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White raven [17]
2 years ago
7

The graph represents the piecewise function

Mathematics
1 answer:
mr Goodwill [35]2 years ago
6 0

The piece-wise function is defined as follows:

  • f(x) = x^2, 0 \leq x < 3.
  • f(x) = -\frac{5}{3}x + 14, 3 < x \leq 6.
  • f(x) = \frac{3}{2}x - 5, 6 \leq x \leq 10.

<h3>What is a piece-wise function?</h3>

A piece-wise function is a function that has multiple definitions, depending on the input.

In this graph, for x at least 0 and less than 3, the parabolic curve passes through (0,0), (1,1), (2,4) and has an open interval at (3,9), hence the definition is:

f(x) = x^2, 0 \leq x < 3

For x greater than 3 and at most 6, it is a line going through (3,9) and (6,4), hence:

m = \frac{9 - 4}{3 - 6} = -\frac{5}{3}

f(x) = -\frac{5}{3}x + b

Goes through (3,9), hence:

9 = -\frac{5}{3}(3) + b

b = 14.

So

f(x) = -\frac{5}{3}x + 14, 3 < x \leq 6

For x between 6 and 10, it is a line going through (6,4) and (10,10), hence:

m = \frac{10 - 4}{10 - 6} = \frac{6}{4} = \frac{3}{2}

Then:

f(x) = \frac{3}{2}x + b

When x = 10, f(x) = 10, hence:

10 = \frac{3}{2}(10) + b

10 = 15 + b

b = -5.

Hence:

f(x) = \frac{3}{2}x - 5, 6 \leq x \leq 10

More can be learned about piece-wise functions at brainly.com/question/24734454

#SPJ1

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Step-by-step explanation:

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3 years ago
W sklepie spożywczym jest 11,73 kg cukierków czekoladowych. Pierwszy klient kupił 48 dag cukierków. Drugi klient kupił 4/5 pozos
Elenna [48]

Answer:

<h2>The last costumers got 2.25 kilograms of chocolate candies.</h2>

Step-by-step explanation:

The question is

<em> There is 11.73 kg of chocolate candies in the grocery store. The first customer bought 48 dag of candies. A second customer bought 4/5 of the remaining quantity. The last three customers bought the same amount of candy. How much chocolate did the last customers get?</em>

<em />

Givens

  • The total amount of chocolate candies is 11.73 kilograms.
  • First costumer bought 48 dag of candies. (1 kg equals 100 dags)
  • Second costumer bougth 4/5 of the remaining.
  • Another three costumers bought the same amount of candy.

Let's transform 48 dag to kilograms.

48dag \times \frac{1kg}{100dag}= 0.48 \ kg

Therefore, the first costumer bought 0.48 kilograms of candies.

The remaining amount is: 11.73kg-0.48kg=11.25kg

Now, we need to multiply the remaining amount of candies with 4/5

\frac{4}{5} \times 11.25kg=9 \ kg

Therefore, the second costumer bought 9 kilograms of chocolate candies.

At last, we need to find the new remaining part of candies, which is

11.25-9=2.25 \ kg

Therefore, the last costumers got 2.25 kilograms of chocolate candies.

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3 years ago
Five players are given $10 each and asked to contribute any portion of it to a group account. They are also told that the total
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Answer:

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Step-by-step explanation:

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Triss [41]

Hola mate

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7 0
3 years ago
According to the Knot, 22% of couples meet online. Assume the sampling distribution of p follows a normal distribution and answe
Ann [662]

Using the <em>normal distribution and the central limit theorem</em>, we have that:

a) The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

b) There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

c) There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

<h3>Normal Probability Distribution</h3>

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

Z = \frac{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.
  • By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1 - p)}{n}}, as long as np \geq 10 and n(1 - p) \geq 10.

In this problem:

  • 22% of couples meet online, hence p = 0.22.
  • A sample of 150 couples is taken, hence n = 150.

Item a:

The mean and the standard error are given by:

\mu = p = 0.22

s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.22(0.78)}{150}} = 0.0338

The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

Item b:

The probability is <u>one subtracted by the p-value of Z when X = 0.25</u>, hence:

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

By the Central Limit Theorem:

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

Z = \frac{0.25 - 0.22}{0.0338}

Z = 0.89

Z = 0.89 has a p-value of 0.8133.

1 - 0.8133 = 0.1867.

There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

Item c:

The probability is the <u>p-value of Z when X = 0.2 subtracted by the p-value of Z when X = 0.15</u>, hence:

X = 0.2:

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

Z = \frac{0.2 - 0.22}{0.0338}

Z = -0.59

Z = -0.59 has a p-value of 0.2776.

X = 0.15:

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

Z = \frac{0.15 - 0.22}{0.0338}

Z = -2.07

Z = -2.07 has a p-value of 0.0192.

0.2776 - 0.0192 = 0.2584.

There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

To learn more about the <em>normal distribution and the central limit theorem</em>, you can check brainly.com/question/24663213

4 0
2 years ago
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