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3 years ago

5. The students are trying to determine the distance between two particular

Mathematics

1 answer:

natali 33 [55]3 years ago

8 0

Answer:

The actual distance between the cities is 211.67 cm.

Step-by-step explanation:

Given:

The map key indicates that 4.5 cm is equivalent to 75 cm.

Now, to find the actual distance between the cities if the cities are 12.7 cm apart on the map.

Let the actual distance be $x$ .

So, we set a proportion to find the actual distance:

4.5:12.7 = 75:x.

Changing into fractions.

$\frac{4.5}{12.7} =\frac{75}{x}$

By cross multiplication we get:

$4.5x=952.5$

Dividing both sides by 4.5 we get:

$x=211.67$

Therefore, the actual distance between the cities is 211.67 cm.

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Marianna [84]

Answer:

Step-by-step explanation:

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3 years ago

HELP PLEASE EQUATION OF PARABOLA

Masteriza [31]

It has a minimum value at x = 3 and f(x) = 4

Vertex form is

f(x) = a(x - 3)^2 + 4 where a is some constant to be found

From the graph when x = 5 f(x) = 15, so

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3 years ago

Solve the following equation. Then place the correct answer in the box provided. Answer in terms of a mixed number. 7(2x + 6) =

BabaBlast [244]

Hey there! :D

7(2x+6)=1

Use the distributive property. Multiply each number by each other.

7*2x= 14x 7*6= 42

14x+42=1

Subtract 42 on both sides.

14x= -41

Divide both sides by 14.

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-2 13/14 (we just need one more part of 14)

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3 years ago

Read 2 more answers

Graph both functions to find the solution(s) to the system.

BaLLatris [955]

f(2) = 2×2 − 1 = 3

The point of intersection is (2, 3).

The example shows that we can find the point of intersection in two ways.
Either graphically, by drawing the two graphs in the same coordinate system, or algebraically by solving the equation such as the one in the above example.

Solving an equation graphically is easy with a graphical calculator or a computer program such as Excel.
Some equations cannot be solved algebraically but we can find solutions that are correct to as many significant figures as we want by using computers and calculators

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3 years ago

The United States Coast Guard assumes the mean weight of passengers in commercial boats is 185 pounds. The previous value was lo

Valentin [98]

Answer:

There is a 5.5% probability that a random sample of passengers will have a mean weight that is as extreme or more extreme (either above or below the mean) than was observed in this sample.

Step-by-step explanation:

To solve this problem, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

In this problem, we have that:

$\mu = 185, \sigma = 26.7, n = 48, s = \frac{26.7}{\sqrt{48}} = 3.85$

The weights of a random sample of 48 commercial boat passengers were recorded. The sample mean was determined to be 177.6 pounds. Find the probability that a random sample of passengers will have a mean weight that is as extreme or more extreme (either above or below the mean) than was observed in this sample.

The probability of an extreme value below the mean.

This is the pvalue of Z when X = 177.6.

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{177.6 - 185}{3.85}$

$Z = -1.92$

$Z = -1.92$ has a pvalue of 0.0274.

So there is a 2.74% of having a sample mean as extreme than that and lower than the mean.

The probability of an extrema value above the mean.

Measures above the mean have a positive z score.

So this probability is 1 subtracted by the pvalue of $Z = 1.92$

$Z = 1.92$ has a pvalue of 0.9726.

So there is a 1-0.9726 = 0.0274 = 2.74% of having a sample mean as extreme than that and above than the mean.

Total:

2*0.0274 = 0.0548 = 0.055

There is a 5.5% probability that a random sample of passengers will have a mean weight that is as extreme or more extreme (either above or below the mean) than was observed in this sample.

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3 years ago