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2 years ago

A. The average yearly salary of a lawyer is $24 thousand less than twice that of an architect.

Mathematics

1 answer:

erastovalidia [21]2 years ago

4 0

Answer:

The average yearly salary of the architect is $62,000 and the lawyer is $100,000.

Step-by-step explanation:

Given:

The average yearly salary of a lawyer is $24 thousand less than twice that of an architect.

Combined, an architect and a lawyer earn $ 210 thousand.

Now, to find the average yearly salary of an architect and a lawyer.

Let the average yearly salary of an architect be $x.$

So, the average yearly salary of the lawyer = $2x-24,000.$

Combined, architect and lawyer earn = $\$210,000.$

Now, to get the average salary of the architect and lawyer:

$x+(2x-24000)=210000$

$x+2x-24000=210000\\\\3x-24000=210000$

Adding both sides 24000 we get:

$3x=186,000\\\\Dividing\ both\ sides\ by\ 3\ we\ get:\\\\x=62000.$

The average salary of architect = $62,000.

Now, to get the average salary of lawyer by substituting the value of $x$ :

$2x-24000\\\\=2(62000)-24000\\\\=124000-24000\\\\=\$100,000.$

The average salary of lawyer = $100,000.

Therefore, the average yearly salary of the architect is $62,000 and the lawyer is $100,000.

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A standard weight known to weigh 10 grams. Some suspect bias in weights due to manufacturing process. To assess the accuracy of

notsponge [240]

Answer:

a) The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams

b) 49 measurements are needed.

Step-by-step explanation:

Question a:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.98}{2} = 0.01$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.01 = 0.99$ , so Z = 2.327.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 2.327\frac{0.0003}{\sqrt{5}} = 0.00031$

The lower end of the interval is the sample mean subtracted by M. So it is 10.0044 - 0.00031 = 10.00409 grams

The upper end of the interval is the sample mean added to M. So it is 10 + 0.00031 = 10.00471 grams

The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams.

(b) How many measurements must be averaged to get a margin of error of +/- 0.0001 with 98% confidence?

We have to find n for which M = 0.0001. So

$M = z\frac{\sigma}{\sqrt{n}}$

$0.0001 = 2.327\frac{0.0003}{\sqrt{n}}$

$0.0001\sqrt{n} = 2.327*0.0003$

$\sqrt{n} = \frac{2.327*0.0003}{0.0001}$

$(\sqrt{n})^2 = (\frac{2.327*0.0003}{0.0001})^2$

$n = 48.73$

Rounding up

49 measurements are needed.

7 0

1 year ago

"Majesty Video Production Inc. wants the mean length of its advertisements to be 30 seconds. Assume the distribution of ad lengt

Westkost [7]

Answer:

a) $\bar X \sim N(\mu=30, \frac{2}{\sqrt{16}})$

b) $Se=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{16}}=0.5$

c) $P(\bar X >31.25)=0.006=0.6\%$

d) $P(\bar X >28.25)=0.9997=99.97\%$

e) $P(28.25$

Step-by-step explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variabl length of advertisements produced by Majesty Video Production Inc. We know from the problem that the distribution for the random variable X is given by:

$X\sim N(\mu =30,\sigma =2)$

We take a sample of n=16 . That represent the sample size.

a. What can we say about the shape of the distribution of the sample mean time?

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is also normal and is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\bar X \sim N(\mu=30, \frac{2}{\sqrt{16}})$

b. What is the standard error of the mean time?

The standard error is given by this formula:

$Se=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{16}}=0.5$

c. What percent of the sample means will be greater than 31.25 seconds?

In order to answer this question we can use the z score in order to find the probabilities, the formula given by:

$z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}$

And we want to find this probability:

$P(\bar X >31.25)=1-P(\bar X$

d. What percent of the sample means will be greater than 28.25 seconds?

In order to answer this question we can use the z score in order to find the probabilities, the formula is given by:

$z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}$

And we want to find this probability:

$P(\bar X >28.25)=1-P(\bar X$

e. What percent of the sample means will be greater than 28.25 but less than 31.25 seconds?"

We want this probability:

$P(28.25$

3 0

2 years ago

David drove a distance (d) of 187 km, correct to 3 significant figures.

Bogdan [553]

Answer: 6.68km/litre

Step-by-step explanation:

Distance(d) = 187km ( to 3 significant figures)

Petrol(p) = 28litres ( to 2 significant figures)

A number, x, rounded to 1 significant figure is 200

Write down the error interval for x

Calculating the lower and upper bound for both 'd' and 'p'

Upper bound 'd' = 187.5km

Lower bound 'd' = 186.5km

Upper bound 'p' = 28.5km

Lower bound 'p' = 27.5km

Therefore,

Petrol consumption (C) = d/p

Upper C = 187.5 / 28.5 = 6.57894

Lower C = 186.5 / 27.5 = 6.781818

Accuracy answer for C = (6.57894 + 6.781818) / 2 = 6.680379

= 6.68km/ litre ( to 3 significant figures)

To achieve a reasonable level of accuracy, the upper and lower bounds of P and D was used to calculate C, which was then averaged.

Using 3 significant figures ensure that we capture the C better, instead of early rounding.

7 0

2 years ago

Next you need to find the critical points. What are the critical points, and how are they found?

m_a_m_a [10]

For x-25/5(10+x) less than or equal to 0 .

The critical points are found by setting the numerator and denominator to equal 0.

The critical points are at -10 and 25.

8 0

2 years ago

Read 2 more answers

Find the quotient: 5) 16 tons 400 pounds

yan [13]

Answer:

after we divide one number by another. dividend ÷ divisor = quotient.

Step-by-step explanation:

12 ÷ 3 = 4, 4 is the quotient. hope this helps you :)

4 0

2 years ago