I need help because i’m not smart :)

During April of 2013, Gallup randomly surveyed 500 adults in the US, and 47% said that they were happy, and without a lot of str

Brilliant_brown [7]

Answer:

number of successes

$k = 235$

number of failure

$y = 265$

The criteria are met

A

The sample proportion is $\r p = 0.47$

B

$E =4.4 \%$

C

What this mean is that for N number of times the survey is carried out that the which sample proportion obtain will differ from the true population proportion will not more than 4.4%

Ci

$r = 0.514 = 51.4 \%$

$v = 0.426 = 42.6 \%$

D

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

E

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

F

Yes our result would support the claim because

$\frac{1}{3 } \ of N < \frac{1}{2} (50\%) \ of \ N , \ Where\ N \ is \ the \ population\ size$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 500$

The sample proportion is $\r p = 0.47$

Generally the number of successes is mathematical represented as

$k = n * \r p$

substituting values

$k = 500 * 0.47$

$k = 235$

Generally the number of failure is mathematical represented as

$y = n * (1 -\r p )$

substituting values

$y = 500 * (1 - 0.47 )$

$y = 265$

for approximate normality for a confidence interval criteria to be satisfied

$np > 5 \ and \ n(1- p ) \ >5$

Given that the above is true for this survey then we can say that the criteria are met

Given that the confidence level is 95% then the level of confidence is mathematically evaluated as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha =0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{ \alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{\r p (1- \r p}{n} }$

substituting values

$E = 1.96 * \sqrt{ \frac{0.47 (1- 0.47}{500} }$

$E = 0.044$

=> $E =4.4 \%$

What this mean is that for N number of times the survey is carried out that the proportion obtain will differ from the true population proportion of those that are happy by more than 4.4%

The 95% confidence interval is mathematically represented as

$\r p - E < p < \r p + E$

substituting values

$0.47 - 0.044 < p < 0.47 + 0.044$

$0.426 < p < 0.514$

The upper limit of the 95% confidence interval is $r = 0.514 = 51.4 \%$

The lower limit of the 95% confidence interval is $v = 0.426 = 42.6 \%$

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

$\frac{1}{3 } < \frac{1}{2} (50\%)$

3 0

3 years ago

Sove -5x + 3 = 2x - 1

DiKsa [7]

Answer:

4/7

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

30 pointss!HELP! Two students get into an argument about the correct answer.

Mamont248 [21]

Both Rhianna and Christopher were correct as they took different reference angles.

Step-by-step explanation:

Step 1:

$tan \theta = \frac{opposite side}{adjacentside} .$

The value of the tan of a reference angle is calculated using the above formula.

The hypotenuse is the longest side of the triangle is always opposite the right angle of the triangle.

In this triangle, the hypotenuse is the side measuring 5 units.

Step 2:

When the reference angle is A, the opposite side measures 3 units and the adjacent side measures 4 units.

So $tan A = \frac{3}{4} .$

When the reference angle is C, the opposite side measures 4 units and the adjacent side measures 3 units.

So $tan C = \frac{4}{3} .$

So both Rhianna and Christopher were right as they took different reference angles i.e. A and C.

7 0

3 years ago

Male and female high school students reported how many hours they worked each week in summer jobs. The data is represented in th

mihalych1998 [28]

Question:

1. The females worked less than the males, and the female median is close to Q1.

2. There is a high data value that causes the data set to be asymmetrical for the males.

3. There are significant outliers at the high ends of both the males and the females.

4. Both graphs have the required quartiles.

Answer:

The correct option is;

1. The females worked less than the males, and the female median is close to Q1

Step-by-step explanation:

Based on the given data, we have;

For males

Minimum = 0

Q1 = 1

Median or Q2 = 20

Q3 = 25

Maximum = 50

For females;

Minimum = 0

Q1 = 5

Median or Q2 = 6

Q3 = 10

Maximum = 18

Therefore, the values of data that affect the statistical measures of spread and center are that

The females worked less than the males as such the statistical data for the females have less variability than the males in terms of interquartile range

Also the female median is very close to Q1, therefore it affects the definition of a measure of center.

5 0

3 years ago

In Math, what does justify your answer using complete sentences mean?

const2013 [10]

'justify your answer using complete sentences' means re-write your answer in a complete sentence. In other words, say your answer in a full and clear sentence.

5 0

3 years ago