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kogti [31]
3 years ago
13

Willy Wonka has 2 candies, Wonka bars and Everlasting Gobstoppers. Both have both natural sugar and sucrose in them. Each Wonka

Bar has 4 grams of natural sugar and 1 gram of sucrose. Each Everlasting Gobstopper has 2 grams of natural sugar and 3 grams of sucrose. Mr. Wonka has 60 grams of natural sugar and 75 grams of sucrose. If each Everlasting Gobstopper has a profit of $1.3 and each Wonka Bar has a profit of $3.2, how many of each candy would give him the maximum profit?
Mathematics
1 answer:
love history [14]3 years ago
8 0

Answer:

Wonka bars=3 and  Everlasting Gobstoppers=24

Step-by-step explanation:

let the wonka bars be X

and everlasting gobstoppers be Y

the objective is to

maximize 1.3x+3.2y=P

subject to constraints

natural sugar

4x+2y=60------1

sucrose

x+3y=75---------2

x>0, y>0

solving 1 and 2 simultaneously we have

4x+2y=60----1

x+3y=75------2

multiply equation 2 by 4 and equation 1 by 1 to eliminate x we have

 4x+2y=60

 4x+12y=300

-0-10y=-240

10y=240

y=240/10

y=24

put y=24 in equation 2 we have'

x+3y=75

x+3(24)=75

x+72=75

x=75-72

x=3

put x=3 and y=24 in the objective function we have

maximize 1.3x+3.2y=P

1.3(3)+3.2(24)=P

3.9+76.8=P

80.7=P

P=$80.9

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Among persons donating blood to a clinic, 85% have Rh+ blood (that is, the Rhesus factor is present in their blood.) Six people
Leona [35]

Answer:

a) There is a 62.29% probability that at least one of the five does not have the Rh factor.

b) There is a 22.36% probability that at most four of the six have Rh+ blood.

c) There need to be at least 8 people to have the probability of obtaining blood from at least six Rh+ donors over 0.95.

Step-by-step explanation:

For each person donating blood, there are only two possible outcomes. Either they have Rh+ blood, or they do not. This means that we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem we have that:

p = 0.85, n = 6.

a) fine the probability that at least one of the five does not have the Rh factor.

Either all six have the factor, or at least one of them do not. The sum of the probabilities of these events is decimal 1. So:

P(X < 6) + P(X = 6) = 1

P(X < 6) = 1 - P(X = 6)

In which:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 6) = C_{6,6}.(0.85)^{6}.(0.15)^{0} = 0.3771

So

P(X < 6) = 1 - P(X = 6) = 1 - 0.3771 = 0.6229

There is a 62.29% probability that at least one of the five does not have the Rh factor.

b) find the probability that at most four of the six have Rh+ blood.

Either more than four have Rh+ blood, or at most four have. So

P(X \leq 4) + P(X > 4) = 1

P(X \leq 4) = 1 - P(X > 4)

In which

P(X > 4) = P(X = 5) + P(X = 6)

P(X = 5) = C_{6,5}.(0.85)^{5}.(0.15)^{1} = 0.3993

P(X = 6) = C_{6,6}.(0.85)^{6}.(0.15)^{0} = 0.3771

P(X > 4) = P(X = 5) + P(X = 6) = 0.3993 + 0.3771 = 0.7764

P(X \leq 4) = 1 - P(X > 4) = 1 - 0.7764 = 0.2236

There is a 22.36% probability that at most four of the six have Rh+ blood.

c) The clinic needs six Rh+ donors on a certain day. How many people must donate blood to have the probability of obtaining blood from at least six Rh+ donors over 0.95?

With 6 donors:

P(X = 6) = C_{6,6}.(0.85)^{6}.(0.15)^{0} = 0.3771

37.71% probability of obtaining blood from at least six Rh+ donors over 0.95.

With 7 donors:

P(X = 6) = C_{7,6}.(0.85)^{6}.(0.15)^{1} = 0.3960

0.3771 + 0.3960 = 0.7764 = 77.64% probability of obtaining blood from at least six Rh+ donors over 0.95.

With 8 donors

P(X = 6) = C_{8,6}.(0.85)^{6}.(0.15)^{2} = 0.2376

0.3771 + 0.3960 + 0.2376 = 1.01 = 101% probability of obtaining blood from at least six Rh+ donors over 0.95.

