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2 years ago

Can Someone Please Help!!!

Mathematics

2 answers:

Brums [2.3K]2 years ago

5 0

Answer: 5 raised to the power of 4 is 625 and 3 raised to the power of 2 is 9

Step-by-step explanation:

1- 5x5x5x5= your answer (625) 5x5=25, 25 x 25 = 625

2- 3x3= your answer (9)

You then add both products, leaving you with 634.

Kobotan [32]2 years ago

3 0

Answer: 634

Step-by-step explanation: All we do is solve the exponents and then add. First we solve 5^4 which is equal to 625

5^4=625

Then we solve for 3^2 which is equal to 9

3^2=9

Finally we add.

625+9=634

And we have our answer.

634

Hope this helps!

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Fill in the gaps in this sequence: -8, -19, -30, _ , _. (Geometric and arithmetic sequences)

gavmur [86]

-8, -19, -30, -49 , -60

Step-by-step explanation:

Here we have the following sequence :

-8, -19, -30, _ , _

First term of sequence is -8 .
Second term of sequence is -19 : $-8-11 = -19$
Third term of sequence is -30 : $-19-11 = -30$
Fourth term of sequence is : $-30-11 = -49$
Fifth term of sequence is : $-49-11 = -60$

Following sequence was an AP( Arithmetic Progression ) with first term as -8 i.e. $a = -8$ and common difference $d = -11$ having general equation as : $-8 -11n$ .

6 0

3 years ago

Simplify by dividing (-3/8) -5/9

Elodia [21]

Answer:

$\frac{-27}{40}$

Step-by-step explanation:

$(\frac{-3}{8})$ ÷ $\frac{5}{9}$

$\frac{-3}{8}$ × $\frac{9}{5}$

$\frac{-27}{40}$

6 0

2 years ago

PLEASE HELPPPP i will give brainliest to whoever answers first please and thanks

FromTheMoon [43]

Answer: I got A. 768 as the answer

8 0

3 years ago

28 is 12 less than k equation

Virty [35]

K-12=28...? I think this is what your question is asking for

5 0

2 years ago

Suppose the number of insect fragments in a chocolate bar follows a Poisson process with the expected number of fragments in a 2

leonid [27]

Answer:

a)The expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b)0.6004

c)19.607

Step-by-step explanation:

Let X denotes the number of fragments in 200 gm chocolate bar with expected number of fragments 10.2

X ~ Poisson(A) where $\lambda = \frac{10.2}{200} = 0.051$

a)We are supposed to find the expected number of insect fragments in 1/4 of a 200-gram chocolate bar

$\frac{1}{4} \times 200 = 50$

50 grams of bar contains expected fragments = \lambda x = 0.051 \times 50=2.55

So, the expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b) Now we are supposed to find the probability that you have to eat more than 10 grams of chocolate bar before ending your first fragment

Let X denotes the number of grams to be eaten before another fragment is detected.

$P(X>10)= e^{-\lambda \times x}= e^{-0.051 \times 10}= e^{-0.51}=0.6004$

c)The expected number of grams to be eaten before encountering the first fragments :

$E(X)=\frac{1}{\lambda}=\frac{1}{0.051}=19.607 gram$ s

7 0

3 years ago