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ycow [4]
2 years ago
15

A sausage company makes two different kinds of hot dogs, regular and all-beef. Each pound of all-beef hot dogs requires 0.75 lb

of beef and 0.2 lb of spices, and each pound of regular hot dogs requires 0.18 lb of beef, 0.3 lb of pork, and 0.2 lb of spices. Suppliers can deliver at most 1020 lb of beef, at most 600 lb of pork, and at least 500 lb of spices. If the profit is $1.50 on each pound of all-beef hot dogs and $1.00 on each pound of regular hot dogs, how many pounds of each should be produced to obtain maximum profit? regular hot dogs all-beef hot dogs What is the maximum profit?

Mathematics
1 answer:
Gre4nikov [31]2 years ago
8 0

Answer:

  • 880 lbs of all-beef hot dogs
  • 2000 lbs of regular hot dogs
  • maximum profit is $3320

Step-by-step explanation:

We can let x and y represent the number of pounds of all-beef and regular hot dogs produced, respectively. Then the problem constraints are ...

  • .75x + 0.18y ≤ 1020 . . . . . . limit on beef supply
  • .30y ≤ 600 . . . . . . . . . . . . . limit on pork supply
  • .2x + .2y ≥ 500 . . . . . . . . . . limit on spice supply

And the objective is to maximize

  p = 1.50x + 1.00y

The graph shows the constraints, and that the profit is maximized at the point (x, y) = (880, 2000).

2000 pounds of regular and 880 pounds of all-beef hot dogs should be produced. The associated maximum profit is $3320.

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Your band is one of 8 bands competing at the battle of the bands. The order of the performances is determined at random. The fir
lorasvet [3.4K]

Answer:

1.79 %.

Step-by-step explanation:

you have to calculate the probability of a series of events before you can calculate the final probability.

The first event is that the band plays on Friday.

If there are 8 bands, but only 5 play on Friday, then the probability that they will play on Friday is

5/8

Now, the second event that they are the last ones is 1/5, since they are 5 bands.

Therefore the probability that they will play on Saturday and last is

5/8 * 1/5 = 1/8

Now, for the rival band it would be, knowing that on Friday there is already a quota assigned and one band less, the probability that they will play that day is

4/7

as they play before the other group, it would be 1/4, as there is already a quota assigned.

The final probability would then be:

4/7 * 1/4 = 1/7

Now the probability that the band gives the last performance on Friday night and your rival band performs immediately before them, is:

1/8 * 1/7 = 0.0179

That is 1.79 %.

3 0
2 years ago
Matthew ran 3/8 mile and then walked 7/10 mile. Which pair of fractions can he use to find how far he went in all? (A 15/40 and
Strike441 [17]
Before we could add these numbers, 3/8 and 7/10 need a common denominator. Both 8 and 10 go into 40.

8 goes into 40 five times
3/8= (3*5)/40 = 15/40

10 goes into 40 four times
7/10= (7*4)/40= 28/40

ANSWER: A) 15/40 and 28/40

Hope this helps! :)
7 0
2 years ago
Read 2 more answers
Suppose that the universal set is U={1,2,3,4,5,6,7,8,9,10}. Express each of the following subsets with bit strings (of length 10
Vladimir [108]

Answer:

0011100000

1010010001

0111001110

Step-by-step explanation:

As the question is not complete, Here is the complete question.

Suppose that the universal set is U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Express each of these sets with bit strings where the ith bit in the string is 1 if i is in the set and 0 otherwise.

a) {3, 4, 5}

b) {1, 3, 6, 10}

c) {2, 3, 4, 7, 8, 9}.

So, we need to express a) b) and c) into bit strings.

Firstly, number of elements in the universal set represent the number of bits in the bit string.

Secondly, 1 = yes element is present in both universal set as well as in sub set.

0 = No, element is not present in sub set but present in universal set.

Hence, we have:

a) Sub set {3,4,5} = 0011100000  (As there are 3 1's which means only 3,4,5 are present in both universal set and subset.

Similarly,

b) Sub set {1, 3, 6, 10} = 1010010001

c) Sub set {2, 3, 4, 7, 8, 9} = 0111001110

5 0
1 year ago
Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation: 22, 18, 1
BigorU [14]

Answer:

a:  Sample mean: 20, Sample standard deviation: 1.732

b:  19.13 <µ < 20.87

c:  18.84< µ < 21.16

d:  as the confidence level increases, the interval becomes wider

Step-by-step explanation:

a:  Sample mean is found by adding up all the individual values of the sample, then dividing by the total number of values in the sample.  

Sample standard deviation is the square root of the variance

See the first attached photo for the calculations of these values.

We need to create a 80% confidence interval for the population.  Since n < 30, we will use a t-value with degree of freedom of 7 (the degree of freedom is always one less than the sample size.  

Look for the column on the t-distribution chart that has "area in two tails" of 0.20 (80%), and row 7 (degree of freedom)

The t-value is 1.415

See the second attached photo for the construction of the confidence interval

We need to create a 90% confidence interval for the population.  Since n < 30, we will use a t-value with degree of freedom of 7 (the degree of freedom is always one less than the sample size.  

Look for the column on the t-distribution chart that has "area in two tails" of 0.10 (90%), and row 7 (degree of freedom)

The t-value is 1.895

See the third attached photo for the construction of the confidence interval

3 0
1 year ago
PLEASE HELP MATH EXPAND LOG
blsea [12.9K]

Answer:

½log3 + ½logx

Step-by-step explanation:

½(log(3x))

½(log3 + logx)

½logx + ½log3

8 0
1 year ago
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