x⁴ + 4x³ + 4x² = -16x
x⁴ + 4x³ + 4x² + 16x = -16x + 16x
x⁴ + 4x³ + 4x² + 16x = 0
x(x³) + x(4x²) + x(4x) + x(16) = 0
x(x³ + 4x² + 4x + 16) = 0
x(x²(x) + x²(4) + 4(x) + 4(4)) = 0
x(x²(x + 4) + 4(x + 4)) = 0
x(x² + 4)(x + 4) = 0
x = 0 U x² + 4 = 0 U x + 4 = 0
<u> - 4 - 4</u> <u> - 4 - 4</u>
x² = -4 x = -4
x = <u>+</u>2i
The solution set is equal to {-4, -2i, 0, 2i}.
Part A: Equation 1: 20x +120
Equation 2: 25x +80
( I think this is the answer)
Part B:
Answer:
1/4
Step-by-step explanation:
1/3 - 1/12 = 4/12 - 1/12 (LCM)
= 3/12
=1/4 (simplified)
1. All the Y values decrease by 3, so this is a linear function.
Answer would be the third choice.
2. First, factor 2 out of the equation. Then divide both sides by 2. After that use the quadratic formula to solve for x.
X = 0.94, -3.19 The third choice is the answer
3. subtract 17 from both sides, now you have a polynomial. Using the quadratic formula to solve for X, the answer is 3.25, -5.24
Answer:
Step-by-step explanation:
Hello!
The variable of interest is
X: the amount a 10-11-year-old spends on a trip to the mall.
Assuming that the variable has a normal distribution, you have to construct a 98% CI for the average of the amount spent by the 10-11 year-olds on one trip to the mall.
For this you have to use a Student-t for one sample:
X[bar] ± *
n= 6
$18.31, $25.09, $26.96, $26.54, $21.84, $21.46
∑X= 140.20
∑X²= 3333.49
X[bar]= ∑X/n= 140.20/6= 23.37
S²= 1/(n-1)*[∑X²-(∑X)²/n]= 1/5[3333.49-(140.2)²/6]= 11.50
S= 3.39
X[bar] ± *
[23.37 ± 3.365 * ]
[18.66;29.98]
With a confidence level of 98%, you'd expect that the interval $[18.66;29.98] will include the population mean of the money spent by 10-11 year-olds in one trip to the mall.
I hope you have a SUPER day!