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polet
1 month ago
15

How much of a 90% alloy must be combined with 70% gold alloy in order to get 60 ounces of an 85% gold alloy?

Mathematics
1 answer:
Leona [12.6K]1 month ago
4 0
Let X be the amount of 90% alloy and Y be the amount of 70% alloy. The equations are: x + y = 60  0.9x + 0.7y = 0.85 * 60 By substituting, we have: 0.9x + 0.7(60 - x) = 0.85 * 60 This simplifies to: (0.9 - 0.7)x = (0.85 - 0.7)*60 Solving for x yields: x = (0.85 - 0.7)*60/(0.9 - 0.7) x = 45 ounces For Y, we find: y = 60 - 45 y = 15 ounces
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To better understand how husbands and wives feel about their finances, Money Magazine conducted a national poll of 1010 married
Svet_ta [12734]

Answer:

  • a. Refer to the table below
  • b. Refer to the table below
  • c. 0.548
  • d. 0.576
  • e. 0.534
  • f) i) 0.201, ii) 0.208

Explanation:

To begin with, organize the data provided:

Table: "Who excels at obtaining deals?"

                       Who Excels?

Respondent      I Am        My Spouse     We are Equal

Husband           278             127                 102

Wife                   290            111                   102

a. Create a joint probability table and utilize it to respond to the ensuing inquiries.

The joint probability table presents identical details expressed as proportions. The values from the table need to be divided by the total number of responses involved.

1. Total responses: 278 + 127 + 102 + 290 + 111 + 102 = 1,010.

2. Determine each proportion:

  • 278/1,010 = 0.275
  • 127/1,010 = 0.126
  • 102/1,010 = 0.101
  • 290/1,010 = 0.287
  • 111/1,010 = 0.110
  • 102/1,010 = 0.101

3. Construct the table containing these values:

Joint probability table:

Respondent      I Am        My Spouse     We Are Equal

Husband           0.275           0.126                 0.101

Wife                   0.287           0.110                  0.101

This table illustrates that the joint probability of identifying as a husband while choosing 'I am' equals 0.275. Each cell conveys the joint probability associated with each gender's response.

Consequently, this delineates the purpose of a joint probability table.

b. Generate marginal probabilities for Who Excels (I Am, My Spouse, We Are Equal). Provide commentary.

Marginal probabilities are computed for each row and column of the table, indicated in the margins, which is their namesake.

For the column titled "I am," it amounts to: 0.275 + 0.287 = 0.562

Similarly, perform calculations for the other two columns.

For the row designated 'Husband,' it would thus be 0.275 + 0.126 + 0.101 = 0.502. Apply the same for the row labeled 'Wife.'

Table Marginal probabilities:

Respondent      I Am        My Spouse     We Are Equal     Total

Husband           0.275           0.126                 0.101             0.502

Wife                   0.287           0.110              0.101             0.498

Total                 0.562           0.236            0.202             1.000

Notably, when summing the marginal probabilities for both rows and columns, the results will always equate to 1. This is a consistent truth for marginal probabilities.

c. Given the respondent is a husband, what is the likelihood that he believes he is better at securing deals than his wife?

This requires the utilization of conditional probability.

The goal here is to ascertain the probability of the response being "I am" when the respondent identifies as a "Husband."

Using conditional probability:

  • P ( "I am" / "Husband") = P ("I am" ∩ "Husband) / P("Husband")

  • P ("I am" ∩ "Husband) = 0.275 (obtained from the intersection of columns "I am" and rows "Husband")

  • P("Husband") = 0.502 (derived from total of row "Husband")

  • P ("I am" ∩ "Husband) / P("Husband") = 0.275 / 0.502 = 0.548

d. In the instance that the respondent is a wife, what probability exists that she believes she is superior to her husband in acquiring deals?

We seek to identify the probability wherein the response claims "I am" while the respondent is labeled a "Wife," applying the conditional probability formula again:

  • P ("I am" / "Wife") = P ("I am" ∩ "Wife") / P ("Wife")

  • P ("I am" / "Wife") = 0.287 / 0.498

  • P ("I am" / "Wife") = 0.576

e. When responding that "My spouse" is better at scoring deals, what is the likelihood that the claim originated from a husband?

We aim to compute: P ("Husband" / "My spouse")

Applying the conditional probability formula:

  • P("Husband" / "My spouse") = P("Husband" ∩ "My spouse")/P("My spouse")

  • P("Husband" / "My spouse") = 0.126/0.236

  • P("Husband" / "My spouse") = 0.534

f. When the response indicates "We are equal," what likelihood exists that this response is from a husband? What is the chance that it hails from a wife?

What is the likelihood that this response came from a husband?

  • P("Husband" / "We are equal") = P("Husband" ∩ "We are equal") / P ("We are equal")

  • P("Husband" / "We are equal") = 0.101 / 0.502 = 0.201

What is the chance the response originated from a wife:

  • P("Wife") / "We are equal") = P("Wife" ∩ "We are equal") / P("We are equal")

  • P("Wife") / "We are equal") = 0.101 / 0.498 = 0.208
6 0
3 months ago
Billy Jo's school is selling tickets to a Fall Festival. On the first day of ticket sales the school sold 3 adult ticket and 8 s
PIT_PIT [12445]
Y = 6 Here’s a step-by-step explanation: On the initial day of ticket sales, the school sold 3 adult tickets and 8 student tickets for a total of $72. On the following day, the school collected $152 from 7 adult tickets and 16 student tickets. What is the price of a student ticket? Day 1: 3x + 8y = 72 Day 2: 7x + 16y = 152 This leads to a system of equations. By manipulating these equations, we find: Ultimately, y = 6. If you wish to continue from this point, substituting y into the equation will give the value for x.
6 0
2 months ago
Round 75391 to the nearest ten
babunello [11817]
The digit 7 in 75390 is in the ten thousand position, 5 is in the thousand position, 3 is in the hundreds position, 9 is in the tens position, and 1 is located in the ones position.
8 0
2 months ago
Read 2 more answers
Stackable polystyrene cups have a height h1=12.5 cm. Two stacked cups have a height of h2=14 cm. Three stacked cups have a heigh
AnnZ [12381]

Answer:

Approximately 59 stacked cups.

Step-by-step explanation:

We have the following measurements:

Height of one cup = 12.5 cm,

Height of two cups stacked = 14 cm,

Height of three cups stacked = 15.5 cm,

...and so on.

This situation can be described by an arithmetic sequence,

12.5, 14, 15.5,....

The first term is defined as a = 12.5,

with a common difference of d = 1.5 cm.

Thus, the height of x stacked cups is given by

h(x) = a+(x-1)d = 12.5 + (x-1)1.5 = 1.5x + 11

As per the problem,

h(x) = 200

⇒ 1.5x + 11 = 200

⇒ 1.5x = 189

⇒ x = 59.3333333333 ≈ 59.

Therefore, you will need approximately 59 stacked cups.

3 0
3 months ago
Read 2 more answers
A bug was running along a number line at a speed of 11 units per minute. It never changed its direction. If at 7:15 pm it was at
lawyer [12517]
The time from 7:15 to 7:20 is 5 minutes
Since each minute equals 11 units, the total is 5x11=55
Starting from point 100, the new position would be 100+55=155
Final answer: point 155
I hope this is accurate and useful:)
3 0
2 months ago
Read 2 more answers
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