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2 years ago

An airplane cruises 1 kilometer in 1/12 of a minute. What is its cruising speed?

Mathematics

1 answer:

Rasek [7]2 years ago

6 0

Answer:

The airplane is cruising at 200 metres per second.

Step-by-step explanation:

The speed of the object can be got by dividing the distance it travelled by the time taken to travel that distance.

For us to get the accurate cruising speed, we will have to convert all units to their S.I form

Distance - 1 kilometer = 1 X 1000 metres = 1000 metres

1/12 of a minute = 1/12 X 60 = 5 seconds.

Therefore the speed = distance / time

= 1000/5 = 200m/s

The airplane is cruising at 200 metres per second.

You might be interested in

Write an equation that represents a horizontal translation 10 units left of the graph of g(x) = |x|

Vika [28.1K]

Answer: h(x) = |x+10|

---------------------------------------------------------

Explanation:

To shift the graph 10 units to the left, we replace x with x+10. What's really going on is that the xy axis shifts 10 units to the right (because x is now x+10; eg, x = 2 ---> x+10 = 2+10 = 12) so it appears that the graph is moving to the left. The general rule is h(x) = g(x+10).

So,
g(x) = |x|
g(x+10) = |x+10| ... every x has been replaced with x+10
h(x) = g(x+10)
h(x) = |x+10|

We can use a graphing tool like GeoGebra to visually confirm we have the right answer (see attached). Note how a point like (0,0) on the green graph moves to (-10,0) on the red graph.

7 0

2 years ago

Read 2 more answers

Suppose that 20% of the adult women in the United States dye or highlight their hair. We would like to know the probability that

Rasek [7]

Answer:

71.08% probability that pˆ takes a value between 0.17 and 0.23.

Step-by-step explanation:

We use the binomial approxiation to the normal to solve this question.

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:

$p = 0.2, n = 200$ . So

$\mu = E(X) = np = 200*0.2 = 40$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.2*0.8} = 5.66$

In other words, find probability that pˆ takes a value between 0.17 and 0.23.

This probability is the pvalue of Z when X = 200*0.23 = 46 subtracted by the pvalue of Z when X = 200*0.17 = 34. So

X = 46

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

$Z = \frac{46 - 40}{5.66}$

$Z = 1.06$

$Z = 1.06$ has a pvalue of 0.8554

X = 34

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

$Z = \frac{34 - 40}{5.66}$

$Z = -1.06$

$Z = -1.06$ has a pvalue of 0.1446

0.8554 - 0.1446 = 0.7108

71.08% probability that pˆ takes a value between 0.17 and 0.23.

6 0

2 years ago

A map of a rectangular park has a length of 4 inches and a width of 6 inches. It uses a scale of 1 inch for every 30 Miles. What

MakcuM [25]

Answer:

The actual area 21600 miles²

Step-by-step explanation:

* <em>Lets explain how to solve the problem</em>

- A drawing that shows a real object with accurate sizes reduced or

enlarged by a certain amount called the scale

- Drawing scale is a ratio between the drawing length and the

actual length

- To find the actual dimensions from the drawing dimensions use the

scale drawing

*<em> Lets solve the problem</em>

- A map of a rectangular park has a length of 4 inches and a width

of 6 inches

∴ The drawing dimensions are:

# length = 4 inches

# width = 6 inches

- The scale of the map is 1 inch foe every 30 miles

∴ The scale drawing is 1 : 30

- To find the actual area find the actual dimensions

∵ The scale drawing is 1 : 30

∵ The length = 4 inches and the width = 6 inches

- By using cross multiplication

∴ 1 : 30

4 : actual length

6 : actual width

∴ Actual length = (4 × 30)/1 = 120

∴ Actual length = 120 miles

∴ Actual width = (6 × 30)/1 = 180

∴ Actual Width = 180 miles

∴ The actual dimensions are 120 miles and 180 miles

∵ The area of any rectangle = length × width

∴ The actual area = 120 × 180 = 21600

∴ The actual area 21600 miles²

5 0

2 years ago

F(x)=x2−x−1f, left parenthesis, x, right parenthesis, equals, x, squared, minus, x, minus, 1 What is the average rate of change

Alexandra [31]

Answer: The average rate of change = -1

Step-by-step explanation: Please find the attached file for the solution

5 0

2 years ago

In choice situations of this type, subjects often exhibit the "center stage effect," which is a tendency to choose the item in t

victus00 [196]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The probability that he or she would choose the pair of socks in the center position is $p =\frac{1}{5}$

The correct answer choice is

X has a binomial distribution with parameters n=100 and p=1/5

The mean is $\mu = 20$

The standard deviation is $\sigma=4$

The probability, $P =0.0002$

The correct answer is

The experiment supports the center stage effect. If participants were truly picking the socks at random, it would be highly unlikely for 34 or more to choose the center pair.

Using the R the probability $Pe = 0.0003$

The probabilities $P \approx Pe$

Step-by-step explanation:

Since the person selects his or her desired pair of socks at random , then the probability that the person would choose the pair of socks in the center position from all the five identical pair is mathematically evaluated as

$p =\frac{1}{5}$

$=0.2$

The mean of this distribution is mathematical represented as

$\mu = np$

substituting the value

$\mu = 100 * 0.2$

$\mu = 20$

The standard deviation is mathematically represented as

$\sigma = \sqrt{np (1-p)}$

substituting the value

$= \sqrt{100 * 0,2 (1-0.2)}$

$\sigma=4$

Applying normal approximation the probability that 34 or more subjects would choose the item in the center if each subject were selecting his or her preferred pair of socks at random would be mathematically represented as

$P=P(X \ge 34 )$

By standardizing the normal approximation we have that

$P(X \ge 34) \approx P(Z \ge z)$

Now z is mathematically evaluated as

$z = \frac{x-\mu}{\sigma }$

Substituting values

$z = \frac{34-20}{4}$

$=3.5$

So using the z table the $P(Z \ge 3.5)$ is 0.0002

The probability P and Pe that 34 or more subject would choose the center pair is very small So

The correct answer is

The experiment supports the center stage effect. If participants were truly picking the socks at random, it would be highly unlikely for 34 or more to choose the center pair.

6 0

2 years ago