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3 years ago

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The data sets represent the five starting players of four basketball teams. Order the teams from least to greatest based on the

mean height of the teams.
69,68,72,69,76

64,74,73,69,73

68,74,69,64,77

72,67,25,60,70

1 answer:

monitta3 years ago

6 0

Answer:

Step-by-step explanation:

Top team=1

Next to top=2

Next to 2=3

Bottom=4

Team 1=69+68+72+69+76=354; 354/5=70.8

Team 2=64+74+73+73+69=353; 353/5=70.6

Team 3=68+74+69+64+77=352; 352/5=70.4

Team 4=72+67+25+60+70=294; 294/5=58.8

Therefore in order

Team 4=58.8

Team 3=70.4

Team 2=70.6

Team 1=70.8

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Answer:

She sells 4 loaves of bread

Step-by-step explanation:

Lets explain how to solve the problem

→ A baker uses $13\frac{1}{2}$ cups of flour to make bread

→ She uses $2\frac{1}{4}$ cups of flour to make each loaf

From these information we can find the number of loaves of bread

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∵ There are $13\frac{1}{2}$ cups of flour

∵ Each loaf of bread needs $2\frac{1}{4}$ cups of flour

∴ The number of loaves = $13\frac{1}{2}$ ÷ $2\frac{1}{4}$

To divide two mixed numbers make them improper fractions and

change the division sign to multiplication sign and reciprocal the

fraction after the division sign

∵ $13\frac{1}{2}$ = $\frac{(13)(2)+1}{2}$

∴ $13\frac{1}{2}$ = $\frac{27}{2}$

∵ $2\frac{1}{4}$ = $\frac{(2)(4)+1}{4}$

∴ $2\frac{1}{4}$ = $\frac{9}{4}$

∴ The number of loaves = $\frac{27}{2}$ × $\frac{4}{9}$

∴ The number of loaves = 6

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→ We need to find the number of loves of bread she sells

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Answer:

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Answer:

Answer from the answer data to choose from

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Step-by-step explanation:

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What is the positive square root of 16?

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Answer:

Dimensions of printed poster are

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Step-by-step explanation:

Let's assume

length of printed poster is x cm

width of printed poster is y cm

now, we can find area of printed poster

so, area of printed poster is

$=xy$

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so, we can set it to 1536

$xy=1536$

now, we can solve for y

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now, we are given

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$A=(x+16)\times (\frac{1536}{x}+24)$

now, we have to minimize A

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$A'=\frac{d}{dx}\left(\left(x+16\right)\left(\frac{1536}{x}+24\right)\right)$

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$A'=\frac{d}{dx}\left(x+16\right)\left(\frac{1536}{x}+24\right)+\frac{d}{dx}\left(\frac{1536}{x}+24\right)\left(x+16\right)$

$=\frac{d}{dx}\left(x+16\right)\left(\frac{1536}{x}+24\right)+\frac{d}{dx}\left(\frac{1536}{x}+24\right)\left(x+16\right)$

now, we can simplify it

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now, we can set it to 0

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$A'=-\frac{24576}{x^2}+24=0$

$-\frac{24576}{x^2}x^2+24x^2=0\cdot \:x^2$

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Since, x is dimension

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now, we can solve for y

$y=\frac{1536}{32}$

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so, dimensions of printed poster are

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