Every 1 minute, you multiply the mass by 27.7% or 0.277
after 1 minute, you multiply once
after 2 minutes, you multiply twice
3, thrice
etc
so after 13 minutes, you multiply 13 times or 0.277^13
so what is 970g *0.277^13 =?
Answer: the equations are
0.02x + 0.07y = 156
y = 300 + x
Step-by-step explanation:
Let x represent the total dollar amount of phone sales that she makes.
Let y represent the the total dollar amount of computer sales that she makes.
Josiah earns a 2% commission on the total dollar amount of all phone sales he makes, and earns a 7% commission on all computer sales. She earned a total of $156 in commission. This means that
0.02x + 0.07y = 156 - - - - - - - - - - -1
Josiah had $300 more in computer sales than in phone sales. This means that
y = 300 + x
Answer:
14t + 58 ≤ 150
Step-by-step explanation:
If she cannot spend more than what she has, which is 150, the inequality sign has to be "less than or equal to". It's ok if she spends less than 150, but not ok if she spends more, because she doesn't have it to spend.
We know the cost of 1 pair of jeans is 58. Now she wants to make up the difference by getting as many $14 shirts as possible (the number of shirts being our unknown).
That means that the cost of the jeans PLUS the unknown number of shirts cannot exceed 150.
Therefore, the inequality is:
14t + 58 ≤ 150
Answer:
The standard deviation of car age is 2.17 years.
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 7.5
(a) If 99.7% of the ages are between 1 year and 14 years, what is the standard deviation of car age?
This means that 1 is 3 standard deviations below the mean and 14 is 3 standard deviations above the mean.
So

I want to find 



The standard deviation of car age is 2.17 years.
Benchmark are numbers that are used as standards to which the rest of the data is compared to. When counting numbers using a number line, the benchmark numbers are the intervals written on the axis. For benchmark numbers of 10, the number line on top of the attached picture is shown. Starting from 170, the tick marks are added by 10, such that the next numbers are 180, 190, 200, and so on and so forth. When you want to find 410, just find the benchmark number 410.
The same applies to benchmark numbers in intervals of 100. If you want to find 170, used the benchmark numbers 100 and 200. Then, you estimate at which point represents 170. For 410, you base on the benchmark numbers 400 and 500.