How many minutes are there in 4 hours and half

Help what’s the perimeter?

KonstantinChe [14]

Answer:

2.8

Step-by-step explanation:

From the similarity ratio we can calculate the missing side length for the smaller rectangle

and when we add all side lengths the answer is 2.8

7 0

3 years ago

(a) has no solutions. (b) has infinitely many solutions. (c) has exactly one solution.

grandymaker [24]

Answer and Step-by-step explanation:

Given

ax + by = c

qx + ry = s

(a) the equation has no solutions if a/q = b/r ≠ c/s, when this happens, we say the system of equations has no solution. For example

x + y = 3

x + y = 5

Subtracting first equation from the second we have:

0 = 2 which is impossible.

(b) the equations have infinite solutions if a/q = b/r = c/s, for example

x + y = 2

Subtracting the first equation from the second we have

0 = 0, since this is always true, then it has infinite solutions.

(c) the equations have unique solutions if a/q ≠ b/r, for example

x + y = 2

x – y = 1

Adding the first and second equation we have

2x = 3, we can get x from here and definitely y, so we have just one solution.

6 0

3 years ago

Question 1

9966 [12]

I know this isn’t a answer or a step by step explanation but can someone please help and explained this it’s so confusing and hard

8 0

2 years ago

Will give Brainliest! if right! Use the box-and-whisker plot find the range of the data.

k0ka [10]

Range = 100 - 50 = 50

answer is 50

3 0

3 years ago

Read 2 more answers

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