On a local swim team 6 members each swam 1,685 meters during the season. How many total kilometers did the 6 members swim during
1 answer:
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Answer:
The approximate total cost for the blanket is $10.08 ⇒ B
Step-by-step explanation:
The formula of the area of a rectangle is A = L × W
The formula of the perimeter of a rectangle is P = 2(L + W)
where:
- L is the length of the rectangle
- W is the width of the rectangle
∵ The baby blanket shaped a rectangle
∵ The dimensions of the blanket are 3 feet and 5 feet
- Find the area of the blanket
∵ A = 3 × 5 = 15 feet²
- Change feet to yard because the cost is by square yard
∵ 1 foot = yard
∴ 1 foot² = yard²
∴ 15 feet² = 15 × =
yards²
∴ The area of the blanket is yards²
∵ The fabric cost of the blanket is $1.25 per square yard
∴ The fabric cost = × 1.25
∴ The fabric cost = 2.08333 dollars
∵ Heather is going to put a ribbon border around it
- The length of the ribbon is equal to the perimeter of the blanket
∵ P = 2(3 + 5) = 16 feet
∴ The length of the ribbon is 16 feet
∵ The cost of the ribbon is $0.5 per foot
∴ The cost of the ribbon = 16 × 0.5
∴ The cost of the ribbon = 8 dollars
Now add the fabric cost to the ribbon cost to find the total cost of the blanket
∵ The total cost = 2.08333 + 8
∴ The total cost = 10.08333
- Round it to the nearest cent
∴ The total cost = 10.08 dollars
The approximate total cost for the blanket is $10.08
Answer:
A 90% confidence interval for the mean number of letter sounds identified in one minute is: (30.84, 37.34).
Step-by-step explanation:
We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of .
That is z with a pvalue of , so Z = 1.645.
Now, find the margin of error M as such
In which is the standard deviation of the population and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 34.09 - 3.25 = 30.84.
The upper end of the interval is the sample mean added to M. So it is 34.09 + 3.25 = 37.34.
A 90% confidence interval for the mean number of letter sounds identified in one minute is: (30.84, 37.34).