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3 years ago

CAN ANYONE HELP ME ANSWER THIS

Mathematics

1 answer:

Lera25 [3.4K]3 years ago

3 0

Answer: My best answer would be independent

Step-by-step explanation: because you would be the one rolling the dice but you don't know what it would be landing on

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Is 50 a rational or irrational number

Illusion [34]

50 is a rational number because it can be expressed in two integers; 50 ÷ 1

4 0

3 years ago

Hey all!

Rainbow [258]

Answer:

65°

Step-by-step explanation:

Use the law of cosines:

cos C = (25² + 27² - 28²)/(2*25*27)
cos C = 0.4222
m∠C = arccos 0.4222
m∠C = 65°

8 0

3 years ago

Simplify (8^4/10^4)^-1/4using rational exponents

Vlad1618 [11]

Answer:

$(\frac{8^4}{10^4} )^{-\frac{1}{4} }$ = $\frac{5}{4}$

Step-by-step explanation:

Explanation

Given

$(\frac{8^4}{10^4} )^{-\frac{1}{4} }$

By using algebra formulas

$(\frac{a}{b} )^{m} = \frac{a^m}{b^n}$

$(\frac{8^4}{10^4} )^{-\frac{1}{4} } = \frac{(8^{4})^{\frac{-1}{4} } }{(10^{4} )^{\frac{-1}{4} } }$

$(a^{m} )^{n} = a^{mn}$

$\frac{(8^{4})^{\frac{-1}{4} } }{(10^{4} )^{\frac{-1}{4} } } = \frac{8^{4X\frac{-1}{4} } }{10^{4 X\frac{-1}{4} } }$

= $\frac{8^{-1} }{10^{-1} }$

$= \frac{10}{8} \\= \frac{5}{4}$

Final answer:-

$(\frac{8^4}{10^4} )^{-\frac{1}{4} }$ = $\frac{5}{4}$

7 0

3 years ago

(a) 127.5 is what percent of 51 ?

ExtremeBDS [4]

To solve this, you need to divide 127.5 by 51:
127.5/51 = 2.5
To find the percent, multiply 2.5 by 100:
2.5 x 100% = 250%
So 127.5 is 250% of 51

8 0

3 years ago

A random sample has been taken from a population. A statistician, using this sample, needs to decide whether to construct a 90 p

AURORKA [14]

Answer:

Step-by-step explanation: a confidence interval is a range that shows how precise the measurement is. if a statistician decides to construct using 90 percent C I over a 95% C I, it tells you that you have 10% chance of being wrong rather than 5% chance of being wrong while using the 95%. Also a 90 percent confidence interval would be narrower than a 95 percent confidence interval

5 0

3 years ago

CAN ANYONE HELP ME ANSWER THIS ​

CAN ANYONE HELP ME ANSWER THIS