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tekilochka [14]
3 years ago
11

What decimal is shown by the shaded area in the model​

Mathematics
1 answer:
Arisa [49]3 years ago
8 0

Answer:

.001

OR

1/100

OR

one hundredth

(remember the "th")

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What is the work and answer to this problem -8(p-8)+8p=-7p+36
natulia [17]
-8p +64 +8p = 7p +36
64 = 7p + 36
64-36 = 7p +36 -36
28=7p
28/7 = 7/7p
4=p

8 0
4 years ago
there were 2,605 peolpe at the basketball game. a reporter rpunded thi mumber to the nearest hundred for a newspaper article. wh
Lostsunrise [7]
The reporter used the number 2,605.

2,605 is closer to 2600 then it is to 2700.
4 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 <em>lognormal </em>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 <u>X when Z = 1.28.</u>

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 <u>p-value of Z when X = 150</u>, 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
3 years ago
In an office building, 93 offices are currently being rented. This represents 60% of the total units. How many offices are there
gavmur [86]

There are 155 offices. 93 is 60% of 155

5 0
3 years ago
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professor190 [17]
Hope this helps!
1,366-1,366
This is = to 0
7 0
2 years ago
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