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3 years ago

5

Phillip deposited 3400 into a savings account that pays 4 percent annual interest. how much will he have in his account at the e

nd of the year

1 answer:

jarptica [38.1K]3 years ago

3 0

$42432 at the end of the year

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Please help me out with this :)

mr Goodwill [35]

Set them equal to each other because they are parallel

3x+7=5x-17

Subtract 3x from both sides

7=2x-17

Add 17 on both sides

24=2x

Divide by 2 on both sides

12=x

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3 years ago

The difference of twelve times a number and nine times a number

inna [77]

Since the problem states "a number" and again "a number", I assume it;'s the same number. Let the number be x.

12x - 9x = 3x

The difference of twelve times a number and nine times a number is three times the number.

6 0

3 years ago

Read 2 more answers

Critical Thinking: Empirical/Quantitative Skills

aliya0001 [1]

Answer:

1. 0.0910 = 9.10% probability that exactly 180 passengers show up for the flight.

2. 0.4522 = 45.22% probability that at most 180 passengers show up for the flight.

3. 0.5478 = 54.78% probability that more than 180 passengers show up for the flight.

Step-by-step explanation:

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.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

Assume that there is a 0.905 probability that a passenger with a ticket will show up for the flight.

This means that $p = 0.905$

Also assume that the airline sells 200 tickets

This means that $n = 200$

Question 1:

Exactly, so we can use the P(X = x) formula, to find P(X = 180).

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

$P(X = 180) = C_{200,180}.(0.905)^{180}.(0.095)^{20} = 0.0910$

0.0910 = 9.10% probability that exactly 180 passengers show up for the flight.

2. When 200 tickets are sold, calculate the probability that at most 180 passengers show up for the flight.

Now we have to use the approximation.

Mean and standard deviation:

$\mu = E(X) = np = 200*0.905 = 181$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.905*0.095} = 4.15$

Using continuity correction, this is $P(X \leq 180 + 0.5) = P(X \leq 180.5)$ , which is the p-value of Z when X = 180.5. Thus

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

$Z = \frac{180.5 - 181}{4.15}$

$Z = -0.12$

$Z = -0.12$ has a p-value of 0.4522.

0.4522 = 45.22% probability that at most 180 passengers show up for the flight.

3. When 200 tickets are sold, calculate the probability that more than 180 passengers show up for the flight.

Complementary event with at most 180 passengers showing up, which means that the sum of these probabilities is 1. So

$p + 0.4522 = 1$

$p = 1 - 0.4522 = 0.5478$

0.5478 = 54.78% probability that more than 180 passengers show up for the flight.

4 0

3 years ago

Please help!!!!!!!!!!

Degger [83]

I hope this helps you

4 0

3 years ago

. In the U.S., from 2004−2015, the correlation coefficient for the relationship between the size of a cell phone data plan, x, a

bonufazy [111]

Answer: strong positive correlafion between data plan size 'x' and number of text messages sent 'y'

Step-by-step explanation:

'R' in statistics is used to denote correlation Coefficient. The correlation Coefficient is a value which ranges between -1 to +1. It tells us the level of relationship or correlation which exists between the relative movement of two variables, in this case the relationship between data plan size and the number of text messages sent in the US. R value of 0 depicts that no relationship exists between the two variables, R value closer the R value is to +1 and - 1 depicts the strength of positive and negative correlation of the two variables respectively.

A R value of +0.97 in the context above, depicts a strong positive correlation between data plan size and number of text messages sent in the US. That is large data size usually corresponds to large number of text messages and vice versa.

4 0

3 years ago