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Triss [41]
3 years ago
7

The improper fraction is equivalent to which of the following decimal numbers?

Mathematics
1 answer:
Alla [95]3 years ago
5 0

Answer:

3.6 is your Answer.

Step-by-step explanation:

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Is this the correct answer?
satela [25.4K]

The answer is 25 5/8ths.

4 0
3 years ago
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Describe how (2 cubed) (2 superscript negative 4) can be simplified.
Varvara68 [4.7K]

<u>Given</u>:

The given expression is (2^3)(2^{-4})

We need to determine how the expression can be simplified.

<u>Simplifying the expression:</u>

Let us determine how the expression can be simplified.

The expression is given by

(2^3)(2^{-4})

Applying the exponent rule that a^{b} \cdot a^{c}=a^{b+c} in the above expression, we get;

2^{3} \cdot 2^{-4}=2^{3-4}

Adding the exponents, we get;

2^{3} \cdot 2^{-4}=2^{-1}

Again, applying the exponent rule a^{-1}=\frac{1}{a}, we get;

2^{3} \cdot 2^{-4}=\frac{1}{2}

Thus, the simplified expression is \frac{1}{2}

Hence, the expression is simplified by adding the exponents and keep the same base. Then find the reciprocal and change the sign of the exponent.

Therefore, Option D is the correct answer.

8 0
3 years ago
Read 2 more answers
PLZ HELP:
UkoKoshka [18]

1. In each graph, the two lines are parallel to each other.

2. Each set of equations have the same slope.

3. Equations that have the same slope as each other will always be parallel to each other.

8 0
2 years ago
I need help on this question
givi [52]

Answer:

Mexico

Step-by-step explanation:

It’s in Spanish

7 0
2 years ago
If a person tosses a coin 23 times, how many ways can he get 11 heads
EastWind [94]

Tossing a coin is a binomial experiment.

Now lets say there are 'n' repeated trials to get heads. Each of the trials can result in either a head or a tail.

All of these trials are independent since the result of one trial does not affect the result of the next trial.

Now, for 'n' repeated trials the total number of successes is given by

_{r}^{n}\textrm{C}

where 'r' denotes the number of successful results.

In our case n=23 and r=11,

Substituting the values we get,

_{11}^{23}\textrm{C}=\frac{23!}{11!\times 12!}

\frac{23!}{11!\times 12!}=1352078

Therefore, there are 1352078 ways to get heads if a person tosses a coin 23 times.


3 0
3 years ago
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