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babunello [35]
3 years ago
5

Describe the graph of the basic rational function, f(x)=1/x. Talk about its domain and range and compare the function to other o

nes, i.e. linear or quadratic or exponential.

Mathematics
1 answer:
Anna007 [38]3 years ago
4 0
F(x) = 1/x is a special function. is called the multiplicative inverse or the reciprocal function. the specific shape of the graph is called a rectangular hyperbola.

negative values of x with produce negative values of f(x) and positive values will produce positive function values. so this hyperbola is symmetric with respect to the origin and all negative values are in Q3 and all positive values are in Q1.

as x gets farther and farther from 0 in either direction, you can see that 1/x will be an ever-increasingly smaller distance from the x-axis. indeed, the limit of this function is zero as x approaches either positive or negative infinity.

if x is between -1 and 1, then the value of f(x) increases until x = 0, the limit of the function as x approaches 0 being positive and negative infinity depending on which direction you are coming from.

the domain is all real numbers with the exception of 0

the range also happens to be all real numbers with the exception of zero. all that really means is that zero has no multiplicative inverse.

strictly speaking, this is an exponential function, as the function is f(x) = x^(-1). 

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One less then five times a number is fewer then twenty four
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5n - 1 < 24 is the inequality in question.  Add 1 to both sides, obtaining:

5n < 25.  Divide both sides by 5, obtaining n < 5  (answer)

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3 years ago
an item is regularly priced at $70. It is on sale for 30% off the regular price. how much in dollars is discounted from the regu
sleet_krkn [62]
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7 0
3 years ago
A random sample of 12 supermarkets from Region 1 had mean sales of 84 with a standard deviation of 6.6. A random sample of 17 su
Sladkaya [172]

Answer:

We conclude that there is no difference in potential mean sales per market in Region 1 and 2.

Step-by-step explanation:

We are given that a random sample of 12 supermarkets from Region 1 had mean sales of 84 with a standard deviation of 6.6.

A random sample of 17 supermarkets from Region 2 had a mean sales of 78.3 with a standard deviation of 8.5.

Let \mu_1 = mean sales per market in Region 1.

\mu_2  = mean sales per market in Region 2.

So, Null Hypothesis, H_0 : \mu_1-\mu_2 = 0      {means that there is no difference in potential mean sales per market in Region 1 and 2}

Alternate Hypothesis, H_A : > \mu_1-\mu_2\neq 0      {means that there is a difference in potential mean sales per market in Region 1 and 2}

The test statistics that will be used here is <u>Two-sample t-test statistics</u> because we don't know about population standard deviations;

                            T.S.  =  \frac{(\bar X_1 -\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+ {\frac{1}{n_2}}} }   ~  t__n_1_+_n_2_-_2

where, \bar X_1 = sample mean sales in Region 1 = 84

\bar X_2 = sample mean sales in Region 2 = 78.3

s_1  = sample standard deviation of sales in Region 1 = 6.6

s_2  = sample standard deviation of sales in Region 2 = 8.5

n_1 = sample of supermarkets from Region 1 = 12

n_2 = sample of supermarkets from Region 2 = 17

Also, s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times  s_2^{2}  }{n_1+n_2-2} }  = s_p=\sqrt{\frac{(12-1)\times 6.6^{2}+(17-1)\times  8.5^{2}  }{12+17-2} } = 7.782

So, <u><em>the test statistics</em></u> =  \frac{(84-78.3)-(0)}{7.782 \times \sqrt{\frac{1}{12}+ {\frac{1}{17}}} }  ~   t_2_7

                                   =  1.943  

The value of t-test statistics is 1.943.

 

Now, at a 0.02 level of significance, the t table  gives a critical value of -2.472 and 2.473 at 27 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so we have<u><em> insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.

Therefore, we conclude that there is no difference in potential mean sales per market in Region 1 and 2.

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3 years ago
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mart [117]

Answer:

qw567

Step-by-step explanation:

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Find the condition that one root of the quadratic equation may be 1 more than the other.
Eddi Din [679]
<span>Let p, np be the roots of the given QE.So p+np = -b/a, and np^2 = c/aOr (n+1)p = -b/a or p = -b/a(n+1)So n[-b/a(n+1)]2 = c/aor nb2/a(n+1)2 = cor nb2 = ac(n+1)2
Which will give can^2 + (2ac-b^2)n + ac = 0, which is the required condition.</span>
4 0
4 years ago
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