Find the following for the function f(x) = 3x2 + 4x - 4.(a) f(0)
(e) - f(x)
(b) f(3)
(f) f(x+3)
(c) f(-3)
(g) f(3x)
(d) f(-x)
(h) f(x+h)
(a) f(0) = (Simplify your answer.)
(b) f(3) = (Simplify your answer.)
(c) f(-3)=(Simplify your answer.)

Answers

Answer 1

Answer:

Answer:

f(0)=2

f(3)=14

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Find the arc length of the curve below on the given interval. y equals one third (x squared plus 2 )Superscript 3 divided by 2y= 1 3x2+23/2 on [00,66]

Answers

Answer

$\int_(0)^(6)√(1+12x^4+8x^2)dx$

Step-by-step explanation:

We are given that

$y=(1)/(3)(3x^2+2)^{(3)/(2)}$

Interval=[0,6]

a=0 and b=6

Differentiate w.r. t x

$(dy)/(dx)=(1)/(3)(3x^2+2)^{(1)/(2)}* 6x=2x(3x^2+2)^{(1)/(2)}$

By using the formula ; $(dx^n)/(dx)=nx^(n-1)$

We know that arc length of curve

$s=\int_(a)^(b)\sqrt{1+((dy)/(dx))^2}dx$

Substitute the values

$s=\int_(0)^(6)\sqrt{1+(2x(3x^2+2)^{(1)/(2)})^2}dx$

$s=\int_(0)^(6)√(1+4x^2(3x^2+2))dx$

$s=\int_(0)^(6)√(1+12x^4+8x^2)dx$

Length of curve,= $s=\int_(0)^(6)√(1+12x^4+8x^2)dx$

Translate into a variable expression.the square of the difference between a number n and fifty

Answers

Answer:

f(n)= (n-50)²

Step-by-step explanation:

The difference between a number n and fifty:

n - 50

Square of this difference:

(n-50)²

It would look like this as variable expression:

f(n)= (n-50)²

What is the value of K?

Answers

The value of K is 10

Zoey invested $230 in an account paying an interest rate of 6.3% compounded daily.Assuming no deposits or withdrawals are made, how much money, to the nearest
hundred dollars, would be in the account after 12 years?

Answers

Answer:

A ≈ $500

General Formulas and Concepts:

Pre-Alg

Order of Operations: BPEMDAS

Algebra I

Compounded Interest Rate: A = P(1 + r/n)ⁿˣ

A is final amount
P is initial (principle) amount
r is rate
n is number of compounds
x is number of years

Step-by-step explanation:

Step 1: Define

P = 230

r = 0.063

n = 365

x = 12

Step 2: Solve for A

Substitute: A = 230(1 + 0.063/365)³⁶⁵⁽¹²⁾
Divide: A = 230(1 + 0.000173)³⁶⁵⁽¹²⁾
Multiply: A = 230(1 + 0.000173)⁴³⁸⁰
Add: A = 230(1.00017)⁴³⁸⁰
Exponents: A = 230(2.1296)
Multiply: A = 489.808

Use the given degree of confidence and sample data to construct a confidence interval for the population mean mu μ. Assume that the population has a normal distribution. A laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was 185 milligrams with s equals =17.6 milligrams. Construct a 95% confidence interval for the true mean cholesterol content of all such eggs A. 175.9 mg less than < mu μ less than <194.1 mg
B. 173.9 mg less than < mu μ less than <196.1 mg
C. 173.8 mg less than < mu μ less than <196.2 mg
D. 173.7 mg less than < mu μ less than <196.3 mg

Answers

Answer:

option (C) 173.8 mg less than < mu μ less than <196.2 mg

Step-by-step explanation:

Data provided ;

number of sample, n = 12

Mean = 185 milligram

standard deviation, s = 17.6 milligrams

confidence level = 95%

α = 0.05 [for 95% confidence level]

df = n - 1 = 12 - 1 = 11

Now,

Confidence interval = Mean ± E

here,

E is the margin of error = $t_(\alpha/2, df)(s)/(√(n))$

also,

$t_(\alpha/2, df)$

= $t_(0.05/2, (11))$

= 2.201 [ from standard t value table]

Thus,

E = $2.201*(17.6)/(√(12))$

E = 11.182 milligrams ≈ 11.2 milligrams

Therefore,

Confidence interval:

Mean - E < μ < Mean + E

185 - 11.2 < μ < 185 + 11.2

173.8 < μ < 196.2

Hence,

the correct answer is option (C) 173.8 mg less than < mu μ less than <196.2 mg

Final answer:

To construct a confidenceinterval for the population mean cholesterol content of all chicken eggs with a 95% confidence level, we use the sample mean, standard deviation, and sample size to calculate the margin of error. The confidence interval is then constructed by subtracting the margin of error from the sample mean and adding it to the sample mean.

Explanation:

To construct a confidenceinterval for the population mean cholesterol content of all chicken eggs, we first need to find the margin of error. The margin of error depends on the samplemean, standard deviation, sample size, and the desired level of confidence. In this case, we have a sample mean of 185 mg, a standard deviation of 17.6 mg, and a sample size of 12. Since we want a 95% confidence interval, we use a z-score of 1.96. The margin of error is then calculated as 1.96 * (17.6/sqrt(12)), which is approximately 9.61 mg. We can then construct the confidenceinterval by subtracting the margin of error from the sample mean and adding it to the sample mean. Therefore, the 95% confidence interval for the true mean cholesterol content of all such eggs is 175.9 mg to 194.1 mg.