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2 years ago

6

The starting line for a race in Fulton is 1.2 miles from the finish line. On a map of the course, the starting and finish lines

are 4 inches apart. What is the scale of the map?

2 answers:

topjm [15]2 years ago

6 0

Answer:

1 inch = 0.3 miles

Step-by-step explanation:

We basically want to know how many miles in real life 1 inch on the map represents. We already know that 4 inches on the map corresponds to 1.2 miles in real life, but we need to find what 1 inch on the map corresponds to. Let's make a proportion:

$\frac{4}{1.2} =\frac{1}{x}$

Cross-multiply:

4x = 1.2

Divide by 4 from both sides:

x = 1.2/4 = 0.3

Thus, the scale is 1 inch = 0.3 miles.

Hope this helps!

WARRIOR [948]2 years ago

4 0

Things to know before we start:

In real life, the starting line of the race in Fultion is 1.2 miles from the finish line.

On a map, the starting line of the race in Fultion is 4 inches from the finish line.

To find the scale, we can set up a proportion.

1.2mi/4in

Divide to find the scale.

1.2 / 4 = 0.3 miles.

Therefore the scale of the map is 0.3 miles = 1 inch.

Best of Luck!

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The 13.50 is a constant.

The total needs to be less than or equal to 40 because that's all the money he has.

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Answer:

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Step-by-step explanation:

Tamika already has $55, which will stay the same independent of how much time passes. Since she deposits $5 per week, we can assign weeks a variable x. The amount of money in her account after x weeks is equal to 55+5x. If she must save over $100, we can create an inequality to solve for how many weeks must pass: 55+5x>100.

The question also asks to solve the inequality:

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2 years ago

Richard has just been given an l0-question multiple-choice quiz in his history class. Each question has five answers, of which o

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Answer:

a) 0.0000001024 probability that he will answer all questions correctly.

b) 0.1074 = 10.74% probability that he will answer all questions incorrectly

c) 0.8926 = 89.26% probability that he will answer at least one of the questions correctly.

d) 0.0328 = 3.28% probability that Richard will answer at least half the questions correctly

Step-by-step explanation:

For each question, there are only two possible outcomes. Either he answers it correctly, or he does not. The probability of answering a question correctly is independent of any other question. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Each question has five answers, of which only one is correct

This means that the probability of correctly answering a question guessing is $p = \frac{1}{5} = 0.2$

10 questions.

This means that $n = 10$

A) What is the probability that he will answer all questions correctly?

This is $P(X = 10)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{10,10}.(0.2)^{10}.(0.8)^{0} = 0.0000001024$

0.0000001024 probability that he will answer all questions correctly.

B) What is the probability that he will answer all questions incorrectly?

None correctly, so $P(X = 0)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.2)^{0}.(0.8)^{10} = 0.1074$

0.1074 = 10.74% probability that he will answer all questions incorrectly

C) What is the probability that he will answer at least one of the questions correctly?

This is

$P(X \geq 1) = 1 - P(X = 0)$

Since $P(X = 0) = 0.1074$ , from item b.

$P(X \geq 1) = 1 - 0.1074 = 0.8926$

0.8926 = 89.26% probability that he will answer at least one of the questions correctly.

D) What is the probability that Richard will answer at least half the questions correctly?

This is

$P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{10,5}.(0.2)^{5}.(0.8)^{5} = 0.0264$

$P(X = 6) = C_{10,6}.(0.2)^{6}.(0.8)^{4} = 0.0055$

$P(X = 7) = C_{10,7}.(0.2)^{7}.(0.8)^{3} = 0.0008$

$P(X = 8) = C_{10,8}.(0.2)^{8}.(0.8)^{2} = 0.0001$

$P(X = 9) = C_{10,9}.(0.2)^{9}.(0.8)^{1} \approx 0$

$P(X = 10) = C_{10,10}.(0.2)^{10}.(0.8)^{0} \approx 0$

So

$P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.0264 + 0.0055 + 0.0008 + 0.0001 + 0 + 0 = 0.0328$

0.0328 = 3.28% probability that Richard will answer at least half the questions correctly

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3 years ago

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Tju [1.3M]

3x-4y=12
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