All categories

3 years ago

What is c3 - 1 when c = 4

Mathematics

2 answers:

den301095 [7]3 years ago

6 0

The answer to your question is 11.
4x3=12-1=11

weeeeeb [17]3 years ago

4 0

You just have to substitute
c3-1
4*3-1
12-1
=11

You might be interested in

Help, write an expression in the simplest form that represents the area (in square feet) of the banner.

k0ka [10]

Answer:

3 (3 + x) is equal to the banner

6 0

3 years ago

Read 2 more answers

Find the margin of error for a 90% confidence interval when the standard deviation is LaTeX: \sigma= 50????=50 and LaTeX: n = 25

Murrr4er [49]

Answer:

The margin of error for a 90% confidence interval is 16.4

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 25

Standard deviation = 50

$z_{critical}\text{ at}~\alpha_{0.10} = \pm 1.64$

Margin of error =

$z_{critical}\times \dfrac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$1.64\times \dfrac{50}{\sqrt{25}} = 16.4$

Thus, the margin of error for a 90% confidence interval is 16.4

8 0

3 years ago

According to a 2018 survey by Bankrate, 20% of adults in the United States save nothing for retirement (CNBC website). Suppose t

nalin [4]

Answer:

0.5981 = 59.81% probability that three or less of the selected adults have saved nothing for retirement

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they save nothing for retirement, or they save something. The probability of an adult saving nothing for retirement is independent of any other adult. This means that the binomial probability distribution is used 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.

20% of adults in the United States save nothing for retirement (CNBC website).

This means that $p = 0.2$

Suppose that sixteen adults in the United States are selected randomly.

This means that $n = 16$

What is the probability that three or less of the selected adults have saved nothing for retirement?

This is:

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

In which

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

$P(X = 0) = C_{16,0}.(0.2)^{0}.(0.8)^{16} = 0.0281$

$P(X = 1) = C_{16,1}.(0.2)^{1}.(0.8)^{15} = 0.1126$

$P(X = 2) = C_{16,2}.(0.2)^{2}.(0.8)^{14} = 0.2111$

$P(X = 3) = C_{16,3}.(0.2)^{3}.(0.8)^{13} = 0.2463$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0281 + 0.1126 + 0.2111 + 0.2463 = 0.5981$

0.5981 = 59.81% probability that three or less of the selected adults have saved nothing for retirement

4 0

3 years ago

Nine actors audition for four different parts in a play in how many ways can the director fill these roles ?

podryga [215]

Answer:126

Step-by-step explanation: your welcome

5 0

3 years ago

Assume that 33.4% of people have sleepwalked. Assume that in a random sample of 1459 adults, 526 have sleepwalked.

lutik1710 [3]

Answer:

Step-by-step explanation:

Given that 33.4% of people have sleepwalked.

Sample size n =1459

Sample favourable persons = 526

Sample proportion p = $\frac{526}{1459} \\=0.361$

Sample proportion p is normal for large samples with mean = 0.334 and

std error = $\sqrt{\frac{pq}{n} } \\=\sqrt{\frac{0.334(1-0.334)}{1459} } \\=0.0123$

a) P(526 or more of the 1459 adults have sleepwalked.)

$=P(p\geq 0.361)\\=P(Z\geq \frac{0.361-0.334}{0.0123} \\=P(Z\geq 2.20)\\=0.5-0.4861\\=0.0139$

b) Yes, because hardly 1.4% is the probability

c) 33.4 is very less compared to the average. Either sample should be improved representing the population or population mean should be increased accordingly.

7 0

3 years ago