Which statement is true about the result of a rigid transformation?

A bank is reviewing its risk management policies with regards to mortgages. To minimize the risk of lending, the bank wants to c

agasfer [191]

1. μ = 306,500

2.σ = 24,500

3. n = 150

4. μ $_x$ = $306,500

5. σ $_x$ = $2000

<h3>What is Standard deviation?</h3>

The standard deviation serves as a gauge for the degree of variation or dispersion among a group of numbers. While a high standard deviation suggests that the values are dispersed throughout a wider range, a low standard deviation suggests that the values tend to be close to the mean of the collection.

With a standard deviation of $24,500 and an average mortgage debt of $306,500, Americans have a median mortgage debt.

<h3>According to the given information:</h3>

The population's average is

μ = 306,500

Standard deviation for the general population is

σ = 24500

Imagine 150 Americans are chosen at random for the sample.

n= 150

The sample mean is roughly normally distributed as a result of the high sample size, which is supported by the central limit theorem.

The population mean and sample mean would be the same,

μ $_n$ = μ =306,500

Here is how to calculate the sample standard deviation:

$\sigma_{x}=\frac{\sigma}{\sqrt{n}}$

the sample size is n, and is the population standard deviation.

$\begin{aligned}&\sigma_{x}=\frac{24,500}{\sqrt{150}} \\&\sigma_{x}=\$ 2,000\end{aligned}$

The necessary conditions are thus:

1. μ = 306,500

2.σ = 24,500

3. n = 150

4. μ $_x$ = $306,500

5. σ $_x$ = $2000

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I understand that the question you are looking for is :

A bank is reviewing its risk management policies with regards to mortgages. To minimize the risk of lending, the bank wants to compare the typical mortgage owed by their clients against other homebuyers. The average mortgage owed by Americans is $306,500, with a standard deviation of $24,500. Suppose a random sample of 150 Americans is selected. Identify each of the following, rounding your answers to the nearest cent when appropriate:

1. $1. $mu=?\\2. $sigma=?\\3. $=n=$\\4. $mu_{overlinex}=$x=?\\5. $sigma_{overlinex}=$x=?$

4 0

2 years ago

Which of the following represents the additive inverse of 125? A. -(-125) B. 125 C. -125 D. 1/25

Elden [556K]

Additive inverse of 125 is – 125, option C is correct.

Solution:

Given that , a number is 125

We have to find the additive inverse for the above given number from the above given set of options.

Now, we know that, sum of a number and its additive inverse equals to 0.

So, let us check option by option.

Option A:

Given number is -(-125)

-(-125) ⇒ 125 + ( - ( - 125 ) )

⇒ 125 + 125

⇒ 250 ⇒ wrong option

Option B:

Given number is 125

⇒ 125 + 125 ⇒ 250 ⇒ wrong option

Option C:

Given number is -125

125 + ( - 125 ) ⇒ 125 – 125 ⇒ 0 ⇒ correct option

Hence, additive inverse of 125 is – 125, option C is correct

7 0

4 years ago

Read 2 more answers

Find the perimeter. Simplify your answer. 4a-10 1 a-3 a-3 4a-10

kaheart [24]

2(a - 3) + 2(4a - 10) =

= 2a - 6 + 8a - 20 =

= 2a + 8a - 6 - 20 = 10a - 26 ← the end

4 0

3 years ago

Read 2 more answers

Iridium -192 has a half life of 74 days after 148 days, how many milligrams of 900 mg sample will remain?

Ad libitum [116K]

$\bf \textit{Amount for Exponential Decay using Half-Life}
\\\\
A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{initial amount}\to &900\\
t=\textit{elapsed time}\to &148\\
h=\textit{half-life}\to &74
\end{cases}
\\\\\\
A=900\left( \frac{1}{2} \right)^{\frac{148}{74}}\implies A=900\left( \frac{1}{2} \right)^2\implies A=225$

8 0

4 years ago

When 3015 adults were surveyed in a poll, 73% said that they use the internet. is it okay for a newspaper reporter to write th

Serjik [45]

Answer:

Claim is rejected

Step-by-step explanation:

Solution:-

- The claim was made by the newspaper reporter " 3 divided by 4 of all adults use the internet " the proportion of people who are claimed to use internet are p = 0.75.

- A random sample was taken of N = 305 individuals were surveyed in a poll.

- We are to test the claim made by the reporter for the sample N.

- State the hypothesis for the effectiveness of medication:

Null Hypothesis: p = 0.75

Alternate hypothesis: p ≠ 0.75

- The conditions of standard normality:

n*p > 5 , 3015*0.75 = 2261.25 > 5 .. ( Ok )

n*(1-p) > 5 , 3015*0.25 = 753.75 > 5 .. ( Ok )

The standard normal test is applicable since normal approximation to binomial distribution for a fairly large sample size N = 3015 adults.

Assuming the population proportion to be normally distributed.

- We will estimate the population proportion with the sample proportion obtained from a poll survey p* = 0.73

- Testing against the claimed population proportion ( p ) = 0.75. The standard normal statistic value is given by:

$Z-test = \frac{p* - p}{\sqrt{p*( 1 -p) / N} } \\\\Z-test = \frac{0.73 - 0.75}{\sqrt{0.75*0.25 / 3015} } \\\\Z-test = -\frac{0.02}{0.00788 } \\\\Z-test = -2.53807$

- We will see whether the Z-test statistic falls in the rejection region defined by the critical value of Z at significance level ( α ) of 0.05.

- The rejection region is defined by the Alternate hypothesis which is not equal to the reporter's claimed value. So, the rejection region defined by the lower and upper tail of the standard normal.

- So for two - tailed test the critical value of statistics is:

P ( Z < ±Z-critical ) = α / 2 = 0.025

Z-critical = ± 1.96

- The rejected values all lie to the left or right of the Z-critical value ±1.96

- The claim test value is compared the rejection region:

-2.53807 < -1.96

Z-test < Z-critical

Hence, Null hypothesis rejected because test lies in the rejection region.

Conclusion:

The Null hypothesis or claim made by the reporter that 3 out of 4 adults use internet i.e 75% use internet is without sufficient evidence. Hence, the claim made is false or has no statistical evidence.

6 0

3 years ago