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andriy [413]
2 years ago
12

What substitution should be used to rewrite 4x12 – 5x6 – 14 = 0 as a quadratic equation? u = x2 u = x3      u = x6      u = x12

Mathematics
2 answers:
VLD [36.1K]2 years ago
8 0

Answer: u=x^6

Explanation:

A quadratic equation is an equation written in the form

au^2 + bu + c=0 (1)

where a,b,c are coefficients and where u is the variable.

Our original equation is:

4x^{12} - 5x^6 -14=0

If we make the substitution u=x^6, we obtain

4u^2 -5u-14 =0

Which corresponds to eq.(1), with coefficients a=4, b=-5, c=-14.

Arada [10]2 years ago
5 0
Sent a picture of the solution to the problem (s).

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Two processes are used to produce forgings used in an aircraft wing assembly. Of 200 forgings selected from process 1, 10 do not
tangare [24]

Answer:

a.

\bar p_1=0.05\\\bar p_2=0.067

b-Check illustration  below

c.(-0.0517,0.0177

Step-by-step explanation:

a.let p_1  \& p_2 denote processes 1 & 2.

For p_1: T1=10,n1=200

For p_2:T2=20,n2=300

Therefore

\bar p_1=\frac{t_1}{N_1}=\frac{10}{200}=0.05\\\bar p_2=\frac{t_2}{N_2}=\frac{20}{300}=0.067

b. To test for hypothesis:-

i.

H_0:p_1=p_2\\H_A=p_1\neq p_2\\\alpha=0.05

ii.For a two sample Proportion test

Z=\frac{\bar p_1-\bar p_2}{\sqrt(\bar p(1-\bar p)(\frac{1}{n_1}+\frac{1}{n_2})}\\

iii. for \frac{\alpha}{2}=(-1.96,+1.96) (0.5 alpha IS 0.025),

reject H_o if|Z|>1.96

iv. Do not reject H_o. The noncomforting proportions are not significantly different as calculated below:

z=\frac{0.050-0.067}{\sqrt {(0.06\times0.94)\times \frac{1}{500}}}

z=-0.78

c.(1-\alpha).100\% for the p1-p2 is given as:

(\bar p_1-\bar p_2)\pm Z_0_._5_\alpha \times \sqrt   \frac{ \bar p_1(1-\bar p_1)}{n_1}+\frac{\bar p_2(1-\bar p_2)}{n_2}\\\\=(0.05-0.067)\pm 1.645  \times \sqrt \ \frac{0.05+0.95}{200}+\frac{0.067+0.933}{300}\\

=(-0.0517,+0.0177)

*CI contains o, which implies that proportions are NOT significantly different.

4 0
2 years ago
The rabbit population on a small island is observed to be given by the function P(t) = 130t − 0.3t^4 + 1100 where t is the time
Montano1993 [528]
The maximum occurs when the derivative of the function is equal to zero.
P(t)=-0.3t^{4}+130t+1100 \\ P'(t)=-1.2t^{3}+130 \\ 0=-1.2t^{3}+130 \\ 1.2t^{3}=130 \\ t^{3}= \frac{325}{3}  \\ t=4.76702
Then evaluate the function for that time to find the maximum population.
P(t)=-0.3t^{4}+130t \\ P(4.76702)=-0.3*4.76702^{4}+130*4.76702+1100 \\ P(4.76702)=1564.79201
Depending on the teacher, the "correct" answer will either be the exact decimal answer or the greatest integer of that value since you cannot have part of a rabbit.
7 0
2 years ago
The volume of a rectangular prism is (x3 – 3x2 + 5x – 3), and the area of its base is (x2 – 2). If the volume of a rectangular p
IrinaK [193]
The first one is the answer because
8 0
2 years ago
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A quality control engineer tests the quality of produced computers. suppose that 5% of computers have defects, and defects occur
11111nata11111 [884]
(a) 0.059582148 probability of exactly 3 defective out of 20

 (b) 0.98598125 probability that at least 5 need to be tested to find 2 defective.

  (a) For exactly 3 defective computers, we need to find the calculate the probability of 3 defective computers with 17 good computers, and then multiply by the number of ways we could arrange those computers. So

 0.05^3 * (1 - 0.05)^(20-3) * 20! / (3!(20-3)!)

 = 0.05^3 * 0.95^17 * 20! / (3!17!)

 = 0.05^3 * 0.95^17 * 20*19*18*17! / (3!17!)

 = 0.05^3 * 0.95^17 * 20*19*18 / (1*2*3)

 = 0.05^3 * 0.95^17 * 20*19*(2*3*3) / (2*3)

 = 0.05^3 * 0.95^17 * 20*19*3

 = 0.000125* 0.418120335 * 1140

 = 0.059582148

  (b) For this problem, let's recast the problem into "What's the probability of having only 0 or 1 defective computers out of 4?" After all, if at most 1 defective computers have been found, then a fifth computer would need to be tested in order to attempt to find another defective computer. So the probability of getting 0 defective computers out of 4 is (1-0.05)^4 = 0.95^4 = 0.81450625.

 The probability of getting exactly 1 defective computer out of 4 is 0.05*(1-0.05)^3*4!/(1!(4-1)!)

 = 0.05*0.95^3*24/(1!3!)

 = 0.05*0.857375*24/6

 = 0.171475

 
 So the probability of getting only 0 or 1 defective computers out of the 1st 4 is 0.81450625 + 0.171475 = 0.98598125 which is also the probability that at least 5 computers need to be tested.
3 0
2 years ago
The number 8.64E9 appears on a calculator. In standard notation, this number is
Lunna [17]
Here are all of your answers:

1) 8,640,000,000
2) 864,000,000,000,000
3) 864,000,000,000
4) 86,400,000,000,000

I hope this helps! If it does, please rate as Brainliest :D
6 0
2 years ago
Read 2 more answers
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