The sum of two numbers is 65. The smaller number is 11 less than the larger number. What are the numbers?

What single decimal multiplier would you use to increase by 4% followed by a 4% decrease?

Triss [41]

Answer:

if you increase by 4%, it means you will have a new total of 104%. In decimals it is 1.04 (divide your percentage by 100) after that you do the same with the 3% increase. you'll have a new total of 103%. In decimals 1.03. If you want to know what single multiplier you can use, you need to multiply the both decimals. So you get 1.04 x 1.03 = 1.0712

Step-by-step explanation:

5 0

3 years ago

3. State banking officials claim that the probability that a person in Newberg has a checking account

Fiesta28 [93]

Answer:

Step-by-step explanation:

g

3 0

3 years ago

The hourly median power (in decibels) of received radio signals transmitted between two cities

trasher [3.6K]

Using the lognormal and the binomial distributions, it is found that:

The 90th percentile of this distribution is of 136 dB.

There is a 0.9147 = 91.47% probability that received power for one of these radio signals is less than 150 decibels.

There is a 0.0065 = 0.65% probability that for 6 of these signals, the received power is less than 150 decibels.

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

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

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.

In this problem:

The mean is of $\mu = 3.5$ .

The standard deviation is of $\sigma = \sqrt{1.22}$

Question 1:

The 90th percentile is X when Z has a p-value of 0.9, hence X when Z = 1.28.

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

$1.28 = \frac{\ln{X} - 3.5}{\sqrt{1.22}}$

$\ln{X} - 3.5 = 1.28\sqrt{1.22}$

$\ln{X} = 1.28\sqrt{1.22} + 3.5$

$e^{\ln{X}} = e^{1.28\sqrt{1.22} + 3.5}$

$X = 136$

The 90th percentile of this distribution is of 136 dB.

Question 2:

The probability is the p-value of Z when X = 150, hence:

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

$Z = \frac{\ln{150} - 3.5}{\sqrt{1.22}}$

$Z = 1.37$

$Z = 1.37$ has a p-value of 0.9147.

There is a 0.9147 = 91.47% probability that received power for one of these radio signals is less than 150 decibels.

Question 3:

10 signals, hence, the binomial distribution is used.

Binomial probability distribution

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

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

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

For this problem, we have that $p = 0.9147, n = 10$ , and we want to find P(X = 6), then:

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

$P(X = 6) = C_{10,6}.(0.9147)^{6}.(0.0853)^{4} = 0.0065$

There is a 0.0065 = 0.65% probability that for 6 of these signals, the received power is less than 150 decibels.

You can learn more about the binomial distribution at brainly.com/question/24863377

5 0

2 years ago

Help please.........

IrinaVladis [17]

Im on sure but probably 0.75

7 0

3 years ago

Read 2 more answers

If $2940 is invested at 3.2% compounded quarterly, how much is in the account after 8 years?

tino4ka555 [31]

Answer:

The future value is $3793.776

Step-by-step explanation:

Given

$PV = 2940$ --- present value

$r = 3.2\%$ --- rate

$t = 8$ --- time

$n = 4$ --- quarterly

Required

Determine the future value

This is calculated using:

$FV = PV(1 + \frac{r}{n})^{nt$

So, we have:

$FV =2940 * (1 + \frac{3.2\%}{4})^{4*8$

$FV =2940 * (1 + \frac{3.2}{400})^{32$

$FV =2940 * (1 + 0.008)^{32$

$FV =2940 * (1.008)^{32$

$FV =2940 * 1.2904$

$FV =3793.776$

3 0

2 years ago