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Leno4ka [110]
2 years ago
10

Which of the following rational numbers is equal to 3 point 1 with bar over 1?

Mathematics
2 answers:
Alecsey [184]2 years ago
8 0

Answer:

28/9

Step-by-step explanation:

22/9 = 2.444444...

24/9 = 2.66666...

26/9 = 2.88888...

28/9 = 3.1111111...

Arisa [49]2 years ago
5 0

Answer:

28/9

Step-by-step explanation:

took the test

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A television station plans to ask a random sample of 400 city residents if they can name the news anchor on the evening news at
hodyreva [135]

Answer:

There is an 8.38% probability that the anchor will be fired.

Step-by-step explanation:

For each resident, there is only two possible outcomes. Either they can name the news anchor on the evening news at their station, or they cannot.

This means that the binomial probability distribution will be used in our solution.

However, we are working with samples that are considerably big. So i am going to aaproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

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 samples 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)}.

In this problem, we have that:

400 city residents are going to be asked. So n = 400.

Suppose that in fact 12% of city residents could name the anchor if asked. This means that p = 0.12.

So,

\mu = E(X) = 400*0.12 = 48

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{400*0.12*0.88} = 6.5.

They plan to fire the news anchor if fewer than 10% of the residents in the sample can do so.

What is the approximate probability that the anchor will be fired?

10% of the residents is 0.10*400 = 40 residents.

Fewer than 10% is 39 residents. So the probability that the anchor will be fired is the pvalue of Z when X = 39.

So:

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

Z = \frac{39 - 48}{6.5}

Z = -1.38

Z = -1.38 has a pvalue of 0.0838. This means that there is an 8.38% probability that the anchor will be fired.

5 0
2 years ago
Suppose f and g are continuous functions such that g(2) = 6 and lim x → 2 [3f(x) + f(x)g(x)] = 36. find f(2).
True [87]

Answer: f(2) = 4

Step-by-step explanation:

F(x) and g(x) are said to be continuous functions

Lim x=2 [3f(x) + f(x)g(x)] = 36

g(x) = 2

Limit x=2

[3f(2) + f(2)g(2)] = 36

[3f(2) + f(2) . 6] = 36

[3f(2) + 6f(2)] = 36

9f(2) = 36

Divide both sides by 9

f(2) = 36/9

f(2) = 4

7 0
2 years ago
A newly drilled water well produces 50,000 quarts of water per week. With no new water feeding the well, the production drops by
Sveta_85 [38]

Answer:

Total amount of water = 5,200,000

Step-by-step explanation:

Given:

water produced = 50,000 quarts of water per week

Production drop = 5% = 0.05 per year

Number of week in year = 52 week

Find:

Total amount of water

Computation:

Sum = a / r

a = 50,000 x 52

a = 2,600,000

Sum = a / [1-r]

Sum = 2,600,000 / 5%

Sum = 2,600,000 / 0.05

Total amount of water = 5,200,000

5 0
1 year ago
Two investors have net worths of $12 million (investor A) and $1 million (investor B). they set their investment limits at 2% an
Varvara68 [4.7K]

Answer:

Investor B

Step-by-step explanation:

Investor A: 2% of 12 million is 240,000

Investor B: 30% of 1 million is 300,000

Thus, Investor B has a greater investment limit

5 0
2 years ago
A disadvantage of the contention approach for LANs, such as CSMA/CD, is the capacity wasted due to multiple stations attempting
sammy [17]

Answer:

The overview of the given problem is outlined in the following segment on the explanation.

Step-by-step explanation:

The proportion of slots or positions that have been missed due to numerous concurrent transmission incidents can be estimated as follows:

Checking a probability of transmitting becomes "p".

After considering two or even more attempts, we get

Slot fraction wasted,

= [1-no \ attempt \ probability-first \ attempt \ probability-second \ attempt \ probability+...]

On putting the values, we get

= 1-no \ attempt \ probability-[N\times P\times probability \ of \ attempts]

= 1-(1-P)^{N}-N[P(1-P)^{N}]

So that the above seems to be the right answer.

8 0
2 years ago
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