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17 days ago

Maya spent her allowance on playing an arcade game a few times and riding the Ferris wheel more than once.

2 answers:

9 0

To determine the expense for the arcade game and Ferris wheel, you would require information about how many times Maya played each and the total amount of money she spent from her allowance.

Inessa [3.9K]17 days ago

6 0

Sample answer: To form a two-variable equation, I would first need to find out Maya’s total allowance. Then, identify the prices for playing the arcade game and for the Ferris wheel ride. The equation could express the total cost as the sum of the cost per arcade game times the number of games plus the cost per Ferris wheel ride times the number of rides. The variables would represent how many times Maya played the arcade game and how many times she rode the Ferris wheel. This would allow solving for the number of Ferris wheel rides given the number of arcade plays.

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−4x+60<72 or 14x+11<−31 solve for x

Leona [4166]

The answer to your question is x = 8

5 0

10 days ago

Read 2 more answers

To better understand how husbands and wives feel about their finances, Money Magazine conducted a national poll of 1010 married

Svet_ta [4321]

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

15 days ago

The equation f = v + at represents the final velocity of an object, f, with an initial velocity, v, and an acceleration rate, a,

PIT_PIT [3919]

Starting with the equation f = v + at

Subtract v on both sides:

f - v = at

Divide both sides by a:

(f - v) / a = t

Swap the sides:

t = (f - v) / a

$t = \dfrac{f - v}{a}$

4 0

14 days ago

Read 2 more answers

Water is poured into a conical paper cup at the rate of 3/2 in3/sec (similar to Example 4 in Section 3.7). If the cup is 6 inche

tester [3916]

Response:

The height of the water when it reaches 4 inches is $\frac{3}{8\times \pi} inch/s$ .

Detailed Explanation:

Flow rate of water from the cone = R= $\frac{3}{2} inch^3/s$

Height of the cup = h = 6 inches

Radius of the cup = r = 3 inches

$\frac{r}{h}=\frac{3 inch}{6 inch}=\frac{1}{2}$

r = h/2

Volume of the cone = $V=\frac{1}{3}\pi r^2h$

$V=\frac{1}{3}\pi r^2h$

$\frac{dV}{dt}=\frac{d(\frac{1}{3}\pi r^2h)}{dt}$

$\frac{dV}{dt}=\frac{d(\frac{1}{3}\pi (\frac{h}{2})^2h)}{dt}$

$\frac{dV}{dt}=\frac{1}{3\times 4}\pi \times \frac{d(h^3)}{dt}$

$\frac{dV}{dt}=\frac{1\pi }{12}\times 3h^2\times \frac{dh}{dt}$

$\frac{3}{2} inch^3/s=\frac{1\pi }{12}\times 3h^2\times \frac{dh}{dt}$

h = 4 inches

$\frac{3}{2} inch^3/s=\frac{1\pi }{12}\times 3\times (4inches )^2\times \frac{dh}{dt}$

$\frac{3}{2} inch^3/s=\pi\times 4\times \frac{dh}{dt} inches^2$

$\frac{dh}{dt}=\frac{3}{8\times \pi} inch/s$

The height of the water when it is 4 inches deep is $\frac{3}{8\times \pi} inch/s$ .

6 0

11 days ago

An air show is scheduled for an airport located on a coordinate system measured in miles. The air traffic controllers have close

Leona [4166]

The system of equations that can determine if the commuter jet’s flight path crosses the restricted airspace is:

y = \frac{1}{4}(x - 10)^2 + 6 (i)
y = \frac{-27}{34}x - \frac{5}{17} (ii)

Here's why:

The closed airspace boundary is defined by points (10, 6) and (12, 7).

The commuter jet’s linear path runs from (-18, 14) to (16, -13).

Equation (i) describes the boundary since it fits both (10, 6) and (12, 7):

For (10, 6):
\frac{1}{4}(10-10)^2 + 6 = 6 (true)

For (12, 7):
\frac{1}{4}(12-10)^2 + 6 = 1 + 6 = 7 (true)

Equation (ii) represents the commuter jet’s path as it fits both (-18, 14) and (16, -13):

For (16, -13):
-13 = \frac{-27}{34} \times 16 - \frac{5}{17} = -13 (true)

For (-18, 14):
14 = \frac{-27}{34} \times (-18) - \frac{5}{17} = 14 (true)

By solving this system, we can confirm that the jet’s flight path intersects the closed airspace.

4 0

13 days ago