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4 years ago

245. 202, 254, 245, 276Write the values from least togreatestMeanMedian:Mode:

Mathematics

1 answer:

Burka [1]4 years ago

7 0

Answer:

The values from least to greatest would be 202, 245, 245, 254, 276

The mean would be 244.4

The median would be 245

The mode would be 245

Step-by-step explanation:

To find the mean, you add all the numbers together and divide that sum by the number of numbers. In this case, you add all the numbers together and you get 1222. Next, you divide that by 5 because there are 5 numbers. To find the median, you order the numbers from least to greatest and find the middle number. For this question, the answer would be 245. Lastly, to find the mode you look for the number that appears the most. In this case it's 245.

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Estimate product using compatible numbers 0.67 x 623

alekssr [168]

It could possible be 417.

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3 years ago

Solve the following and explain your steps. Leave your answer in base-exponent form. (3^-2*4^-5*5^0)^-3*(4^-4/3^3)*3^3 please st

Naily [24]

Answer:

$\boxed{2^{\frac{802}{27}} \cdot 3^9}$

Step-by-step explanation:

I will try to give as many details as possible.

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$(3^{-2} \cdot 4^{-5} \cdot 5^0)^{-3} \cdot (4^{-\frac{4}{3^3} })\cdot 3^3$

Note that

$\boxed{a^{-b} = \dfrac{1}{a^b}, a\neq 0 }$

The denominator can't be 0 because it would be undefined.

So, we can solve the expression inside both parentheses.

$\left(\dfrac{1}{3^2} \cdot \dfrac{1}{4^5} \cdot 5^0 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{3^3} } }\right)\cdot 3^3$

Also,

$\boxed{a^{0} = 1, a\neq 0 }$

$\left(\dfrac{1}{9} \cdot \dfrac{1}{1024} \cdot 1 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27$

Note

$\boxed{\dfrac{1}{a} \cdot \dfrac{1}{b}= \frac{1}{ab} , a, b \neq 0}$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

Note

$\boxed{\dfrac{1}{\dfrac{1}{a} } = a}$

$9216^3\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)$

$\left(\dfrac{ 9216^3\cdot 27}{4^{\frac{4}{27} } }\right)$

Once

$9216=2^{10}\cdot 3^2 \implies 9216^3=2^{30}\cdot 3^6$

$\boxed{(a \cdot b)^n=a^n \cdot b^n}$

And

$4^{\frac{4}{27}} = 2^{\frac{8}{27}$

We have

$\left(\dfrac{ 2^{30} \cdot 3^6\cdot 27}{2^{\frac{8}{27} } }\right)$

Also, once

$\boxed{\dfrac{c^a}{c^b}=c^{a-b}}$

$2^{30-\frac{8}{27}} \cdot 3^6\cdot 27$

$30-\dfrac{8}{27} = \dfrac{30 \cdot 27}{27}-\dfrac{8}{27} =\dfrac{802}{27}$

$2^{30-\frac{8}{27}} \cdot 3^6\cdot 27 = 2^{\frac{802}{27}} \cdot 3^6 \cdot 3^3$

$2^{\frac{802}{27}} \cdot 3^9$

4 0

3 years ago

‼️‼️‼️‼️‼️‼️ PLEASE HELP FIRST ANSWER GETS BRAINLIEST NO LINK NO LINKS OR BANNED

vodka [1.7K]

Answer:

The true statements are (D) and (E). This is a square with sides equal to 5 units so that means that its perimeter is 4×(5 units) or 20 units and its area is (5 units)×(5 units) or 25 units^2.

5 0

3 years ago

Find x if eg is a median of triangle DEF.. 5x-17 & 3x+1

GREYUIT [131]

If eg is the median, that means
5x-17=3x+12x=18x=9

4 0

3 years ago

I need help with this

olga2289 [7]

Answer:

x = 24.52

Step-by-step explanation:

Since this is a right triangle, use Pythagorean Theorem to solve for the hypotenuse.

Pythagorean Theorem: a² + b² = c²

24² + 5² = x²

576 + 25 = x²

601 = x²

√601 = x

4 0

3 years ago

245. 202, 254, 245, 276Write the values from least togreatestMeanMedian:Mode:​

245. 202, 254, 245, 276Write the values from least togreatestMeanMedian:Mode: