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3 years ago

Which of the functions have a range of all real numbers? Check all that apply.

Mathematics

2 answers:

____ [38]3 years ago

8 0

<h2>Hello!</h2>

The answer is: The correct options are B. Tan(x) and C. Cot(x)

The range is the output of a function when we evaluate the function with inputs (domain), not all functions have a range of all real numbers, however it's more common to find functions without range or domain restrictions.

Both options B. Tan(x) and C. Cot(x) have a range of all real numbers while the functions Sec(x) and Csc (x) have a restricted range.

Sec(x) and Csc(x):

Domain, all real numbers.

Range, from -1 to 1 or [-1,1]

Tan(x) and Cot(x):

Domain, all real numbers except $\frac{\pi }{2}+-n\pi$

Range, all real numbers.

Hence, the correct options are B. Tan(x) and C. Cot(x)

Have a nice day!

Svetach [21]3 years ago

6 0

Answer:

B. y = tan x and C. y = cot x

Step-by-step explanation:

Range:

range is defined as the set of values that any given function y=f(x), can take for which the x values are defined. It is the value that y can take for the values of x.

Given:

From the given options

the range of tanx and cotx are all real numbers

as the period of tanx and cotx are π

where as range of secx is all real numbers except π/2 + n*π i.e.

(-∞ , -1] U [1 , + ∞)

And range of cscx is all real numbers except n*π i.e.

(-∞ , -1] U [1 , + ∞)

So from the given option:

option B and C are correct while A and D are not !

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The recursive formula would be:
$a_1=10
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This is because the first term, a₁, is 10. A recursive rule is dependent upon the term before it (a(n-1)) and the common ratio. The common ratio, or the number that each term is multiplied by, is -8.

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3 years ago

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Find an equation for the perpendicular Bisector of the line segment whose endpoints are (-9,-8) and (7,-4

Alina [70]

equation for the perpendicular Bisector of the line segment whose endpoints are (-9,-8) and (7,-4)

Perpendicular bisector lies at the midpoint of a line

Lets find mid point of (-9,-8) and (7,-4)

midpoint formula is

$\frac{x_1+x_2}{2} ,\frac{y_1+y_2}{2}$

$\frac{-9+7}{2} ,\frac{-8+-4}{2}$

midpoint is (-1, -6)

Now find the slope of the given line

(-9,-8) and (7,-4)

$slope = \frac{y2-y1}{x2-x1} = \frac{-4-(-8)}{7-(-9)}$

$slope = \frac{-4+8}{7+9}= \frac{4}{16} =\frac{1}{4}$

Slope of perpendicular line is negative reciprocal of slope of given line

So slope of perpendicular line is -4

slope = -4 and midpoint is (-1,-6)

y - y1 = m(x-x1)

y - (-6) = -4(x-(-1))

y + 6 = -4(x+1)

y + 6 = -4x -4

Subtract 6 on both sides

y = -4x -4-6

y= -4x -10

equation for the perpendicular Bisector y = -4x - 10

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3 years ago

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.

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3 years ago

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hram777 [196]

Of 85% of cars sold where not black then 100-85% where black = 15%
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then 15%*y=300
y= 300 divided by 15%
= (300*100)/15
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3 years ago