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Mademuasel [1]
2 years ago
8

Graph the functions and approximate an x-value in which the exponential function exceeds the polynomial function. y = 4x y = 7x2

+ 4x - 2

Mathematics
1 answer:
nordsb [41]2 years ago
5 0

Answer:

The Exponential function is

y=4^x

And the polynomial function is

y=7 x^2 +4 x -2

And, we have to find the value of x for which, exponential function exceeds the polynomial function which can be written as

4^x> 7 x^2 +4 x -2

1. When , x= -1

LHS

4^{-1}=\frac{1}{4}=0.25

RHS

=7*(-1)^2 +4 *(-1)-2\\\\=7-4-2=1

2. When , x=0

L HS

 4^0=1

RHS

7*0+4*0-2= -2

3. When ,x= 0.5

L HS

 4^{0.5}=2

RHS

=7*0.25 +4*0.5 -2\\\\= 1.75+2-2\\\\=1.75

4. When , x=2

LHS

 4^{2}=16

RHS

=7*4^2+4*4-2\\\\=112+16-2\\\\=126

The Minimum value for which  exponential function exceeds the polynomial function is , x= 0.5

But,there is other value for which  exponential function exceeds the polynomial function is , x=2.

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2x^2-7=74\\
2x^2=81\\
x^2=\dfrac{81}{2}\\
x=-\sqrt{\dfrac{81}{2}} \vee x=\sqrt{\dfrac{81}{2}}\\
x=-\dfrac{9}{\sqrt2} \vee x=\dfrac{9}{\sqrt2}}\\
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8 0
3 years ago
A highway department executive claims that the number of fatal accidents which occur in her state does not vary from month to mo
den301095 [7]

This question is incomplete, the complete question;

A highway department executive claims that the number of fatal accidents which occur in her state does not vary from month to month. The results of a study of 193 fatal accidents were recorded. Is there enough evidence to reject the highway department executive's claim about the distribution of fatal accidents between each month? Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Fatal Accidents 11 13 22 11 15 29 15 9 13 15 22 18

Step 10 of 10 : State the conclusion of the hypothesis test at the 0.1 level of significance.

Answer: H₀ : The distribution of fatal accidents between each month is same

Step-by-step explanation:

H₀ : The distribution of fatal accidents between each month is same

H₁ : The distribution of fatal accidents between each month is different

Let the los be alpha 0.10

given data ;

       Observed    Expected  

Mon   Freq (Oi)   Freq Ei       (Oi-Ei)^2 /Ei

Jan      11    16.0833        1.606649

Feb      13    16.0833        0.591105

Mar      22    16.0833        2.176598

Apr       11    16.0833        1.606649

May     15    16.0833        0.072971

Jun     29    16.0833        10.373489

Jul     15    16.0833        0.072971

Aug       9    16.0833        3.119603

Sep      13    16.0833        0.591105

Oct      15    16.0833        0.072971

Nov      22    16.0833        2.176598

Dec      18    16.0833       0.228411

Total:   193     93                22.689119

Expected Freq ⇒ 193 / 12 = 16.08333

Test Statistic, x² : 22.6891

Num Categories: 12

Degrees of freedom: 12 - 1 = 11

Critical value X² : 24.725

P-Value: 0.0195

since Chi-square value < Chi-square critical value

and P-value > alpha 0.01 so we accept H₀

Therefore we conclude that the distribution of fatal accidents between each month is same

7 0
3 years ago
Can somebody help me do this question?
Sonja [21]
Is this math homework?
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3 years ago
1 Simplify the expression
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Where’s the expression
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2 years ago
The sales price of a car is $12,590, which is 20% off the original price. What is the original price?
qaws [65]
We know that the final price after a discount of 20% is $12,590. If the original price was x, the price after discount is 80%<span> of x (</span>100%<span> - 20%). OP (original price) multiplied by (1-20/100) equals $15,737.5

I hope this helps.</span>
8 0
3 years ago
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