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2 years ago

How do I write parametric equations of the line 2x+3y=11

Mathematics

1 answer:

vfiekz [6]2 years ago

3 0

This line is in standard form, and the first thing we need to do is convert it into point - slope form or simply isolate y -

2x + 3y = 11, Subtract 2x from either side

3y = - 2x + 11, Divide either side by 3

y = ( - 2x + 11 ) / 3

To derive the parametric equation say x = t -

y = ( - 2t + 11 ) / 3,

or in other words

<u><em>Solution = y = ( 11 - 2t ) / 3</em></u>

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One week, Claire earned $272.00 at her job when she worked for 17 hours. If she is paid the same hourly wage, how many hours wou

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Answer: 448 hours

Step-by-step explanation:

272 divided by 17 = hourly wage

hourly wage= 16$

448 divided by 16 = 28 weeks

28 x 16= 448 hours

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What's 362, 584 rounded to the nearest hundred thousand

hammer [34]

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2 years ago

The accompanying data are lengths (inches) of bears. Find the percentile corresponding to 65.5 in.

Jlenok [28]

Complete question is missing, so i have attached it.

Answer:

Percentile is 74th percentile

Step-by-step explanation:

All the lengths given are;

Bear Lengths 36.5 37.5 39.5 40.5 41.5 42.5 43.0 46.0 46.5 46.5 48.5 48.5 48.5 49.5 51.5 52.5 53.0 53.0 54.5 56.8 57.5 58.5 58.5 58.5 59.0 60.5 60.5 61.0 61.0 61.5 62.0 62.5 63.5 63.5 63.5 64.0 64.0 64.5 64.5 65.5 66.5 67.0 67.5 69.0 69.5 70.5 72.0 72.5 72.5 72.5 72.5 73.0 76.0 77.5

The number of lengths (inches) of bears given are 54 in number.

We are looking for the percentile corresponding to 65.5 in.

Looking at the lengths given, since they are already arranged from smallest to highest, let's locate the position of 65.5 in.

The position of 65.5 in is the 40th among 54 lengths given.

If the percentile is P, then;

P% x 54 = 40

P = (40 × 100)/54

P ≈ 74

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2 years ago

Rocky Mountain National Park is a popular park for outdoor recreation activities in Colorado. According to U.S. National Park Se

Ugo [173]

Answer:

a) 0.6628 = 66.28% probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance

b) 0.5141 = 51.41% probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance

c) 0.5596 = 55.96% probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance.

d) 0.9978 = 99.78% probability that more than 55 visitors have no recorded point of entry

Step-by-step explanation:

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:

$n = 175$

(a) What is the probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance?

46.7% of visitors to Rocky Mountain National Park in 2018 entered through the Beaver Meadows. This means that $p = 0.467$ . So

$\mu = E(X) = np = 175*0.467 = 81.725$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.467*0.533} = 6.6$

This probability, using continuity correction, is $P(X \geq 85 - 0.5) = P(X \geq 84.5)$ , which is 1 subtracted by the pvalue of Z when X = 84.5. So

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

$Z = \frac{84.5 - 81.725}{6.6}$

$Z = 0.42$

$Z = 0.42$ has a pvalue of 0.6628.

66.28% probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance.

(b) What is the probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance?

Using continuity correction, this is $P(80 - 0.5 \leq X < 90 - 0.5) = P(79.5 \leq X \leq 89.5)$ , which is the pvalue of Z when X = 89.5 subtracted by the pvalue of Z when X = 79.5. So

X = 89.5

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

$Z = \frac{89.5 - 81.725}{6.6}$

$Z = 1.18$

$Z = 1.18$ has a pvalue of 0.8810.

X = 79.5

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

$Z = \frac{79.5 - 81.725}{6.6}$

$Z = -0.34$

$Z = -0.34$ has a pvalue of 0.3669.

0.8810 - 0.3669 = 0.5141

51.41% probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance

(c) What is the probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance?

6.3% of visitors entered through the Grand Lake park entrance, which means that $p = 0.063$

$\mu = E(X) = np = 175*0.063 = 11.025$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.063*0.937} = 3.2141$

This probability, using continuity correction, is $P(X < 12 - 0.5) = P(X < 11.5)$ , which is the pvalue of Z when X = 11.5. So

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

$Z = \frac{11.5 - 11.025}{3.2141}$

$Z = 0.15$

$Z = 0.15$ has a pvalue of 0.5596.

55.96% probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance.

(d) What is the probability that more than 55 visitors have no recorded point of entry?

22.7% of visitors had no recorded point of entry to the park. This means that $p = 0.227$

$\mu = E(X) = np = 175*0.227 = 39.725$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.227*0.773} = 5.54$

Using continuity correction, this probability is $P(X \leq 55 + 0.5) = P(X \leq 55.5)$ , which is the pvalue of Z when X = 55.5. So

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

$Z = \frac{55.5 - 39.725}{5.54}$

$Z = 2.85$

$Z = 2.85$ has a pvalue of 0.9978

0.9978 = 99.78% probability that more than 55 visitors have no recorded point of entry

8 0

2 years ago

How do I write parametric equations of the line 2x+3y=11​

How do I write parametric equations of the line 2x+3y=11