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4 years ago

Create a word problem that involves speed

1 answer:

6 0

A hockey puck is sliding on frictionless ice on an infinite hockey rink.
Its speed is 36 km/hour. How far does the puck slide in 10 seconds ?

(36 km/hr) x (1,000 m/km) x (1 hr/3600 sec) x (10sec) =

(36 x 1,000 x 10 / 3,600) meters = <em>100 meters</em>

What is the puck's speed in miles per hour ?

(36 km/hr) x (0.6214 mi/km) = <em>22.37 mi/hr</em>

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The table shows the price of applause in the local market what is the cost of 12 pounds of apples

pishuonlain [190]

The answer is: B 18$

4 0

4 years ago

Read 2 more answers

What us the answer for this question $64 jacket; 20% discount

Sav [38]

$51.20 is the answer just use a calculator

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3 years ago

In a certain country, the average age is 31 years old and the standard deviation is 4 years. If we select a simple random sample

KengaRu [80]

Answer: 0.0062

Step-by-step explanation:

Given : Mean : $\mu=\ 31$

Standard deviation : $\sigma= 4$

Sample size : $n=100$

Assume that age of people in the country is normally distributed.

The formula to calculate the z-score :-

$z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

For x = 32

$z=\dfrac{32-31}{\dfrac{4}{\sqrt{100}}}=5$

The p-value = $P(x\geq32)=P(z\geq5)$

$=1-P(z$

Hence, the the probability that the average age of our sample is at least =0.0062

3 0

3 years ago

The weight distribution of parcels sent in a certain manner is normal with mean value 15 lb and standard deviation 3.3 lb. The p

disa [49]

Answer: 23.7 lb

Step-by-step explanation:

Mean m = 15 lb

Standard deviation d = 3.3 lb

To determine the surcharge weight, we need to know the highest weight of 99% of the parcels.

P(z<x) = 0.99 = ¢(Z)

Z = 2.33

Since Z = (x - m)/d

x = dZ + m

x = 3.3*2.33 + 15

x = 22.689 lb approximately

x = 22.7 lb

Therefore, the highest weight for 99% if the parcels is 22.7 lb.

That is, the surcharge weight = 22.7 + 1 = 23.7 lb

3 0

3 years ago

Consider the following sample set of scores. Assume these scores are from a discrete distribution. 21 29 32 38 38 45 50 64 72 10

ASHA 777 [7]

Answer:

Original data: 21 29 32 38 38 45 50 64 72 100

$\bar X = \frac{21+29+32+38+38+45+50+64+72+100}{10}=48.9$

Mode =38.

$Median = \frac{38+45}{2}=41.5$

Change 1: 2 29 32 38 38 45 50 64 72 100

$\bar X = \frac{2+29+32+38+38+45+50+64+72+100}{10}=47$

Mode =38.

$Median = \frac{38+45}{2}=41.5$

Change 2: 29 32 38 38 45 50 64 72 100

$\bar X = \frac{29+32+38+38+45+50+64+72+100}{9}=52$

Mode =38.

$Median = 45$

Step-by-step explanation:

Original data: 21 29 32 38 38 45 50 64 72 100

The mean is calculated with the following formula:

$\bar X = \frac{\sum_{i=1}^{10} X_i}{10}$

And if we replace we got:

$\bar X = \frac{21+29+32+38+38+45+50+64+72+100}{10}=48.9$

The mode is the most repeated value in the sample and on this case is Mode =38.

Since we have an even number of points the median is calculated as the average between the observations 5 and 6 from the dataset ordered.

$Median = \frac{38+45}{2}=41.5$

Change 1: 2 29 32 38 38 45 50 64 72 100

The mean is calculated with the following formula:

$\bar X = \frac{\sum_{i=1}^{10} X_i}{10}$

And if we replace we got:

$\bar X = \frac{2+29+32+38+38+45+50+64+72+100}{10}=47$

The mode is the most repeated value in the sample and on this case is Mode =38.

Since we have an even number of points the median is calculated as the average between the observations 5 and 6 from the dataset ordered.

$Median = \frac{38+45}{2}=41.5$

Change 2: 29 32 38 38 45 50 64 72 100

Now the sample size is 9 instead of 10

The mean is calculated with the following formula:

$\bar X = \frac{\sum_{i=1}^{9} X_i}{9}$

And if we replace we got:

$\bar X = \frac{29+32+38+38+45+50+64+72+100}{9}=52$

The mode is the most repeated value in the sample and on this case is Mode =38.

Since we have odd number of points the median is calculated from the 5 position of the dataset ordered.

$Median = 45$

7 0

4 years ago