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3 years ago

In which section of the number line is √55?

Mathematics

1 answer:

Luden [163]3 years ago

4 0

The answer is [A]: "Section A: "7.0 - 7.5" — in the figure shown" .
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Explanation:
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The question asks:
________________________________________________________
"In which section of the number line is √55?"
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 We know that: 7² = 49 ; 8² = 64 .

So, 7 < √55 < 8 ; So, √55 will be in between 7 and 8.

So, √49 < √55 < √64 .

55 - 49 = 6 .

64 - 55 = 9 .

 So, there is greater difference between the radicands in

 √64 and √55 ; compared to the difference in radicands between

 √49 and √55 .
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 This tells us that the correct answer is:
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 [A] { in the figure shown: "7.0 - 7.5" } ;
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 rather than: [B]: { "7.5 - 8.0" } .
________________________________________________

You might be interested in

Drag and drop the answer into the box to identify each function as linear, exponential, or neither.

katrin2010 [14]

x 2 3 4
f(x) 5.5 7 8.5 

The first function is linear. When you subtract 5.5 from 7, you get 1.5 and when you subtract 7 from 8.5 you also get 1.5. That's how we know it's a linear function because there's clearly pattern. HOWEVER, if you had 5.5, 7, and 9 that would not be a linear function. Why? Because when you take away 7 from 9 you get 2. In order for it to be a linear function you have to get the same exact number when you subtract.

x 0 3 6
f(x) 1 8 64

The second function is exponential because when you divide you get the exact same number. 8/1 = 8 and 64/8 = 8. So this is exponential. But if you had 72 instead of 64 it would not be an exponential function because there has to be a pattern.

I hope this makes sense. Btw I had this question on my test too and this is correct.

4 0

3 years ago

Read 2 more answers

Lenovo uses the zx-81 chip in some of its laptop computers. the prices for the chip during the last 12 months were as follows:

Stella [2.4K]

Given the table below of the prices for the Lenovo zx-81 chip during the last 12 months

$\begin{tabular}
{|c|c|c|c|}
Month&Price per Chip&Month&Price per Chip\\[1ex]
January&\$1.90&July&\$1.80\\
February&\$1.61&August&\$1.83\\
March&\$1.60&September&\$1.60\\
April&\$1.85&October&\$1.57\\
May&\$1.90&November&\$1.62\\
June&\$1.95&December&\$1.75
\end{tabular}$

The forcast for a period $F_{t+1}$ is given by the formular

$F_{t+1}=\alpha A_t+(1-\alpha)F_t$

where $A_t$ is the actual value for the preceding period and $F_t$ is the forcast for the preceding period.

Part 1A:
Given α = 0.1 and the initial forecast for october of $1.83, the actual value for october is $1.57.

Thus, the forecast for period 11 is given by:

$F_{11}=\alpha A_{10}+(1-\alpha)F_{10} \\ \\ =0.1(1.57)+(1-0.1)(1.83) \\ \\ =0.157+0.9(1.83)=0.157+1.647 \\ \\ =1.804$

Therefore, the foreast for period 11 is $1.80

Part 1B:

Given α = 0.1 and the forecast for november of $1.80, the actual value for november is $1.62

Thus, the forecast for period 12 is given by:

$F_{12}=\alpha
 A_{11}+(1-\alpha)F_{11} \\ \\ =0.1(1.62)+(1-0.1)(1.80) \\ \\ 
=0.162+0.9(1.80)=0.162+1.62 \\ \\ =1.782$

Therefore, the foreast for period 12 is $1.78

Part 2A:

Given α = 0.3 and the initial forecast for october of $1.76, the actual value for October is $1.57.

