Answer:
(a) What will his age be after 5 years?
"5 + y" years old
(b) What was his age 6 years ago?
"y - 6" years old
(c) If his grandfather’s age is five times his, how old is his grandfather?
"5y" years old
(d) His father is 6 years older than three times his age. How old is his father?
"6 + 3y" years old
Note: Disregard the quotation marks, ""
Response:
2:3 is
✔ lesser than
5:6
10:12 is
✔ greater than
6:9
The ratios shown in Table 1 are
✔ smaller than
those in Table 2
Step-by-step explanation:
Answer:
The anticipated number of tests required to identify 680 acceptable circuits is 907.
Step-by-step explanation:
For any circuit, there are two potential results: it either passes the test or it fails. The likelihood of passing is independent between circuits. Therefore, we apply the binomial probability distribution to address this scenario.
Binomial probability distribution
This distribution calculates the chance of obtaining exactly x successes across n trials, where x has only two possible outcomes.
To find the expected number of trials to achieve r successes with a probability p, the formula is given by:

Circuits from a specific factory pass a certain quality evaluation with a probability of 0.75.
Thus, to determine the expected number of tests needed for 680 acceptable circuits, let’s denote this as E where r = 680.



The expected number of tests necessary to find 680 acceptable circuits is 907.
Answer:
Step-by-step explanation:
The prices he received quotes for are as follows: $663, $273, $410, $622, $174, $374
To begin, we will find the average.
Average = total of data points/ number of data points.
Total of data points =
663 + 273 + 410 + 622 + 174 + 374
= 2516
Total count = 6
Average = 2516/6 = 419.33
Standard deviation = √summation(x - m)^2/n
summation(x - m)^2/n = (663 - 419.33)^2 + (273 - 419.33)^2 + (410 - 419.33)^2 + (622 - 419.33)^2 + (174 - 419.33)^2 + (374 - 419.33)^2
= 179417.9334/6 = 29902.9889
Standard deviation = √29902.9889
= 172.9
There were 30 adults and 10 children, totaling 40 attendees. Just sum them together.