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Irina18 [472]
2 years ago
13

HELP PLEASE!! Based on the table, which best predicts the end behavior of the graph of f(x)?

Mathematics
2 answers:
Mrac [35]2 years ago
6 0
The answer would be D, it's pretty self explanatory if you look at the choices. Hope this helps!
adelina 88 [10]2 years ago
4 0

Solution: The correct option is option D.

Explanation:

The end behavior is the behavior of function as the input of the function approaches to very large value and very small value.

If a function is f(x), then at x\rightarrow \infty and x\rightarrow -\infty  the value of f(x) approaches to either infinity or negative infinity which defines the end behavior.

From the given table it is noticed that the as x decreases and approaches to large negative values then the values of the function is also approaches to the large negative values. According to the table as x\rightarrow -\infty, f(x)\rightarrow -\infty.

From the given table it is noticed that the as x increases and approaches to large positive values then the values of the function is approaches to the large negative values. According to the table as x\rightarrow \infty, f(x)\rightarrow -\infty .

Since only option D satisfy the both conditions, therefore the correct option is option D.

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The answer is 224.7 you got to turn 30% into decimal which is 0.30 
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A rice merchant purchased 50 bags of rice at rupees 300 per back he spent 500 rupees towards transportation due to lack of deman
Alex777 [14]

Answer:

10500 rupees

Step-by-step explanation:

So 50 bags 300 per pack so

300*50= 15000

15000 is what he spent on the 50 bags of rice

So he spent 500 rupees on transportation

15000 + 500 = 15500 total spent which is not needed but extra information

30% loss is 30% of 15000 so

30% of 15000 = 4500

15000 - 4500 = 10500 rupees

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2 years ago
The GPA of accounting students in a university is known to be normally distributed. A random sample of 20 accounting students re
Vladimir79 [104]

Answer:

The 95% of confidence intervals

(2.84 ,2.99)

Step-by-step explanation:

A random sample of 20 accounting students results in a mean of 2.92 and a standard deviation of 0.16

given small sample size n =20

sample mean x⁻ =2.92

sample standard deviation 'S' =0.16

level of significance ∝ =  0.95

The 95% of confidence intervals

x^{-}  ± t_{\alpha } \frac{S}{\sqrt{n} }

the degrees of freedom γ=n-1 =20-1=19

t-table 2.093

(x^{-}  - t_{\alpha } \frac{S}{\sqrt{n} },x^{-}  + t_{\alpha } \frac{S}{\sqrt{n} })

(2.92 - 2.093(\frac{0.16}{\sqrt{20} } ,2.92+2.093(\frac{0.16}{\sqrt{20} } )

(2.92-0.0748,2.92+0.0748)

(2.84 ,2.99)

Therefore the 95% of confidence intervals

(2.84 ,2.99)

4 0
2 years ago
Les is measuring the border of her bulletin board. She measures around the entire outside of the bulletin board and finds the di
vagabundo [1.1K]

Answer:

Perimeter.

Step-by-step explanation:

Lee measures around the entire outside of the bulletin board and finds the distance is 32 units.

She is measuring the border of her bulletin board.

The sum of outer covering of any object is called its perimeter. Here, 32 unit shows the perimeter of the bulletin board.

7 0
2 years ago
Darren invests $4,500 into an account that earns 5% annual interests. How much will be in the account after 10 years if the inte
xz_007 [3.2K]

We have been given that Darren invests $4,500 into an account that earns 5% annual interests. We are asked to find the amount in his account after 10 years, if the interest rate is compounded annually, quarterly, monthly, or daily.

We will use compound interest formula to solve our given problem.  

A=P(1+\frac{r}{n})^{nt}, where,

A = Final amount,

P = Principal amount,

r = Annual interest rate in decimal form,

n = Number of times interest is compounded per year,

t = Time in years.

5\%=\frac{5}{100}=0.05  

When compounded annually, n=1:

A=4500(1+\frac{0.05}{1})^{1\cdot 10}

A=4500(1.05)^{10}

A=4500(1.6288946267774414)

A=7330.025820498\approx 7330.03

When compounded quarterly, n=4:

A=4500(1+\frac{0.05}{4})^{4\cdot 10}

A=4500(1.0125)^{40}

A=4500(1.6436194634870132)

A=7396.28758569\approx 7396.29

When compounded monthly, n=12:

A=4500(1+\frac{0.05}{12})^{12\cdot 10}

A=4500(1.00416666)^{120}

A=4500(1.64700949769)

A=7411.542739605\approx 7411.54

When compounded daily, n=365:

A=4500(1+\frac{0.05}{365})^{365\cdot 10}

A=4500(1.0001369863013699)^{3650}

A=4500(1.6486648137656943695)

A=7418.9916619456\approx 7419.00

Since amount earned will be maximum, when interest is compounded daily, therefore, Darren should use compounded daily interest rate.

4 0
2 years ago
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