Thus, the forecast for period 11 is given by:

$F_{11}=\alpha
 A_{10}+(1-\alpha)F_{10} \\ \\ =0.3(1.57)+(1-0.3)(1.76) \\ \\ 
=0.471+0.7(1.76)=0.471+1.232 \\ \\ =1.703$

Therefore, the foreast for period 11 is $1.70


Part 2B:

Given α = 0.3 and the forecast for November of $1.70, the actual value for november is $1.62

Thus, the forecast for period 12 is given by:

$F_{12}=\alpha
 A_{11}+(1-\alpha)F_{11} \\ \\ =0.3(1.62)+(1-0.3)(1.70) \\ \\ 
=0.486+0.7(1.70)=0.486+1.19 \\ \\ =1.676$

Therefore, the foreast for period 12 is $1.68


Part 3A:

Given α = 0.5 and the initial forecast for october of $1.72, the actual value for October is $1.57.

Thus, the forecast for period 11 is given by:

$F_{11}=\alpha
 A_{10}+(1-\alpha)F_{10} \\ \\ =0.5(1.57)+(1-0.5)(1.72) \\ \\ 
=0.785+0.5(1.72)=0.785+0.86 \\ \\ =1.645$

Therefore, the forecast for period 11 is $1.65


Part 3B:

Given α = 0.5 and the forecast for November of $1.65, the actual value for November is $1.62

Thus, the forecast for period 12 is given by:

$F_{12}=\alpha
 A_{11}+(1-\alpha)F_{11} \\ \\ =0.5(1.62)+(1-0.5)(1.65) \\ \\ 
=0.81+0.5(1.65)=0.81+0.825 \\ \\ =1.635$

Therefore, the forecast for period 12 is $1.64

Part 4:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using α = 0.3, we obtained that the forcasted values of october, november and december are: $1.83, $1.80, $1.78

Thus, the mean absolute deviation is given by:

$\frac{|1.57-1.83|+|1.62-1.80|+|1.75-1.78|}{3} = \frac{|-0.26|+|-0.18|+|-0.03|}{3} \\ \\ = \frac{0.26+0.18+0.03}{3} = \frac{0.47}{3} \approx0.16$

Therefore, the mean absolute deviation using exponential smoothing where α = 0.1 of October, November and December is given by: 0.157

Part 5:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using α = 0.3, we obtained that the forcasted values of october, november and december are: $1.76, $1.70, $1.68

Thus, the mean absolute deviation is given by:

$
 \frac{|1.57-1.76|+|1.62-1.70|+|1.75-1.68|}{3} = 
\frac{|-0.17|+|-0.08|+|-0.07|}{3} \\ \\ = \frac{0.17+0.08+0.07}{3} = 
\frac{0.32}{3} \approx0.107$

Therefore, the mean absolute deviation using exponential smoothing where α = 0.3 of October, November and December is given by: 0.107


Part 6:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using α = 0.5, we obtained that the forcasted values of october, november and december are: $1.72, $1.65, $1.64

Thus, the mean absolute deviation is given by:

$
 \frac{|1.57-1.72|+|1.62-1.65|+|1.75-1.64|}{3} = 
\frac{|-0.15|+|-0.03|+|0.11|}{3} \\ \\ = \frac{0.15+0.03+0.11}{3} = 
\frac{29}{3} \approx0.097$

Therefore, the mean absolute deviation using exponential smoothing where α = 0.5 of October, November and December is given by: 0.097

5 0

3 years ago

Find the length indicated. find CE

torisob [31]

Okay, to find length CE, your going to know the value of x. Length BC + CE = BD + DE.
3x+47+x+26=27+x+10
Simplify the equation to get
4x+73=37+x
you can choose one of four ways to continue, but I will choose to subtract x
3x+73=37
Subtract 73 from both sides of the equal sign
3x=-36
divide by 3 on both sides of the equal sign to get the value of x
x=-12

Now, plug in -12 for x in length CE to get -12+26=14

6 0

3 years ago

Please help !!!!!!!!!

Ainat [17]

Answer:

Step-by-step explanation:

3x -12 = -36

-36 ÷ -3 = 12

-7 + 12 = 5

7 0

3 years ago

Read 2 more answers

Can someone help please

11Alexandr11 [23.1K]

I get it now!!!!
After some reasearch, I believe the answer is ~q --> p
Can I get brainliest answer pls????

4 0

3 years ago