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Sydney bought 5.65 lbs of roast beef, which was on sale for $4.50 per pound. if she divides the roast beef evenly among 6.5, how

Sever21 [200]

Okay. So, she buoyant 5.65 lbs of roast beef for $4.50. Multiply both of those numbers to get 25.425 or 25.43 when rounded to the nearest hundredth. So the total cost was $25.43. However, we now have to divide the amount by 6.5. 25.43/6.5 is 3.912 and other odd numbers or 3.91 when rounded to the nearest hundredth. The cost per sandwich was $3.91.

6 0

3 years ago

Find the equation of the line with slope = 2 and y-intercept = -8.

elena-s [515]

Answer: d

Step-by-step explanation:

Using the linear equation y=mx+b, you are given m (the slope) and b (the y-intercept). m=2 and b=-8. Therefore, y=2x-8

6 0

3 years ago

A quality analyst of a tennis racquet manufacturing plant investigates if the length of a junior's tennis racquet conforms to th

Burka [1]

Answer:

Confidence interval : $21.506$ to $24.493$

Step-by-step explanation:

A quality analyst selects twenty racquets and obtains the following lengths:

21, 25, 23, 22, 24, 21, 25, 21, 23, 26, 21, 24, 22, 24, 23, 21, 21, 26, 23, 24

So, sample size = n =20

Now we are supposed to find Construct a 99.9% confidence interval for the mean length of all the junior's tennis racquets manufactured at this plant.

Since n < 30

So we will use t-distribution

Confidence level = 99.9%

Significance level = α = 0.001

Now calculate the sample mean

X=21, 25, 23, 22, 24, 21, 25, 21, 23, 26, 21, 24, 22, 24, 23, 21, 21, 26, 23, 24

Sample mean = $\bar{x}=\frac{\sum x}{n}$

Sample mean = $\bar{x}=\frac{21+25+23+22+24+21+25+21+23+ 26+ 21+24+22+ 24+23+21+ 21+ 26+23+ 24}{20}$

Sample mean = $\bar{x}=23$

Sample standard deviation = $\sqrt{\frac{\sum(x-\bar{x})^2}{n-1}}$

Sample standard deviation = $\sqrt{\frac{(21-23)^2+(25-23)^2+(23-23)^2+(22-23)^2+(24-23)^2+(21-23)^2+(25-23)^2+(21-23)^2+(23-23)^2+(26-23)^2+(21-23)^2+(24-23)^2+(22-23)^2+(24-23)^2+(23-23)^2+(21-23)^2+(21-23)^2+(26-23)^2+(23-23)^2+(24-23)^2}{20-1}}$