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pashok25 [27]
3 years ago
8

If A = {(x, y)|x + 2y = 7}, is Set A a function?

Mathematics
1 answer:
valentina_108 [34]3 years ago
6 0
<h3>Answer: Yes it is a function</h3>

===============================================

Here are some reasons why

  1. We can solve x+2y = 7 for y to get y = -0.5x+3.5; this equation is in slope intercept form y = mx+b. All equations in the form y = mx+b are linear and also a function.
  2. For any input x, there is exactly only one y value that pairs with it.
  3. The vertical line test can be used to effectively verify item #2 above. The idea is that it is impossible to draw a vertical line through more than one point on the graph. So it is said to pass the vertical line test.
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rice that costs $480 which is for 8 members is used 20 days. What is the cost of rice which is required for 12 members for 15 da
Ber [7]

Answer:

$540

Step-by-step explanation:

1 member 1 day :$480/80/20 = $3

12 members 15 day = $3 * 12 * 15 = $540

Hope this helps!!!

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3 years ago
Andre has 4 times as many model cars as Peter, and Peter has one-third as many model cars as Jade. Andre has 36 model cars. Writ
vovikov84 [41]
First solve what is probably 'part A.'
Peter has 36 ÷ 4; Peter has 9 model cars. Jade has 3 times as many as Peter; Peter has 1/3 as many as Jade, so 9 x 3 = 27, so Jade has 27 cars.

1st equation: 36 ÷ 4 = 9 -OR- 36 x 1/4 = 9
2nd equation: 9 x 3 = 27 -OR- (36 x 1/4) x 3 = 9
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4 years ago
Assume that females have pulse rates that are normally distributed with a mean of mu equals 72.0 beats per minute and a standard
larisa86 [58]

Answer:

a) 36.88% probability that her pulse rate is between 66 beats per minute and 78 beats per minute

b) 66.30% probability that they have pulse rates with a mean between 66 beats per minute and 78 beats per minute

c)

C.

Since the original population has a normal​ distribution, the distribution of sample means is a normal distribution for any sample size.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are 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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 72, \sigma = 6.5

a. If 1 adult female is randomly​ selected, find the probability that her pulse rate is between 66 beats per minute and 78 beats per minute.

The probability is?

This is the pvalue of Z when X = 78 subtracted by the pvalue of Z when X = 66. So

X = 78

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

Z = \frac{78 - 72}{12.5}

Z = 0.48

Z = 0.48 has a pvalue of 0.6844

X = 66

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

Z = \frac{66 - 72}{12.5}

Z = -0.48

Z = -0.48 has a pvalue of 0.3156

0.6844 - 0.3156 = 0.3688

36.88% probability that her pulse rate is between 66 beats per minute and 78 beats per minute

b. If 4 adult females are randomly​ selected, find the probability that they have pulse rates with a mean between 66 beats per minute and 78 beats per minute

The probability is?

Now we have n = 4, s = \frac{12.5}{\sqrt{4}} = 6.25

So

X = 78

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

By the Central Limit Theorem

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

Z = \frac{78 - 72}{6.25}

Z = 0.96

Z = 0.96 has a pvalue of 0.8315

X = 66

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

Z = \frac{66 - 72}{6.25}

Z = -0.96

Z = -0.96 has a pvalue of 0.1685

0.8315 - 0.1685 = 0.6630

66.30% probability that they have pulse rates with a mean between 66 beats per minute and 78 beats per minute

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

The condition for the sample size exceeding 30 is when the population is skewed. If it is normally distributed, the size is not a condition.

So the correct answer is:

C.

Since the original population has a normal​ distribution, the distribution of sample means is a normal distribution for any sample size.

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