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2 years ago

Write 18.4 as a fraction in lowest terms

Mathematics

2 answers:

Lapatulllka [165]2 years ago

5 0

Answer:

Divide each side of 18/4 by 2 to get <u>9/2.</u>

Elanso [62]2 years ago

4 0

Answer:

184/10

Step-by-step explanation:

You have to delete the comma (,) so, by doing ×10 you have 18,4 -> 184. By doing ÷10 you have a fraction: 184/10 that is equal to 18.4

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An alloy is composed of nickel, zinc, and copper in a 7:2:9 ratio. How many kilograms of each metal is needed to make 3.8 kg of

Tamiku [17]

Ni : Zn: Cu = 7:2:9 =7x : 2x : 9x

7x+2x+9x = 3.8
18x=3.8
x=3.8/18=1.9/9≈0.21

Ni: 7x=7*0.21=1.47 ≈1.5 kg
Zn: 2x=2*0.21=0.42≈0.4 kg
Cu: 9x=9*0.21=1.89 ≈1.9 kg

Check 1.5+0.4+1.9 =3.8 True

6 0

2 years ago

There are 8 soccer matches per day at Greenville's annual soccer tournament, which lasts 17 days. Estimate the number of soccer

Gnesinka [82]

Answer:

140

Step-by-step explanation:

8 per day

17 days

8*17 = 136 soccer matches in the tournament

136 rounded to the nearest 10 = 140

7 0

2 years ago

which of the sets shown includes the elements of set Z that both negative numbers and multiples of ?. Z={-34,-28,-16,-2,4,8,12,2

Ronch [10]

Answer:

multiples of 2

Step-by-step explanation:

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