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2 years ago

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This fall, 126 students participated in the soccer program, while 54 played volleyball. What was the ratio of soccer players to

volleyball players?

1 answer:

Over [174]2 years ago

6 0

...….................. 56

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Elliot has been running a lawn care business since 2000. He cuts grass, trims, and weed whacks yards for his customers throughou

Gnesinka [82]

Df/dy=(1350-750)/(2010-2000)

df/dy=60

f(y)=750+60(y-2000) or neatened up a bit...

f(y)=60y-119250 (note: y is the actual year, ie 2005, not year like 2 years from start)

6 0

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

Draw hundred boxes ten sticks and circles to show a number between 100 and 200

zaharov [31]

I chose the number 155

6 0

2 years ago

Read 2 more answers

a person on a tour has rupees 12000 for his daily expenses. In order to extend his journey for 2 more days he had to cut down hi

myrzilka [38]

Answer:

Duration of the tour he planned first is 8 days.

Step-by-step explanation:

Given that a person has 12000 rupees for his daily expenses.

Let x be the number of days.

Then daily expenses per day $=\frac{12000}{x}$

Given that number of day is increased by 2 more days, that is number of days is x+2.

New daily expense per day $\frac{12000}{x+2}$

Given that this new daily expenses are 300 less than original.

That is $\frac{12000}{x}-\frac{12000}{x+2}=300$

$\frac{12000(x+2)-12000x}{x(x+2)}=300$

$\frac{24000}{x(x+2)}=300$

$x(x+2)=\frac{24000}{300}$

$x^{2}+2x-80=0$

(x-8)(x+10)=0

x=8 or -10.

Since number of days cannot be negative, duration of the tour planned=8.

7 0

2 years ago

One way to measure whether the trees in the Wade Tract are uniformly distributed is to examine the average location in the north

tatyana61 [14]

Answer:

Null hypothesis be H₀ : μ = 100

Alternative hypothesis Hₐ : μ < 100

There is sufficient statistical evidence to suggest that the average location of Wade Tract is 100

Step-by-step explanation:

Here we have;

Let our null hypothesis be H₀ : μ = 100 Average location of trees in the Wade Tract is 100

Our alternative hypothesis becomes Hₐ : μ < 100 at 95% confidence level

Proposed average location in Wade Tract, μ = 100

Sample mean, $\bar x$ = 99.74

Standard deviation, s = 58

Sample size, n = 584

The t test formula is therefore;

$t=\frac{\bar{x}-\mu }{\frac{s }{\sqrt{n}}}$

Therefore, with df = 584 -1 = 583, and α = (1 - 0.95)/2 = 0.025

We have $t_{\alpha /2}$ = -1.65

Plugging in the values into the t test formula, we have t = -0.108338, from which the p-value is given as p = 0.4569 which is much more than the value of α, therefore, we accept the null hypothesis as follows;

There is sufficient statistical evidence to suggest that the average location of Wade Tract = 100.

7 0

2 years ago