Answer:
The 80% confidence interval for the the population mean nitrate concentration is (0.144, 0.186).
Critical value t=1.318
Step-by-step explanation:
We have to calculate a 80% confidence interval for the mean.
The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.
The sample mean is M=0.165.
The sample size is N=25.
When σ is not known, s divided by the square root of N is used as an estimate of σM:
The degrees of freedom for this sample size are:
The t-value for a 80% confidence interval and 24 degrees of freedom is t=1.318.
The margin of error (MOE) can be calculated as:
Then, the lower and upper bounds of the confidence interval are:
The 80% confidence interval for the population mean nitrate concentration is (0.144, 0.186).
The x and y axis so that’s why it’s true
Answer:
Class 7A collected 4.8 ounces of pure gold.
Step-by-step explanation:
<u>Key skills required are: Percentages, Multiplication</u>
<u />
- We only need the information about Class 7A. They collected 40 oz that contains 12% gold.
- In other words, this means that 12% of that 40 oz gold sand is pure gold.
Here we have to do 12% x 40 to find the number of oz of pure gold. We first have to convert 12% into a decimal. Divide it by a 100 (or move the decimal point 2 places to the left) and you will get 0.12.
<em>Do 0.12 x 40 and you will get 4.8</em>
Therefore, there are 4.8 oz of pure gold in Class 7A's gold sand
You would need exactly 9 correct questions to get a 50%.
We can find this by multiplying the total number of questions by the percentage we are looking for.
18*50% = 9