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Amiraneli [1.4K]
4 years ago
13

What is the average of two opposite numbers?

Mathematics
1 answer:
Kryger [21]4 years ago
8 0

Answer:

0

Step-by-step explanation:

So lets imagine x as our number

Let's make an equation:

Average=\frac{x-x}{2} \\Average=\frac{o}{2} \\Average=0

So, This would mean that the average of two opposite numbers is always 0

Hope this helps!

Stay Safe!

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The measurement of angle x is 40°. these two triangle are congruent so their sides share the same length and their angles share the same degrees
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J(1)²,
Natali [406]

The table when completed looks like this:

x   -2   -1   0   1   2

f(x) 4   2   1    2   4.

We plot these points on the graph: (-2, 4), (-1, 2), (0, 1), (1, 2), (2, 4).

<h3>What is a function?</h3>

A function exists as a relation between a dependent variable (f(x)) and an expression of an independent variable (x), utilized to define the value of a dependent variable from a provided value of an independent variable.

We have been given a function,

$f(x)=\left \{ {{(1/2)^{x},x\leq 0 } \atop {2^x > 0}} \right.

We have to estimate the value of f(x) when x = {-2, -1, 0, 1, 2}

If the values of x ≤ 0, take f(x) = (1/2)ˣ then

f(-2) = (1/2)⁻² = 2² = 4

f(-1) = (1/2)⁻¹ = 2¹ = 2

f(0) = (1/2)⁰ = 1

If the values of x > 0,  take f(x) = 2ˣ then

f(1) = 2¹ = 2

f(2) = 2² = 4

The table exists as follows

x   -2   -1   0   1   2

f(x) 4   2   1    2   4.

We plot these points on the graph: (-2, 4), (-1, 2), (0, 1), (1, 2), (2, 4).

To learn more about independent and dependent variable refer to:

brainly.com/question/11719274

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7 0
1 year ago
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Suppose twenty-two communities have an average of = 123.6 reported cases of larceny per year. assume that σ is known to be 36.8
Delvig [45]
We are given the following data:

Average = m = 123.6
Population standard deviation = σ= psd = 36.8
Sample Size = n = 22

We are to find the confidence intervals for 90%, 95% and 98% confidence level.

Since the population standard deviation is known, and sample size is not too small, we can use standard normal distribution to find the confidence intervals.

Part 1) 90% Confidence Interval
z value for 90% confidence interval = 1.645

Lower end of confidence interval = m-z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Lower end of confidence interval=123.6-1.645* \frac{36.8}{ \sqrt{22}}=110.69

Upper end of confidence interval = m+z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Upper end of confidence interval=123.6+1.645* \frac{36.8}{ \sqrt{22}}=136.51

Thus the 90% confidence interval will be (110.69, 136.51)

Part 2) 95% Confidence Interval
z value for 95% confidence interval = 1.96

Lower end of confidence interval = m-z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Lower end of confidence interval=123.6-1.96* \frac{36.8}{ \sqrt{22}}=108.22

Upper end of confidence interval = m+z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Upper end of confidence interval=123.6+1.96* \frac{36.8}{ \sqrt{22}}=138.98

Thus the 95% confidence interval will be (108.22, 138.98)

Part 3) 98% Confidence Interval
z value for 98% confidence interval = 2.327

Lower end of confidence interval = m-z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Lower end of confidence interval=123.6-2.327* \frac{36.8}{ \sqrt{22}}=105.34
Upper end of confidence interval = m+z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Upper end of confidence interval=123.6+2.327* \frac{36.8}{ \sqrt{22}}=141.86

Thus the 98% confidence interval will be (105.34, 141.86)


Part 4) Comparison of Confidence Intervals
The 90% confidence interval is: (110.69, 136.51)
The 95% confidence interval is: (108.22, 138.98)
The 98% confidence interval is: (105.34, 141.86)

As the level of confidence is increasing, the width of confidence interval is also increasing. So we can conclude that increasing the confidence level increases the width of confidence intervals.
3 0
4 years ago
125% of wat number is 45
Masteriza [31]

Answer:

36

Step-by-step explanation:

125% of what number equals 45? Answer: 36

4 0
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