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zavuch27 [327]
3 years ago
7

Ira described the data on a dot plot as having a cluster from 0 to 3, a gap at 4, a peak at 1, and being skewed right. Which dot

plot shows the data Ira is describing?
A dot plot titled Math Assignments completed going from 0 to 5. 0 has 2 dots, 1 has 5 dots, 2 has 4 dots, 3 has 2 dots, 4 has 0 dots, and 5 has 1 dot.
A dot plot titled English Assignments Completed going from 0 to 5. 0 has 1 dot, 1 has 3 dots, 2 has 4 dots, 3 has 0 dots, 4 has 3 dots, and 5 has 1 dot.
A dot plot titled Science Assignments Completed going from 0 to 5. 0 has 1 dot, 1 has 0 dots, 2 has 3 dots, 3 has 4 dots, 4 has 2 dots, and 5 has 1 dot.
Mathematics
2 answers:
Aleksandr-060686 [28]3 years ago
6 0

Answer:

a

Step-by-step explanation:

took the test

rjkz [21]3 years ago
5 0

Answer:

it would be a

Step-by-step explanation:

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Sally has 15 quarts of milk. One is equal to 1 4 of a gallon. How many gallons of milk does Sally have?
Ede4ka [16]
3 and three fourths of a gallon
3 0
3 years ago
Can someone help me find the equivalent expressions to the picture below? I’m having trouble
miss Akunina [59]

Answer:

Options (1), (2), (3) and (7)

Step-by-step explanation:

Given expression is \frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}.

Now we will solve this expression with the help of law of exponents.

\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}=\frac{\sqrt[3]{(2^3)^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}

           =\frac{\sqrt[3]{2\times 3} }{3\times2^{\frac{1}{9}}}

           =\frac{2^{\frac{1}{3}}\times 3^{\frac{1}{3}}}{3\times 2^{\frac{1}{9}}}

           =2^{\frac{1}{3}}\times 3^{\frac{1}{3}}\times 2^{-\frac{1}{9}}\times 3^{-1}

           =2^{\frac{1}{3}-\frac{1}{9}}\times 3^{\frac{1}{3}-1}

           =2^{\frac{3-1}{9}}\times 3^{\frac{1-3}{3}}

           =2^{\frac{2}{9}}\times 3^{-\frac{2}{3} } [Option 2]

2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2 [Option 1]

2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2

                =(2^2)^{\frac{1}{9}}\times (3^2)^{-\frac{1}{3} }

                =\sqrt[9]{4}\times \sqrt[3]{\frac{1}{9} } [Option 3]

2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(2^2)^{\frac{1}{9}}\times (3^{-2})^{\frac{1}{3} }

               =\sqrt[9]{2^2}\times \sqrt[3]{3^{-2}} [Option 7]

Therefore, Options (1), (2), (3) and (7) are the correct options.

6 0
3 years ago
Curt and Melanie are mixing blue and yellow paint to make seafoam green paint. 1.5 quarts of seafoam paint is made of 70% blue,
omeli [17]

Answer:

0.45 quarts

Step-by-step explanatio

4 0
3 years ago
Help with 33 and 34 please!!!
hjlf
33. 15x10=150x9
the answer is b

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3 0
4 years ago
Read 2 more answers
Gravel is being dumped from a conveyor belt at a rate of 30 ft3/min, and its coarseness is such that it forms a pile in the shap
pogonyaev

Answer:

Rate of increase in height =\frac{dh}{dt}=0.3156ft/min

Step-by-step explanation:

we know that volume of a cone is given by

V=\frac{1}{12}\pi d^{2}h

It is Given that diameter equals height thus we have

V=\frac{1}{12}\pi h^{2}h\\\\V=\frac{1}{12}\pi h^{3}

Differentiating both sides with respect to time we get

\frac{dV}{dt}=\frac{1}{12}\pi \frac{dh^{3}}{dt}\\\\\frac{dV}{dt}=\frac{1}{12}\pi(3h^{2}\frac{dh}{dt})

Applying values and solving for \frac{dh}{dt} we get

\frac{dh}{dt}=\frac{12\frac{dV}{dt}}{3\pi h^{2}}\\\\\frac{dh}{dt}=0.3156ft/min

8 0
3 years ago
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