Answer:
a
Step-by-step explanation:
1∪ 2/3.....................
The answer is $412.
Let's first calculate simple interest. Simple interest (I) can be expressed as:
I = P * r * t
P - principal
r - rate
t - time period
It is given:
I = ?
P = $400
r = 3% = 0.03
t = 1 year
Therefore:
I = P * r * t = 400 * 0.03 * 1 = 12
The total amount Kate will repay is the principal amount (P) plus 3% simple interest (I):
P + I = 400 + 12 = $412
Answer:
![12800 + 100x \le 24000](https://tex.z-dn.net/?f=12800%20%2B%20100x%20%5Cle%2024000)
A maximum of 112 number of 100 - kilograms can be loaded in the container.
Step-by-step explanation:
Given that:
Weight of each crate = 100 kg
The greatest weight that can be loaded in the container = 24000 kg
Weight already loaded in the container = 12800 kg
To find:
The inequality to determine the value
i.e. number of 100 - kilograms that can be loaded in the shipping container?
Solution:
Weight already loaded = 12800 kg
Let the number of 100 - kilograms that can be loaded in the container = ![x](https://tex.z-dn.net/?f=x)
Weight of
= 100
kg
This combined weight nor be greater than the capacity of the container.
OR we can say, it must be lesser than or equal to greatest weight that can be loaded into the container.
![12800 + 100x \le 24000](https://tex.z-dn.net/?f=12800%20%2B%20100x%20%5Cle%2024000)
![100x \le 24000 - 12800\\\Rightarrow 100x \le 11200\\\Rightarrow x \le 112](https://tex.z-dn.net/?f=100x%20%5Cle%2024000%20-%2012800%5C%5C%5CRightarrow%20100x%20%5Cle%2011200%5C%5C%5CRightarrow%20x%20%5Cle%20112)
i.e. a maximum of <em>112</em> number of 100 - kilograms can be loaded in the container.