For section 3.01 black, number 1 is correct, but number 2 is wrong.
When you raise an exponent to an exponent, you multiply the 2 exponents.
(x^4)^5 is x^20.
Number 3 is also right.
For 3.02, you use the interest formula. (1 + i/100)^t times x
x is the amount of money you have originally. i is the interest rate, t is the time.
1,500(1.03)^5 = 1738.91111145
$1738.91
For section 3.01 red in fractional exponents the numerator are the powers and the denominator is the root.
![\sqrt[4]{a^{3} } = a^{\frac{3}{4} }](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Ba%5E%7B3%7D%20%7D%20%20%3D%20a%5E%7B%5Cfrac%7B3%7D%7B4%7D%20%7D)
Answer:
One thing that must be true is that the objects will cause some leaning because their masses are unequal, judging from the volume that the objects can hold. One thing that could be true is that the triangle could be heavier than the square, leading to a leaning toward the left. The second pic cannot be true, for there are too much mass concentrated on one side to maintain equilibrium.
Answer:

Step-by-step explanation:
we know that
To find out the scale factor divide the corresponding dimensions of the model by the corresponding dimensions of the actual
so

simplify

That means
1 inch in the model represent 9 feet in the actual
Answer:
13.92
Step-by-step explanation:
We have that the critical z-score associated with 85% to the left is 1.04, we know that by table.
So we have to:
m + z * (sd) = 16
where m is the mean, z is the critical z-scor and sd is the standard deviation, if we replace we are left with:
m + 1.04 * (2) = 16
m = 16 - 1.04 * (2)
m = 13.92
Therefore, the average weight if 85% of cucumbers weigh less than 16 ounces is 13.92
Answer: 
Step-by-step explanation:
Let
be the sample proportion.
As per given , we have
n= 411

Standard error : se=0.016
Critical value for 99% confidence : 
Confidence interval for population proportion :-

Hence, the 99% confidence interval for the proportion of all US adults ages 55 to 64 to use online dating : 