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2 years ago

Patrick is driving form Sydney to Canberra. He travels for 288km trip in 4 hours. What's his average speed?

Mathematics

1 answer:

nika2105 [10]2 years ago

3 0

Answer:

v = 72 km/h

Step-by-step explanation:

Given that,

The distance covered by the Patrick, d = 288 km

Time taken, t = 4 hours

We need to find the average speed of Patrick. We know that the average speed of an object is equal to the total distance covered divided by total time taken. Let it is v. So,

$v=\dfrac{d}{t}\\\\v=\dfrac{288\ km}{4\ h}\\\\v=72\ km/h$

So, his average speed is equal to 72 km/h.

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1/2 (n + 4) = 6 whats n?

madreJ [45]

Answer:

n=8

Step-by-step explanation:

1/2 (n + 4) = 6

n+4=12

n=8

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3 years ago

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3 years ago

Normal Distribution. Cherry trees in a certain orchard have heights that are normally distributed with mu = 112 inches and sigma

Lubov Fominskaja [6]

Answer:

The probability that a randomly chosen tree is greater than 140 inches is 0.0228.

Step-by-step explanation:

Given : Cherry trees in a certain orchard have heights that are normally distributed with $\mu = 112$ inches and $\sigma = 14$ inches.

To find : What is the probability that a randomly chosen tree is greater than 140 inches?

Solution :

Mean - $\mu = 112$ inches

Standard deviation - $\sigma = 14$ inches

The z-score formula is given by, $Z=\frac{x-\mu}{\sigma}$

Now,

$P(X>140)=P(\frac{x-\mu}{\sigma}>\frac{140-\mu}{\sigma})$

$P(X>140)=P(Z>\frac{140-112}{14})$

$P(X>140)=P(Z>\frac{28}{14})$

$P(X>140)=P(Z>2)$

$P(X>140)=1-P(Z$

The Z-score value we get is from the Z-table,

$P(X>140)=1-0.9772$

$P(X>140)=0.0228$

Therefore, the probability that a randomly chosen tree is greater than 140 inches is 0.0228.

5 0

3 years ago

For hw plss help!!!!

yarga [219]

Answer:

4th term is 274.4 cm

5th term is 192.08 cm

Step-by-step explanation:

Here, we want to get the 4th and 5th term

Let us get the common ratio;

That would be ;

392/560 = 560/800 = 0.7

The 4th term will be ar^3

= 800 * 0.7^3 = 274.4

The 5th term will be ar^4

= 800 * 0.7^4 = 192.08

6 0

2 years ago

A cellular phone company monitors monthly phone usage. The following data represent the monthly phone use in minutes of one part

Sergeu [11.5K]

Answer:

The standard deviation increased but there was no change in the interquantile range

Step-by-step explanation:

We are given the following data in the question:

320, 411, 348, 537, 420, 449, 462, 403, 454, 517, 515, 358, 438, 541, 387, 368, 502, 437, 431, 428.

n = 20

a) Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{8726}{20} = 436.3$

Sum of squares of differences =

13525.69 + 640.09 + 7796.89 + 10140.49 + 265.69 + 161.29 + 660.49 + 1108.89 + 313.29 + 6512.49 + 6193.69 + 6130.89 + 2.89 + 10962.09 + 2430.49 + 4664.89 + 4316.49 + 0.49 + 28.09 + 68.89 = 75924.2

$S.D = \sqrt{\frac{75924.2}{19}} = 63.21$

Sorted Data = 320, 348, 358, 368, 387, 403, 411, 420, 428, 431, 437, 438, 449, 454, 462, 502, 515, 517, 537, 541

$IQR = Q_3 - Q_1\\Q_3 = \text{upper median},\\Q_1 = \text{ lower median}$

$Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}$

$Median = \frac{431 + 437}{2} = 434$

$Q_1 = \frac{387 + 403}{2} = 395\\\\Q_3 = \frac{462 + 502}{2} = 482$

IQR = $482 - 395 = 87$

b) After changing the observation

0, 411, 348, 537, 420, 449, 462, 403, 454, 517, 515, 358, 438, 541, 387, 368, 502, 437, 431, 428

$Mean =\displaystyle\frac{8406}{20} = 420.3$

Sum of squares of differences =

176652.09 + 86.49 + 5227.29 + 13618.89 + 0.09 + 823.69 + 1738.89 + 299.29 + 1135.69 + 9350.89 + 8968.09 + 3881.29 + 313.29 + 14568.49 + 1108.89 + 2735.29 + 6674.89 + 278.89 + 114.49 + 59.29 = 247636.2

$S.D = \sqrt{\frac{247636.2}{19}} = 114.16$

Sorted Data = 0, 348, 358, 368, 387, 403, 411, 420, 428, 431, 437, 438, 449, 454, 462, 502, 515, 517, 537, 541

$Median = \frac{431 + 437}{2} = 434$

$Q_1 = \frac{387 + 403}{2} = 395\\\\Q_3 = \frac{462 + 502}{2} = 482$

IQR = $482 - 395 = 87$

Thus. the standard deviation increased but there was no change in the interquantile range.

5 0

3 years ago