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svlad2 [7]
3 years ago
14

The ratio of runners to swimmers was 3:2. When 28 runners dropped out of the competition, the ratio became 5:6. How many total s

wimmers are competing?
Mathematics
1 answer:
xz_007 [3.2K]3 years ago
7 0

The total of 42 swimmers are competing

Step-by-step explanation:

The given is:

  • The ratio of runners to swimmers was 3 : 2
  • When 28 runners dropped out of the competition, the ratio became 5 : 6

We need to find how many total swimmers are competing

∵ The ratio of runners to swimmers was 3 : 2

- Multiply each term of the ratio by x

∴ There are 3x runners and 2x swimmers

∵ 28 runners dropped out of the competition

∴ The number of runners who complete = 3x - 28

∵ The ratio became 5 : 6

- Equate the ratio between the new number of runners and the

   number of swimmers by 5 : 6

∵ The new number of runners = 3x - 28

∵ The number of swimmers = 2x

∴ \frac{3x-28}{2x}=\frac{5}{6}

- By using cross multiplication

∴ 6(3x - 28) = 5(2x)

- Simplify both sides

∴ 18x - 168 = 10x

- Add 168 to both sides

∴ 18x = 10x + 168

- Subtract 10x from both sides

∴ 8x = 168

- Divide both sides by 8

∴ x = 21

∵ The number of swimmers is 2x

∵ x = 21

∴ The number of swimmers = 2(21) = 42

The total of 42 swimmers are competing

Learn more:

You can learn more about the ratio in brainly.com/question/2707032

#LearnwithBrainly

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Answer: half pound

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2 years ago
Perform the following:-1/3 + 6/715/16 - 3/47/6x + -4/3y
Wittaler [7]

SOLUTION

\begin{gathered} a\text{. }\frac{-1}{3}\text{ + }\frac{6}{7} \\  \\ \frac{-7+18}{21} \\  \\ \frac{11}{21} \end{gathered}

b.

\begin{gathered} \frac{15}{16}-\frac{3}{4} \\  \\ \frac{15-12}{16} \\  \\ \frac{3}{16} \end{gathered}

c.

\begin{gathered} \frac{7}{6x}+\frac{-4}{3y} \\  \\ \frac{7y-8x}{6xy} \end{gathered}

So we just have to find the LCM of the denominators, then we perform the rule of addition of fraction

6 0
1 year ago
In a sample of 1200 U.S.​ adults, 191 dine out at a resaurant more than once per week. Two U.S. adults are selected at random
allochka39001 [22]

Answer:

a) The probability that both adults dine out more than once per week = 0.0253

b) The probability that neither adult dines out more than once per week = 0.7069

c) The probability that at least one of the two adults dines out more than once per week = 0.2931

d) Of the three events described, the event that can be considered unusual because of its low probability of occurring, 0.0253 (2.53%), is the event that the two randomly selected adults both dine out more than once per week.

Step-by-step explanation:

In a sample of 1200 U.S. adults, 191 dine out at a restaurant more than once per week.

Assuming this sample.is a random sample and is representative of the proportion of all U.S. adults, the probability of a randomly picked U.S. adult dining out at a restaurant more than once per week = (191/1200) = 0.1591666667 = 0.1592

Now, assuming this probability per person is independent of each other.

Two adults are picked at random from the entire population of U.S. adults, with no replacement, thereby making sure these two are picked at absolute random.

a) The probability that both adults dine out more than once per week.

Probability that adult A dines out more than once per week = P(A) = 0.1592

Probability that adult B dines out more than once per week = P(B) = 0.1592

Probability that adult A and adult B dine out more than once per week = P(A n B)

= P(A) × P(B) (since the probability for each person is independent of the other person)

= 0.1592 × 0.1592

= 0.02534464 = 0.0253 to 4 d.p.

b) The probability that neither adult dines out more than once per week.

Probability that adult A dines out more than once per week = P(A) = 0.1592

Probability that adult A does NOT dine out more than once per week = P(A') = 1 - P(A) = 1 - 0.1592 = 0.8408

Probability that adult B dines out more than once per week = P(B) = 0.1592

Probability that adult B does NOT dine out more than once per week = P(B') = 1 - P(B) = 1 - 0.1592 = 0.8408

Probability that neither adult dines out more than once per week = P(A' n B')

= P(A') × P(B')

= 0.8408 × 0.8408

= 0.70694464 = 0.7069 to 4 d.p.

c) The probability that at least one of the two adults dines out more than once per week.

Probability that adult A dines out more than once per week = P(A) = 0.1592

Probability that adult A does NOT dine out more than once per week = P(A') = 1 - P(A) = 1 - 0.1592 = 0.8408

Probability that adult B dines out more than once per week = P(B) = 0.1592

Probability that adult B does NOT dine out more than once per week = P(B') = 1 - P(B) = 1 - 0.1592 = 0.8408

The probability that at least one of the two adults dines out more than once per week

= P(A n B') + P(A' n B) + P(A n B)

= [P(A) × P(B')] + [P(A') × P(B)] + [P(A) × P(B)]

= (0.1592 × 0.8408) + (0.8408 × 0.1592) + (0.1592 × 0.1592)

= 0.13385536 + 0.13385536 + 0.02534464

= 0.29305536 = 0.2931 to 4 d.p.

d) Which of the events can be considered unusual? Explain.

The event that can be considered as unusual is the event that has very low probabilities of occurring, probabilities of values less than 5% (0.05).

And of the three events described, the event that can be considered unusual because of its low probability of occurring, 0.0253 (2.53%), is the event that the two randomly selected adults both dine out more than once per week.

Hope this Helps!!!

6 0
3 years ago
4 In
Ad libitum [116K]

Answer:

32 in

Step-by-step explanation:

Volume= 1/2 a•c•h

=1/2•4•4•4

= 32 in

3 0
2 years ago
If p = 4/5, what is the smallest value of n that satisfies the requirements for a normally distributed p?
adell [148]

Answer:

The smallest value of n that satisfies the requirements for a normally distributed p is 30

Step-by-step explanation:

The Central Limit Theorem says that if we have a random sample and the sample size is large enough (usually bigger than 30), then the sample proportion will follow a normal distribution with mean p, where p is the true population proportion.

8 0
3 years ago
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