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maks197457 [2]
3 years ago
14

Yeah thanks m8 for this

Mathematics
1 answer:
nikitadnepr [17]3 years ago
4 0

Step-by-step explanation:

the process is shown in the picture.

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You have two exponential functions. One has the formula h(x) = 2 x + 3. The other function, g(x), has the graph shown below.
Alex17521 [72]
To answer, we substitute each of the numbers in parentheses to x of h(x) and get the value of g(x).

A. h(2) = 2(2) + 3 = 7  ;   g(2) < 4               (neither is equal to 7)

B. h(7) = 2(7) + 3 = 17;    g(7) is not shown in the graph          (not true)

C. h(2) = 7   ; g(2) < 4   Thus, g(2) < h(2)   (choice is not true)

D. From C, g(2) < h(2)   (TRUE)

Thus, the answer to this item is letter D. 
5 0
3 years ago
Read 2 more answers
Solve for x: 3(x - 2) = -9
Airida [17]

Answer:

X=-1

Step-by-step explanation:

3x-6=-9

3x=-9+6

3x=-3

x=-1

5 0
3 years ago
The Wall Street Journal reports that 33% of taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deduction
Len [333]

Answer:

(a) <em>                             </em><em>n</em> :      20           50          100         500

P (-200 < <em>X</em> - <em>μ </em>< 200) : 0.2886    0.4444    0.5954    0.9376

(b) The correct option is (b).

Step-by-step explanation:

Let the random variable <em>X</em> represent the amount of deductions for taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return.

The mean amount of deductions is, <em>μ</em> = $16,642 and standard deviation is, <em>σ</em> = $2,400.

Assuming that the random variable <em>X </em>follows a normal distribution.

(a)

Compute the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean as follows:

  • For a sample size of <em>n</em> = 20

P(\mu-200

                                           =P(-0.37

  • For a sample size of <em>n</em> = 50

P(\mu-200

                                           =P(-0.59

  • For a sample size of <em>n</em> = 100

P(\mu-200

                                           =P(-0.83

  • For a sample size of <em>n</em> = 500

P(\mu-200

                                           =P(-1.86

<em>                                  n</em> :      20           50          100         500

P (-200 < <em>X</em> - <em>μ </em>< 200) : 0.2886    0.4444    0.5954    0.9376

(b)

The law of large numbers, in probability concept, states that as we increase the sample size, the mean of the sample (\bar x) approaches the whole population mean (\mu_{x}).

Consider the probabilities computed in part (a).

As the sample size increases from 20 to 500 the probability that the sample mean is within $200 of the population mean gets closer to 1.

So, a larger sample increases the probability that the sample mean will be within a specified distance of the population mean.

Thus, the correct option is (b).

8 0
3 years ago
What are the solutions to the equation |x – 10| – 4 = 2x?
Yuliya22 [10]

|x – 10| – 4 = 2x

x - 10 - 4 = 2x

x - 14 = 2x

x - 2x = 14

-x = 14

x = -14

7 0
3 years ago
 (6 pts) The average age of CEOs is 56 years. Assume the variable is normally distributed. If the SD is four years, find the pr
Alenkinab [10]

Answer:

The probability that the age of a randomly selected CEO will be between 50 and 55 years old is 0.334.

Step-by-step explanation:

We have a normal distribution with mean=56 years and s.d.=4 years.

We have to calculate the probability that a randomly selected CEO have an age between 50 and 55.

We have to calculate the z-value for 50 and 55.

For x=50:

z=\frac{x-\mu}{\sigma}=\frac{50-56}{4}=\frac{-6}{4}=   -1.5

For x=55:

z=\frac{x-\mu}{\sigma}=\frac{55-56}{4}=\frac{-1}{4}=-0.25

The probability of being between 50 and 55 years is equal to the difference between the probability of being under 55 years and the probability of being under 50 years:

P(50

5 0
3 years ago
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