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Katyanochek1 [597]

3 years ago

6

A scale drawing of Joshua's living room is shown below: width 4 length 6 If each 2 cm on the scale drawing equals 4 feet, what a

re the actual dimensions of the room?

1 answer:

krok68 [10]3 years ago

3 0

8ft by 12ft

the real question is why they randomly switched from metric to standard system. :P

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393828294×3838383838229

kirza4 [7]

Answer:

I think this is the answer

Step-by-step explanation:

1511664158726899051326

6 0

3 years ago

10. Contacts. Assume that 30% of students at a university

jonny [76]

Using the <em>normal distribution and the central limit theorem</em>, we have that:

a) A normal model with mean 0.3 and standard deviation of 0.0458 should be used.

b) There is a 0.2327 = 23.27% probability that more than one third of this sample wear contacts.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.

After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

By the Central Limit Theorem, for a <u>proportion p in a sample of size n</u>, the sampling distribution of sample proportion is approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1 - p)}{n}}$ , as long as $np \geq 10$ and $n(1 - p) \geq 10$ .

In this problem:

30% of students at a university wear contact lenses, hence p = 0.3.

We randomly pick 100 students, hence n = 100.

Item a:

$np = 100(0.3) = 30 \geq 10$

$n(1 - p) = 100(0.7) = 70 \geq 10$

Hence a normal model is appropriated.

The mean and the standard deviation are given as follows:

$\mu = p = 0.3$

$s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.3(0.7)}{100}} = 0.0458$

Item b:

The probability is <u>1 subtracted by the p-value of Z when X = 1/3 = 0.3333</u>, hence:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.3333 - 0.3}{0.0458}$

$Z = 0.73$

$Z = 0.73$ has a p-value of 0.7673.

1 - 0.7673 = 0.2327.

0.2327 = 23.27% probability that more than one third of this sample wear contacts.

To learn more about the <em>normal distribution and the central limit theorem</em>, you can take a look at brainly.com/question/24663213

5 0

2 years ago

Jake had $4382 each month from his dad for 36 months how much money did Jake receive in those 36 months

Aloiza [94]

$157,752
$4382 times 36 = $157,752

8 0

3 years ago

Read 2 more answers

Eight percent of all college graduates hired by companies stay with the same company for more than five years. The probability,

Maurinko [17]

Answer:

0.1662 = 16.62% probability that exactly 2 will stay with the same company for more than five years

Step-by-step explanation:

For each graduate, there are only two possible outcomes. Either they stay for the same company for more than five years, or they do not. Graduates are independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Eight percent of all college graduates hired by companies stay with the same company for more than five years.

This means that $p = 0.08$

Sample of 11 college graduates:

This means that $n = 11$

The probability, rounded to four decimal places, that in a random sample of 11 such college graduates hired recently by companies, exactly 2 will stay with the same company for more than five years is: the absolute tolerance is +-0.0001

The tolerance means that the answer is rounded to four decimal places.

The probability is P(X = 2). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{11,2}.(0.08)^{2}.(0.92)^{11} = 0.1662$

0.1662 = 16.62% probability that exactly 2 will stay with the same company for more than five years

4 0

3 years ago

The simple interest formula is I=Prt , where I is the interest, P is the principal, r is the interest rate, and t is the time.

DaniilM [7]

t = 5.3 years
(about 5 years 4 months)

Equation:
t = (1/r)(A/P - 1)
Calculation:
First, converting R percent to r a decimal
r = R/100 = 7.2%/100 = 0.072 per year,
then, solving our equation

t = (1/0.072)((4835.6/3500) - 1) = 5.3
t = 5.3 years

6 0

2 years ago