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sladkih [1.3K]
3 years ago
10

All of the following are equivalent, except _____.

Mathematics
1 answer:
Jet001 [13]3 years ago
3 0
A. (12 + 3x)x isn’t equivalent!
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Mrs.scott has 30 pairs of shoes.Out of the 30 pairs 24 of them are black.what percent of shoes are black
pychu [463]
Do 24/30, which is 0.8. Convert this to a percentage

==============> 80%
4 0
3 years ago
To find 10, Beau found 32 = 9 and 42 = 16. He said that since 10 is between 9 and 16, 10 is between 3
AlexFokin [52]

Given that:

Consider it is \sqrt{10} instead of 10 on two places.

\sqrt{10} is between 3  and 4. So, Beau thinks a good estimate for \sqrt{10} is = 3.5.

Solution:

To find \sqrt{10}, Beau found 3² = 9 and 4² = 16.

He said that since 10 is between 9 and 16.

Since 10 is close to 9, therefore \sqrt{10} must be close to 3. So, Beau's estimate is high.

Now,

(3.1)^2=9.61

(3.2)^2=10.24

Since, 10 lies between 9.61 and 10.24, therefore \sqrt{10} must be lies between 3.1 and 3.2.

\dfrac{3.1+3.2}{2}=3.15

Therefore, the estimated value of \sqrt{10} is 3.15.

4 0
3 years ago
An equation with no solutions<br> 3x + 7 = X + 3
max2010maxim [7]

Answer:

X = -2

Step-by-step explanation:

Hope this helps

7 0
3 years ago
An experiment requires 500 mL. One of the ingredients to be mixed with 1,000 mL of another. How many liters is the solution
shepuryov [24]
1.5 liters all together! 

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6 0
3 years ago
Read 2 more answers
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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