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3 years ago

Write a ratio to express the number of people who preferred WANR to those who preferred WCLM.

Mathematics

2 answers:

Anika [276]3 years ago

6 0

WANR - 22%
WCLM - 24%
WWCN - 41%
WKOD - 13%

The ratio of WANR to WCLM is 22:24 OR 11:12
Ratio of WANR to the whole is 22:100 OR 11:50
Ratio of WCLM to the whole is 24:100 OR 12:50

DENIUS [597]3 years ago

5 0

Answer: 11/12 is the ratio to express the number of people who preferred WANR to those who preferred WCLM.

Step-by-step explanation:

Let the total number of people= x

So, According to the given diagram,

The number of people who preferred WANR = 22% of x

While, The number of people who preferred WCLM = 24% of x

Therefore, The number of people who preferred WANR/The number of people who preferred WCLM = 22% of x / 24% of x

= $\frac{22/100\times x}{24/100\times x}$ = $\frac{22}{24} =\frac{11}{12}$

Which is the required answer.

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What best describes the asymptote of an exponential function of the form f(x)=b^x?

horrorfan [7]

It depends on the value of b

Exponential functions are defined for positive values of the base [ŧex] b [/tex]. Also, they're not defined if $b=1$ , or at least let's say that this is a trivial case, since it is the constant function 1.

Case 0<b<1:

If the base sits between 0 and 1, the exponential function is constantly decreasing. The explanation is quite intuitive, assume for the sake of simpleness that b is rational. If it sits between 0 and 1, it means that it can be written as a fraction p/q, with p<q. So, if you give a large exponent to b, you obtain

$\frac{p^x}{q^x},\quad p^x \ll q^x \text{ as } x \to \infty$

On the other hand, if you consider negative exponents, you switch numerator and denominator and then raise to the same exponent the fraction q/p, which gets larger and larger.

So, if 0<b<1, we have

$\lim_{x \to -\infty} b^x = \infty,\qquad \lim_{x \to \infty} b^x = 0$

and thus 0 is a horizontal asymptote as x tends to (positive) infinity.

Case b>1:

This case is very similar, except all roles are inverted. Now you start with a fraction p/q where p>q. So, with positive, large exponents you get

$\frac{p^x}{q^x},\quad p^x \gg q^x \text{ as } x \to \infty$

And as before, negative exponents switch numerator and denominator, so the fraction becomes q/p and thus you have

$\lim_{x \to -\infty} b^x = 0,\qquad \lim_{x \to \infty} b^x = \infty$

So, again, 0 is a horizontal asymptote, but this time for x tending towards negative infinite.

6 0

3 years ago

Read 2 more answers

If r = 4 and a6 = 192, what is the first term of the sequence?

blagie [28]

Answer:

Answer: Option c. 0.1875 is the first term of the sequence.

Step-by-step explanation:

In this question sixth term of a sequence is given as A(6)= 192

Ratio of terms is given as 4

And we have to calculate the first term of the sequence.

As per formula A(n) = A(1)× $r^{n-1}$

A(6) = $A(1)(r)^{6-1}$

192 = $A(1)4^{5}$

A(1) = $\frac{192}{4^{5} }$

= 192/1024

=0.1875

So the first term of the sequence will be 0.1875

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3 years ago

5y+30 5 complete the factoring ...?

krek1111 [17]

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</u> 5
If it is 5y+30 it would be just in the same process, Divide the equation by 5 and the result will be 5(y+6).

8 0

3 years ago

Mary and ann made drinks for a picnic. one used 3 cans juice concentrate and 8 cans water. the other 5 cans juice concentrate an

Sveta_85 [38]

I would think that it would be the 3:8 one because if you do the math, 8-3 is 5 and 12-5 is 7. The only reason Im doing it this way is because its the only way I can set it up.

so you have more of a difference with the 12:5 one and less of a difference with the 3:8.

I hope I helped a bit

4 0

3 years ago

The graphs of the function f (given in blue) and g (given in red) are plotted above. Suppose that u(x)=f(x)g(x) and v(x)=f(x)/g(

Rus_ich [418]

U(x) = f(x).(gx)
v(x) = f(x) / g(x)

Use chain rule to find u(x) and v(x).

u '(x) = f '(x) g(x) + f(x) g'(x)

v ' (x) = [f '(x) g(x) - f(x) g(x)] / [g(x)]^2

The functions given are piecewise.

You need to use the pieces that include the point x = 1.

You can calculate f '(x) and g '(x) at x =1, as the slopes of the lines that define each function.

And the slopes can be calculated graphycally as run / rise of each graph, around the given point.

f '(x) = slope of f (x); at x = 1, f '(1) = run / rise = 1/1 = 1

g '(x) = slope of g(x); at x = 1, g '(1) = run / rise = 1.5/ 1 = 1.5

You also need f (1) = 1 and g(1) = 2

Then:

u '(1) = f '(1) g(1) + f(1) g'(1) = 1*2 + 1*1.5 = 2 + 1.5 = 3.5

v ' (x) = [f '(1) g(1) - f(1) g(1)] / [g(1)]^2 = [1*2 - 1*1.5] / (2)^2 = [2-1.5]/4 =

= 0.5/4 = 0.125

Answers:
u '(1) = 3.5
v '(1) = 0.125

3 0

3 years ago