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miskamm [114]
3 years ago
6

In Lewis County, there were 2,277 student athletes competing in spring sports in 2014. That was 110% of the number from 2013, wh

ich was 90% of the number from the year before. How many student athletes signed up for a spring sport in 2012.
Mathematics
1 answer:
nasty-shy [4]3 years ago
5 0

In 2014 , 2277 was 110% of 2013.

Number in 2013 = 90% of the year before (2012)

Let the number in 2013 be x. From the first statement,

2277 = 110% of x

2277 = (110/100)x

2277 = 1.10x

x = 2277/1.1 = 2070

Number in 2013 is 2070

Number in 2013 = 90% Number in 2012

2070 = (90/100) * Number in 2012

2070 = 0.90 * Number in 2012

2070 / 0.90    = Number in 2012

2300  = Number in 2012

Number in 2012  =  2300

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Bad gums may mean a bad heart. Researchers discovered that 79% of people who have suffered a heart attack had periodontal diseas
zysi [14]

Answer:

(A) 0.297

(B) 0.595

Step-by-step explanation:

Let,

H = a person who suffered from a heart attack

G = a person has the periodontal disease.

Given:

P (G|H) = 0.79, P(G|H') = 0.33 and P (H) = 0.15

Compute the probability that a person has the periodontal disease as follows:

P(G)=P(G|H)P(H)+P(G|H')P(H')\\=P(G|H)P(H)+P(G|H')(1-P(H))\\=(0.79\times0.15)+(0.33\times(1-0.15))\\=0.399

(A)

The probability that a person had periodontal disease, what is the probability that he or she will have a heart attack is:

P(H|G)=\frac{P(G|H)P(H)}{P(G)}\\=\frac{0.79\times0.15}{0.399} \\=0.29699\\\approx0.297

Thus, the probability that a person had periodontal disease, what is the probability that he or she will have a heart attack is 0.297.

(B)

Now if the probability of a person having a heart attack is, P (H) = 0.38.

Compute the probability that a person has the periodontal disease as follows:

P(G)=P(G|H)P(H)+P(G|H')P(H')\\=P(G|H)P(H)+P(G|H')(1-P(H))\\=(0.79\times0.38)+(0.33\times(1-0.38))\\=0.5048

Compute the probability of a person having a heart attack given that he or she has the disease:

P(H|G)=\frac{P(G|H)P(H)}{P(G)}\\=\frac{0.79\times0.38}{0.5048}\\ =0.59469\\\approx0.595

The probability of a person having a heart attack given that he or she has the disease is 0.595.

4 0
3 years ago
1) 2/7×1/2 <br>2) 5/8×4/5 <br>3) 1/6×2/3<br>4) 1/3×3/4<br>5) 1/2×2/3<br>6) 1/3×2/5​
nignag [31]
1) 1/7
2) 1/2
3) 1/9
4) 1/4
5) 1/3
6) 2/15
hope this helped:)
5 0
3 years ago
Serena uses chalk to draw a straight line on the sidewalk. The line is 12 ft long. She wants to divide the line into sections th
ElenaW [278]
<h3><u>Question:</u></h3>

Serena uses chalk to draw a straight line on the sidewalk. The line is 1/2 ft long. She wants to divide the line into sections that are each 1/8 ft long. How many sections will the line be divided into?

<h3><u>Answer:</u></h3>

The number of sections that the line is divided is 4

<h3><u>Solution:</u></h3>

Given that, Serena uses chalk to draw a straight line on the sidewalk

The line is 1/2 ft long. She wants to divide the line into sections that are each 1/8 ft long

From given,

Total\ length\ of\ line = \frac{1}{2} \text{ feet }\\\\Length\ of\ each\ section = \frac{1}{8} \text{ feet }

To find: Number of sections can be made

The number of sections that can be made is found by dividing the total length of line by length of each section

\text{Number of sections } = \frac{\text{Total length of line}}{\text{length of each section}}

Substituting the values, we get,

\text{Number of sections } = \frac{\frac{1}{2}}{\frac{1}{8}}\\\\\text{Number of sections } = \frac{1}{2} \times \frac{8}{1}\\\\\text{Number of sections } = 4

Thus number of sections that the line is divided is 4

8 0
3 years ago
Please explain how to do and set up
In-s [12.5K]

A

using the Cosine rule in ΔSTU

let t = SU, s = TU and u = ST, then

t² = u² + s² - (2us cos T )

substitute the appropriate values into the formula

t² = 5² + 9² - (2 × 5 × 9 × cos68° )

   = 25 + 81 - 90cos68°

   = 106 - 33.71 = 72.29

⇒ t = \sqrt{72.29} ≈ 8.5 in → A


6 0
3 years ago
What is the first quartile of the set of data shown below?
____ [38]
The answer would be B 23
7 0
3 years ago
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