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pantera1 [17]
3 years ago
7

Find the area of 7cm 20cm 14cm 11cm

Mathematics
1 answer:
FinnZ [79.3K]3 years ago
3 0
Multiply all 4 an the answer is gonna be the area
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10x + .0085x = 280000
GaryK [48]

Answer:

x = 27976.22021

Step-by-step explanation:

10x + .0085x = 280000

Combine like terms

10.0085x = 280000

Divide by 10.0085 on each side

10.0085x /10.0085= 280000/10.0085

x = 27976.22021

6 0
3 years ago
Are they simular and why​
S_A_V [24]

Answer:

YES

Step-by-step explanation:

because there the same shape but the right one is moved to the side a little bit

4 0
3 years ago
Do #6 for me pls and thank you!​
emmainna [20.7K]

Answer: function a

Hope this helped!

4 0
3 years ago
Graph a system of equations to approximate the
SIZIF [17.4K]

Answer: B

Step-by-step explanation:

8 0
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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