Answer:
3^3×3
81
3^7/3^3
Explanation:
First you should solve the expression you've got there.
(3^6×3^-4)^2
3^6= 3×3×3×3×3×3 the answer is 729
3^-4 A negative exponent means repeated division by the base. So for example 2^-4=1/(2^4)
=1/(2×2×2×2)= 1/16
3^-4= 1/(3^4)
=1/(3×3×3×3)
= 1/81 If you look up the answer in your calculator the answer that will show up will be 0.01234567901
So!
(729×0.01234567901)^2= 9^2
9^2= 81
Now you can look at the following expressions and whichever gives the answer 81 will be correct.
hope it helps!
Answer:
a) X ~ 
b) μ = 100/3
c) 
d) A battery is expected to last 100/3 months (33 months and 10 days approximately).
e) For seven batteries, i would expect them to last 700/3 months (approximately 19 years, 5 months and 10 days).
Step-by-step explanation:
a) The life of a battery is usually modeled with an exponential distribution X ~ 
b) The mean of X is μ = 1/0.03 = 100/3
c) The standard deviation is 
d) The expected value of the bateery life is equal to its mean, hence it is 100/3 months.
e) The expected value of 7 (independent) batteries is the sum of the expected values of each one, hence it is 7*100/3 = 700/3 months.
Answer:

Step-by-step explanation:
Step 1: Define
Difference Quotient: 
f(x) = -x² - 3x + 1
f(x + h) means that x = (x + h)
f(x) is just the normal function
Step 2: Find difference quotient
- <u>Substitute:</u>
![\frac{[-(x+h)^2-3(x+h)+1]-(-x^2-3x+1)}{h}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B-%28x%2Bh%29%5E2-3%28x%2Bh%29%2B1%5D-%28-x%5E2-3x%2B1%29%7D%7Bh%7D)
- <u>Expand and Distribute:</u>
![\frac{[-(x^2+2hx+h^2)-3x-3h+1]+x^2+3x-1}{h}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B-%28x%5E2%2B2hx%2Bh%5E2%29-3x-3h%2B1%5D%2Bx%5E2%2B3x-1%7D%7Bh%7D)
- <u>Distribute:</u>

- <u>Combine like terms:</u>

- <u>Factor out </u><em><u>h</u></em><u>:</u>

- <u>Simplify:</u>

Answer:
4
Step-by-step explanation:
ok
Answer:
It has no real solution.
Step-by-step explanation:
6p^2 − 7p + 5 = 0
For this equation: a = 6, b = -7, c = 5
6p^2+ −7p + 5 = 0
Step 1: Use quadratic formula with a=6, b=-7, c=5.
p = -b ± √b^2 - 4ac/2a
p = -(-7) ± √(-7)^2 - 4(6)(5)/2(6)
p = 7 ± √-71/12