $4A+8B\leq 6$

Non-negative values: $A,B\geq0$

b) The feasible region is attached.

c) We have 3 corner points. In one of them lies the optimal solution.

Corner A=0 B=0.75

$C=0.50*0+0.20*0.75=0.15$

Corner A=0.5 B=0.5

$C=0.50*0.5+0.20*0.5=0.35$

Corner A=0.75 B=0

$C=0.50*0.75+0.20*0=0.375$

The optimum solution (minimum cost) is 0 pounds of ingredient A and 0.75 pounds of ingredient B. The cost is $0.15 per ration.

d) If the company requires only 5 units of vitamins per pound rather than 6, one of the restrictions change.

The feasible region changes two of its three corners:

Corner A=0 B=0.625

$C=0.50*0+0.20*0.625=0.125$

Corner A=0.583 B=0.333

$C=0.50*0.583+0.20*0.333=0.358$

Corner A=0.75 B=0

$C=0.50*0.75+0.20*0=0.375$

The optimum solution changes. The cost is now 0 pounds of ingredient A and 0.625 pounds of ingredient B. The cost is $0.125 per ration.

3 0

3 years ago

Exercise 1.A committee of five people is chosen randomly from four men and six women. Find the probability that:a) Exactly four

MatroZZZ [7]

Answer with explanation:

Given : Number of men = 4

Number of women = 6

Total people = $6+4=10$

Total number of ways to make a committee of five people from 10 persons :-

$^{10}C_{4}=\dfrac{10!}{4!(10-4)!}=210$

a) Number of ways to make committee that has exactly four women :

$^6C_4\times ^4C_1=\dfrac{6!}{4!(6-4)!}\times4=60$

The probability that committee has exactly four women :

$\dfrac{60}{210}=\dfrac{2}{7}$

b) Number of ways to make committee that has at-least four women :

$^6C_4\times ^4C_1+^6C_5=\dfrac{6!}{4!(6-4)!}\times4+6=66$

The probability that committee at-least four women :

$\dfrac{66}{210}=\dfrac{22}{70}$

c) Number of ways that committee has more than 4 women :-

$^6C_5\times^4C_0=6$

The probability that committee has more than 4 women :-

$\dfrac{6}{210}$

Now, the probability that committee has at most four women :-

$1-\dfrac{6}{210}=\dfrac{204}{210}=\dfrac{102}{105}$

5 0

3 years ago

If m<1=4x+2and m<2=110 what is x

Rina8888 [55]

Answer:

4x2+110=180

4x=68

x=17

4 0

3 years ago