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zhannawk [14.2K]
2 years ago
6

Based on the graph, which statements are correct? check all that apply. Voters age 55–64 were more likely to use nontraditional

voting methods than people age 45–54. Voters age 65 and older were more likely to use nontraditional voting methods than all other age groups. Men were more likely to use nontraditional voting methods than women. Voters with less than a high school education were less likely to use nontraditional voting methods than voters in all other educational groups
SAT
2 answers:
SVETLANKA909090 [29]2 years ago
8 0

Based on the graph, the correct statements are;

A: Voters age 55–64 were more likely to use nontraditional voting methods than people age 45–54.

B: Voters age 65 and older were more likely to use nontraditional voting methods than all other age groups.

<h3>What is the correct statement about the graph?</h3>

The attached graph shows us that the likelihood of the 55–64 aged group using non-traditional voting methods is 69.7 as against 65.9 scored by the 45–54 age group.  

Secondly, on the graph, the age group with the highest likelihood to use non-traditional voting methods is voters aged 65 and older with 77.7.  

Finally, the graph shows us that age has a great influence on voters in choosing non-traditional voting methods over traditional methods.

Read more about Correct graph statement at; brainly.com/question/24992348

Morgarella [4.7K]2 years ago
4 0

Answer:

A. Voters age 55–64 were more likely to use nontraditional voting methods than people age 45–54.

B. Voters age 65 and older were more likely to use nontraditional voting methods than all other age groups.

Explanation:

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Antique Accents tracks their daily profits and has found that the distribution of profits is approximately normal with a mean of
mezya [45]

Answer;

a) 0.434

b) 0.983

c) 0.367

Explanation:

The exact question with the given parameters wasn't obtained online, but the same question, albeit with different parameters is then obtained. Hopefully, this Helps to solve the complete question with the required parameters.

Antique Accents tracks their daily profits and has found that the distribution of profis is approximately normal with a mean of $17,700.00 and a standard deviation of about $900.00. Using this information, answer the following questions For full marks your answer should be accurate to at least three decimal places. Compute the probability that tomorrow's profit will be

a) less than $16,791 or greater than $18,231

b) greater than $15,783

c) between $17,997 and $20,130

Solution

This is a normal distribution problem with

Mean = μ = $17,700

Standard deviation = σ = $900

a) less than $16,791 or greater than $18,231. P(x < 16,791) or P(X > 18,231) = P(X < 16,791) + P(x > 18,231)

We first standardize 16,791 and 18,231

The standardized score for any value is the value minus the mean then divided by the standard deviation.

For 16791

z = (x - μ)/σ = (16791 - 17700)/900 = - 1.01

For 18231

z = (x - μ)/σ = (18231 - 17700)/900 = 0.59

To determine the required probability

P(X < 16,791) + P(x > 18,231) = P(z < -1.01) + P(z > 0.59)

We'll use data from the normal probability table for these probabilities

P(X < 16,791) + P(x > 18,231) = P(z < -1.01) + P(z > 0.59)

P(z < -1.01) = 0.15625

P(z > 0.59) = 1 - (z ≤ 0.59) = 1 - 0.7224 = 0.2776

P(X < 16,791) + P(x > 18,231) = P(z < -1.01) + P(z > 0.59) = 0.15625 + 0.2776 = 0.43385 = 0.434 to 3 d.p

b) greater than $15,783. P(x > 15783)

We standardize 15783

z = (x - μ)/σ = (15783 - 17700)/900 = -2.13

To determine the required probability

P(x > 15783) = P(z > -2.13)

We'll use data from the normal probability table for this probability

P(x > 15783) = P(z > -2.13) = 1 - P(z ≤ - 2.13)

= 1 - 0.01659 = 0.98341 = 0.983 to 3 d.p.

c) between $17,997 and $20,130.

P(17,997 < x < 20,130)

We first standardize 17,997 and 20,130

The standardized score for any value is the value minus the mean then divided by the standard deviation.

For 17,997

z = (x - μ)/σ = (17,997 - 17700)/900 = 0.33

For 20,130

z = (x - μ)/σ = (20,130 - 17700)/900 = 2.70

To determine the required probability

P(17,997 < x < 20,130) = P(0.33 < x < 2.70)

We'll use data from the normal probability table for these probabilities

P(17,997 < x < 20,130) = P(0.33 < x < 2.70)

= P(z < 2.70) - P(z < 0.33)

= 0.99653 - 0.62930

= 0.36723 = 0.367 to 3 d.p.

Hope this Helps!!!

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