Answer:
n > 96
Therefore, the number of samples should be more than 96 for the width of their confidence interval to be no more than 10mg
Step-by-step explanation:
Given;
Standard deviation r= 25mg
Width of confidence interval w= 10mg
Confidence interval of 95%
Margin of error E = w/2 = 10mg/2 = 5mg
Z at 95% = 1.96
Margin of error E = Z(r/√n)
n = (Z×r/E)^2
n = (1.96 × 25/5)^2
n = (9.8)^2
n = 96.04
n > 96
Therefore, the number of samples should be more than 96 for the width of their confidence interval to be no more than 10mg
Answer:
Below!
Step-by-step explanation:
Solve for y.
Rewrite in slope-intercept form.
Use the slope-intercept form to find the slope and y-intercept.
Any line can be graphed using two points. Select two x values, and plug them into the equation to find the corresponding y values.
Graph the line using the slope and the y-intercept, or the points.
Answer: $28,752
Explanation:
The<u> definition of bimonthly</u> is an event that happens once in two months.
represents "two", while
represents "event that happens by month.
Given that Glenda receives a salary of <u>$4,792 bi-monthly</u>, we are required to find the earnings per year.
There are <u>12 months in each year</u>, and if the given value is a 2-month value, then we shall divide 12 by 2 to find that there is in total 6 of the 2-month value.

Finally, we shall <u>multiply the total number with the salary</u>, which will be $4,792 times 6.

Hope this helps!! :)
Please let me know if you have any questions
So first create and define your variables:
Z = amount of zebra fish
N = amount of neon tetras
Now create your equations:
2z+2.15n=31.20
z+n=15
This is your system. There are multiple methods to use but in this problem I’ll use the substitution method by simplifying the bottom equation.
2z+2.15n=31.20
z=15-n
Now I’ll plug the bottom equation into the top one.
2(15-n)+2.15n=31.20
And just solve from here.
30-2n+2.15n=31.20
0.15n=1.20
n=8
So he bout 8 neon tetras, and 15-8= 7, so he bought 7 zebra fish
Answer:
165.6
Step-by-step explanation:
165.6 is 72% of 230