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azamat
3 years ago
10

Simplify the function. Than determine the key aspects of the function

Mathematics
1 answer:
Tom [10]3 years ago
8 0

To simplify the function, we need to know some basic identities involving exponents.


1. b^(ax)=(b^x)^a=(b^a)^x

2. b^(x/d) = (b^x)^(1/d) = ((b^(1/d)^x)


Now simplify f(x), where

f(x)=(1/3)*(81)^(3*x/4)

=(1/3)(3^4)^(3*x/4) [ 81=3^4 ]

=(1/3)(3^(4*3*x/4) [ rule 1 above ]

=(1/3) (3^(3*x)

=(1/3)(3^(3x)) [ or (1/3)(27^x), by rule 1 ]



(A) Initial value is the value of the function when x=0, i.e.

initial value

= f(0)

=(1/3)(3^(3x))

=(1/3)(3^(3*0))

=(1/3)(3^0)

=(1/3)(1)

=1/3


(B) the simplified base base is 3 (or 27 if the other form is used)


(C) The domain for an exponential function is all real values ( - ∞ , + ∞ ).


(D) The range of an exponential function with a positive coefficient and without vertical shift is ( 0, + ∞ ).

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Answer:

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8 0
3 years ago
While conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modem
Kitty [74]

Answer:

We conclude that this is an unusually high number of faulty modems.

Step-by-step explanation:

We are given that while conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modems.

The probability of obtaining this many bad modems (or more), under the assumptions of typical manufacturing flaws would be 0.013.

Let p = <em><u>population proportion</u></em>.

So, Null Hypothesis, H_0 : p = 0.013      {means that this is an unusually 0.013 proportion of faulty modems}

Alternate Hypothesis, H_A : p > 0.013      {means that this is an unusually high number of faulty modems}

The test statistics that would be used here <u>One-sample z-test</u> for proportions;

                             T.S. =  \frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }  ~  N(0,1)

where, \hat p = sample proportion faulty modems= \frac{10}{367} = 0.027

           n = sample of modems = 367

So, <u><em>the test statistics</em></u>  =  \frac{0.027-0.013}{\sqrt{\frac{0.013(1-0.013)}{367} } }

                                     =  2.367

The value of z-test statistics is 2.367.

Since, we are not given with the level of significance so we assume it to be 5%. <u>Now at 5% level of significance, the z table gives a critical value of 1.645 for the right-tailed test.</u>

Since our test statistics is more than the critical value of z as 2.367 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u><em>we reject our null hypothesis</em></u>.

Therefore, we conclude that this is an unusually high number of faulty modems.

6 0
3 years ago
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Alexxx [7]

Answer:

10:40

Step-by-step explanation:

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Step-by-step explanation:

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