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const2013 [10]
3 years ago
14

Is the function given by, (Picture Provided)

Mathematics
1 answer:
Sever21 [200]3 years ago
7 0

Answer:

a. yes

Step-by-step explanation:

The given function is

f(x)=\left \{ {{3x-2,if\:x\le3} \atop {10-x,if\:x\:>\:3}} \right.

f(3)=\left \{ {{3(3)-2,if\:x\le3} \atop {10-3,if\:x\:>\:3}} \right.=7

\lim_{x \to 3^-} f(x)=3(3)-2=7

\lim_{x \to 3^+} f(x)=10-3=7

\Rightarrow \lim_{x \to 3} f(x)=7

Since;

\lim_{x \to 3} f(x)=f(3)

The function is continuous at x=3

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9) If m/1 = 45° and 21 and 22 are complementary angles. Find m22.​
Allisa [31]

❄ Hi there,

keeping in mind that the sum of complementary angles is 90°,

set up an equation, letting \angle2  be x  –

\triangleright \ \sf{\angle1+x=90}                 {and we know that \boxed{\angle1=45}}

\triangleright \ \sf{45+x=90}

\triangleright \ \sf{x=90-45}

\triangleright \ \sf{x=45}

\triangleright \ \sf{\angle2=45\textdegree}

      __________

Keeping in mind that a right angle is 90°,

set up an equation, letting \angle1 be x:

\triangleright \ \sf{x+\angle2=90}                       {and we know that \boxed{\angle2=63}}

\triangleright \ \sf{x+63=90}

\triangleright \ \sf{x=27}

\triangleright \ \sf{\angle1=47\textdegree}

❄

5 0
2 years ago
What is 2/5 E2? (the E is the exponent of ur wondering i couldn't type it in :)​
tigry1 [53]

Answer:

when you do 2 over 5 exponent 5 you get 0.08. I hope this helps you.

4 0
3 years ago
Explain the circumstances for which the interquartile range is the preferred measure of dispersion. What is an advantage that th
KatRina [158]

Answer:

Explain the circumstances for which the interquartile range is the preferred measure of dispersion

Interquartile range is preferred when the distribution of data is highly skewed (right or left skewed) and when we have the presence of outliers. Because under these conditions the sample variance and deviation can be biased estimators for the dispersion.

What is an advantage that the standard deviation has over the interquartile​ range?

The most important advantage is that the sample variance and deviation takes in count all the observations in order to calculate the statistic.

Step-by-step explanation:

Previous concepts

The interquartile range is defined as the difference between the upper quartile and the first quartile and is a measure of dispersion for a dataset.

IQR= Q_3 -Q_1

The standard deviation is a measure of dispersion obatined from the sample variance and is given by:

s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

Solution to the problem

Explain the circumstances for which the interquartile range is the preferred measure of dispersion

Interquartile range is preferred when the distribution of data is highly skewed (right or left skewed) and when we have the presence of outliers. Because under these conditions the sample variance and deviation can be biased estimators for the dispersion.

What is an advantage that the standard deviation has over the interquartile​ range?

The most important advantage is that the sample variance and deviation takes in count all the observations in order to calculate the statistic.

8 0
3 years ago
Line A is represented by the following equation:
Vesnalui [34]
- for line B so the set of equations to have no solutions bc. x+y=2 and so y=2-x 
so from x+y=4 will result y=4-x

so this is the right equation for line B so the set of equations has no solutions 
6 0
3 years ago
Read 2 more answers
Please help this is 5th Grade Math also show work 15 Points awarded
Dimas [21]

Answer:

The only one that would be reasonable is the first one...and maybe the third one too

Step-by-step explanation:


8 0
3 years ago
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