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1 year ago

What is a decimal number that is larger than 0.0467 but smaller than 0.0468?

Mathematics

2 answers:

Jlenok [28]1 year ago

8 0

0.04671

you gotta add a number smaller than 0.0001

dmitriy555 [2]1 year ago

7 0

Answer:

0.04671

Step-by-step explanation:

Basically, add a number smaller than 0.0001

You might be interested in

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

Mr. Snow bought 90 grams of Christmas candy for each of his 14 grandchildren. How many total kilograms of candy did he buy?

nignag [31]

Answer:

1.26 kg.

Step-by-step explanation:

We have been given that Mr. Snow bought 90 grams of Christmas candy for each of his 14 grandchildren. We are asked to find the amount of candy bought by Mr. Snow in kilograms.

First of all, we will find the amount of candy in grams by multiplying 90 grams by 14 as:

$\text{Amount of candy bought in grams}=14\times 90$

$\text{Amount of candy bought in grams}=1260$

We know that 1 kilogram is equal to 1000 grams. To convert 1260 grams into kg, we will divide 1260 by 1000 as:

$\text{Amount of candy bought in kilograms}=\frac{1260}{1000}$

$\text{Amount of candy bought in kilograms}=1.26$

Therefore, Mr. Snow bought 1.26 kilograms of candy.

7 0

2 years ago

Point G is on segment FH such that it partitions segment FH into a ratio of 2:3.Point F is located at (5, 7) and point H is loca

Margarita [4]

Answer:

G = (9.4, 9,4)

Step-by-step explanation:

The ratio is applied in the x-distance and the y-distance. The ratio is 2:3 so you have to divide the distances by 5 and 2/5 correspond to FG and 3/5 to GH

x-distance:

x2 - x1 = 16 - 5 = 11

11/5 = 2.2

y-distance:

y2 - y1 = 13 - 7 = 6

6/5 = 1.2

Point G = Point F + (2.2*2, 1.2*2)

Point G = (5, 7) + (4.4, 2.4)

= (9.4, 9.4)

8 0

2 years ago

a(0.3−y)+1.1+2.4x (y−1.2) =0 =−1.2(x−0.5) Consider the system of equations above, where aaa is a constant. For which value o

Veseljchak [2.6K]

Answer:

For a = 1.22 there is one solution where y = 1.3

Step-by-step explanation:

Hi there!

Let´s write the system of equations:

a(0.3 - y) + 1.1 +2.4x(y-1.2) = 0

-1.2(x-0.5) = 0

Let´s solve the second equation for x:

-1.2(x-0.5) = 0

x- 0.5 = 0

x = 0.5

Now let´s repalce x = 0.5 and y = 1.3 in the first equation and solve it for a:

a(0.3 - y) + 1.1 +2.4x(y-1.2) = 0

a(0.3 - 1.3) + 1.1 + 2.4(0.5)(1.3 -1.2) = 0

a(-1) + 1.1 + 1.2(0.1) = 0

-a + 1.22 = 0

-a = -1.22

a = 1.22

Let´s check the solution and solve the system of equations with a = 1.22. Let´s solve the first equation for y:

1.22(0.3 - y) + 1.1 +2.4(0.5)(y-1.2) = 0

0.366 - 1.22y + 1.1 + 1.2 y - 1.44 = 0

-0.02y +0.026 = 0

-0.02y = -0.026

y = -0.026 / -0.02

y = 1.3

Then, the answer is correct.

Have a nice day!

4 0

2 years ago

Identify the 12th term of the geometric sequence in which a2 = −6 and a6 = −96

timurjin [86]

$a_2=-6;\ a_6=-96\\\\a_6:a_2=r^4\\\\r^4=-96:(-6)\\\\r^4=16\\\\r=\sqrt[4]{16}\\\\r=2\\\\a_{12}=a_6r^6\\\\a_{12}=-96\cdot2^6=-96\cdot64=-6,144$

6 0

2 years ago

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