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makkiz [27]
3 years ago
11

can someone show me an example of the product that two irrational nembers will not equal irrational number.....is this possible

Mathematics
1 answer:
kobusy [5.1K]3 years ago
5 0
\sqrt{2} *\sqrt{2}=\sqrt{4}=2

the square root of two is irrational

the sqare root of four is rational
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Jake rode his roller skated 2 1/4 miles for 3/4 of an hour. Find his skating speed in miles/hour
alisha [4.7K]

Answer:

3 mph

Step-by-step explanation:

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2 years ago
Two passenger train, A and B, 450 km apart, star to move toward each other at the same time and meet after 2hours.
likoan [24]

Let v be the speed of train A, and let's set the origin in the initial position of train A. The equations of motion are

\begin{cases}s_A(t) = vt\\s_B(t) = -\dfrac{8}{7}vt+450\end{cases}

where s_A,\ s_B are the positions of trains A and B respectively, and t is the time in hours.

The two trains meet if and only if s_A=s_B, and we know that this happens after two hours, i.e. at t=2

\begin{cases}s_A(2) = 2v\\s_B(2) = -\dfrac{16}{7}v+450\end{cases}\implies 2v = -\dfrac{16}{7}v+450

Solving this equation for v we have

2v = -\dfrac{16}{7}v+450 \iff \dfrac{30}{7}v=450 \iff v=\dfrac{450\cdot 7}{30} = 105

So, train A is travelling at 105 km/h. This implies that train B travels at

105\cdot \dfrac{8}{7} = 15\cdot 8=120 \text{ km/h}

5 0
3 years ago
Help with his plz ASAP!!
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Answer:

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2 years ago
IF I HAVE A LAPTOP COMPUTER THAT MEASURES 30.8×6.6×39.8 CENTIMETERS...WHAT SIZE CASE DOES IT REQUIRE
soldi70 [24.7K]

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3 years ago
Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 75.6 Mbp
olga55 [171]

Answer:

a) 59.98

b) 2.99

c) 2.99

d) Significantly High

Step-by-step explanation:

Part a)

Highest speed measured = x = 75.6 Mbps

Average/Mean speed = \overline{x} = 15.62 Mbps

Standard Deviation = s = 20.03 Mbps

We need to find the difference between carrier's highest data speed and the mean of all 50 data​ speeds i.e. x - \overline{x}

x - \overline{x} = 75.6 - 15.62 = 59.98 Mbps

Thus, the difference between​ carrier's highest data speed and the mean of all 50 data​ speeds is 59.98 Mbps

Part b)

In order to find how many standard deviations away is the difference found in previous part, we divide the difference by the value of standard deviation i.e.

\frac{59.98}{20.03}=2.99

This means, the difference is 2.99 standard deviations or in other words we can say, the Carrier's highest data speed is 2.99 standard deviations above the mean data speed.

Part c)

A z score tells us that how many standard deviations away is a value from the mean. We calculated the same in the previous part. Performing the same calculation in one step:

The formula for the z score is:

z=\frac{x-\overline{x}}{s}

Using the given values, we get:

z=\frac{75.6-15.62}{20.03}=2.99

Thus, the Carriers highest data is equivalent to a z score of 2.99

Part d)

The range of z scores which are neither significantly low nor significantly​ high is -2 to + 2. The z scores outside this range will be significant.

Since, the z score for carrier's highest data speed is 2.99 which is well outside the given range, i.e. greater than 2, we can conclude that the  carrier's highest data speed​ is significantly higher.

3 0
3 years ago
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