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frozen [14]
3 years ago
12

Solve the system of linear equations. (enter your answers as a comma-separated list. if there is no solution, enter no solution.

if the system has an infinite number of solutions, set w
Mathematics
1 answer:
Neporo4naja [7]3 years ago
7 0
There is no system to solve for
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While Preparing For A Morning Conference, Principal Corsetti Is Laying Out 8 Dozen Bagels On Square Plates. Each Plate Can Hold
USPshnik [31]
We have 8 dozen bagels, or 8*12=96 bagels.  Each plate can hold 14 bagels, so we have enough bagels to fill 96/14=about 6.86 plates.  However, we cannot have a fraction of a plate, so we round up to have a total of seven plates.  To fill all seven plates fully, 7*14=98 bagels would be needed, which is two more than we have.

To summarize, Mr. Corsetti has seven plates of bagels, and would need two more bagels to fill the last one up.
6 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
2 years ago
Yassar has purchased a $30,000 car for his business. The car depreciates at 30%
fiasKO [112]

Answer:

A

Step-by-step explanation:

30% of 30000 is 9000, 30000-9000=21000 (1 year)

30% of 21000 is 6300, 21000-6300=14700 (2 years)

30% of 14700 is 4410, 14700-4410= 10290 (3 years)

30% of 10290 is 3087, 10290-3097= 7203 (4 years)

30% of 7203 is 2160.9, 7203-2160.9= 5042.1 (5 years)

5 0
2 years ago
Which should come next at the end of this row of letters: a b d g?
Sav [38]
Refer to the figure shown below.

From a to b, we skip 1 letter.
From b to d, we skip 2 letters.
From d to g, we skip 3 letters.

Inductively, we should skip 4 letters to arrive at the next letter, which is k.

Answer: k

5 0
3 years ago
Read 2 more answers
1/6 + 2/3 tell me wat it is pls​
Nutka1998 [239]

Answer: 5/6

Step-by-step explanation:

2/3=4/6

1/6 + 4/6 = 5/6

5 0
3 years ago
Read 2 more answers
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