There need to be at least 8 people to have the probability of obtaining blood from at least six Rh+ donors over 0.95.

5 0
3 years ago
100 points
Pavlova-9 [17]
STEP
1
:

y
Simplify —
3
Equation at the end of step
1
:

y
(((18•(x5))•(y3))+((6•(x2))•(y4)))+((((24•(x6))•—)•x)•y)
3
STEP
2
:

Equation at the end of step
2
:

y
(((18•(x5))•(y3))+((6•(x2))•(y4)))+((((23•3x6)•—)•x)•y)
3
STEP
3
:

Canceling Out:

3.1 Canceling out 3 as it appears on both sides of the fraction line


Equation at the end of step
3
:

(((18•(x5))•(y3))+((6•(x2))•(y4)))+((8x6y•x)•y)
STEP
4
:

Equation at the end of step
4
:

(((18•(x5))•(y3))+((2•3x2)•y4))+8x7y2
STEP
5
:

Equation at the end of step
5
:

(((2•32x5) • y3) + (2•3x2y4)) + 8x7y2
STEP
6
:

STEP
7
:
Pulling out like terms

7.1 Pull out like factors :

8x7y2 + 18x5y3 + 6x2y4 = 2x2y2 • (4x5 + 9x3y + 3y2)

Trying to factor a multi variable polynomial :

7.2 Factoring 4x5 + 9x3y + 3y2

Try to factor this multi-variable trinomial using trial and error

Factorization fails

Final result :
2x2y2 • (4x5 + 9x3y + 3y2)
3 0
2 years ago
Prove each of the following statements below using one of the proof techniques and state the proof strategy you use.
pochemuha

Answer:

See below

Step-by-step explanation:

a) Direct proof: Let m be an odd integer and n be an even integer. Then, there exist integers k,j such that m=2k+1 and n=2j. Then mn=(2k+1)(2j)=2r, where r=j(2k+1) is an integer. Thus, mn is even.

b) Proof by counterpositive: Suppose that m is not even and n is not even. Then m is odd and n is odd, that is, m=2k+1 and n=2j+1 for some integers k,j. Thus, mn=4kj+2k+2j+1=2(kj+k+j)+1=2r+1, where r=kj+k+j is an integer. Hence mn is odd, i.e, mn is not even. We have proven the counterpositive.

c) Proof by contradiction: suppose that rp is NOT irrational, then rp=m/n for some integers m,n, n≠. Since r is a non zero rational number, r=a/b for some non-zero integers a,b. Then p=rp/r=rp(b/a)=(m/n)(b/a)=mb/na. Now n,a are non zero integers, thus na is a non zero integer. Additionally, mb is an integer. Therefore p is rational which is contradicts that p is irrational. Hence np is irrational.

d) Proof by cases: We can verify this directly with all the possible orderings for a,b,c. There are six cases:

a≥b≥c, a≥c≥b, b≥a≥c, b≥c≥a, c≥b≥a, c≥a≥b

Writing the details for each one is a bit long. I will give you an example for one case: suppose that c≥b≥a then max(a, max(b,c))=max(a,c)=c. On the other hand, max(max(a, b),c)=max(b,c)=c, hence the statement is true in this case.

e) Direct proof: write a=m/n and b=p/q, with m,q integers and n,q nonnegative integers. Then ab=mp/nq. mp is an integer, and nq is a non negative integer. Hence ab is rational.

f) Direct proof. By part c), √2/n is irrational for all natural numbers n. Furthermore, a is rational, then a+√2/n is irrational. Take n large enough in such a way that b-a>√2/n (b-a>0 so it is possible). Then a+√2/n is between a and b.

g) Direct proof: write m+n=2k and n+p=2j for some integers k,j. Add these equations to get m+2n+p=2k+2j. Then m+p=2k+2j-2n=2(k+j-n)=2s for some integer s=k+j-n. Thus m+p is even.

7 0
3 years ago
Evaluate when x = -3 and y= -1:<br><br><br> x-xy=
LenKa [72]
-3-(-3)(-1)= 0, so 0 is the solution
5 0
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anygoal [31]

Answer:

the answer is 1/2 and then 3/4

3 0
3 years ago
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