8 out of 300 were defective.
80 pairs were defective
Answer:
A. 4.26 in^2
Step-by-step explanation:
Step 1: Find the area of the sector DBC. Here we have to use the formula.
Area of a sector = Central Angle/360 *π
The area of the sector DBC = (54/360)*3.14*
= 30.16
Step 2: Area of segment CFD = Area of the sector DBC - Area of the ΔCBD
= 30.17 - 25.9
Area of segment CFD = 4.27in^2
Answer:
reflected
Step-by-step explanation:
ΔXYZ was dilated, then reflected, to create ΔQAG.
Answer:
The center/ mean will almost be equal, and the variability of simulation B will be higher than the variability of simulation A.
Step-by-step explanation:
Solution
Normally, a distribution sample is mostly affected by sample size.
As a rule, sampling error decreases by half by increasing the sample size four times.
In this case, B sample is 2 times higher the A sample size.
Now, the Mean sampling error is affected and is not higher for A.
But it's sample is huge for this, Thus, they are almost equal
Variability of simulation decreases with increase in number of trials. A has less variability.
With increase number of trials, variability of simulation decreases, so A has less variability.
Answer:
Total, T = (300s+200b+42d) mg
Step-by-step explanation:
Given that,
Kayson mixes 300 mL spinach, 200 mL of berries, and 42 mL of dressing to make a salad.
There are s mg of vitamin C per mL of spinach, b mg per mL of berries, and d mg per mL of dressing.
In 300 mL of spinach vitamin C is = (300 s)mg
In 200 mL of berries vitamin C is = (200 b)mg
In 42 mL of salad vitamin C is = (42 d)mg
It means that, total mg of vitamin C is :
Total, T = (300s+200b+42d) mg
Hence, this is the required solution.
