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2 years ago

11

Which expressions are equivalent to the one below? Check all that apply. 16^x/4^x

2 answers:

lesya692 [45]2 years ago

7 0

Answer:

C D anf F

Step-by-step explanation:

c : by expanding and canceling 4^x .

d : its the expanded form.

f : its just another way of expressing the expression

aliya0001 [1]2 years ago

6 0

Answer : C , D , F
Explanation :

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NEED HELP FAST PLEASE<br> Evaluate the limit.<br><br> a.4<br> c.1<br> b.2<br> d.limit does not exist

BlackZzzverrR [31]

We have the following limit:
(8n2 + 5n + 2) / (3 + 2n)
Evaluating for n = inf we have:
(8 (inf) 2 + 5 (inf) + 2) / (3 + 2 (inf))
(inf) / (inf)
We observe that we have an indetermination, which we must resolve.
Applying L'hopital we have:
(8n2 + 5n + 2) '/ (3 + 2n)'
(16n + 5) / (2)
Evaluating again for n = inf:
(16 (inf) + 5) / (2) = inf
Therefore, the limit tends to infinity.
Answer:
d.limit does not exist

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3 years ago

The sum of three consecutive odd integers is 17 less than four times the smallest integer. Find the integers.

lisabon 2012 [21]

Answer:

23, 25, 27

Step-by-step explanation:

n + (n+2) + (n+4) = 4n - 17

3n + 6 = 4n - 17

3n - 4n = -17 - 6

-n = -23

n = 23

therefore,

23, 25, 27 are the 3 consecutive odd integers

6 0

2 years ago

what multiplication expression can be broken apart into (3×4) + (3×4) using the distributive property? (1-4)

jok3333 [9.3K]

3 x(4+4)= (3x4)+(3x4)
I am not completely sure, but I hope I helped!! (:

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2 years ago

The annual salaries of all employees at a financial company are normally distributed with a mean of $34,000 and a standard devia

IRISSAK [1]

$z=\frac{X-\mu}{\sigma}=\frac{54000-34000}{4000}=5$

5 0

3 years ago

Read 2 more answers

At a certain high school, 15 students play stringed instruments and

Aleks [24]

Answer:

<em>0</em> is the probability that a randomly selected student plays both a stringed and a brass instrument.

Step-by-step explanation:

Given that:

Number of students who play stringed instruments, N(A) = 15

Number of students who play brass instruments, N(B) = 20

Number of students who play neither, N( $A \cup B$ )' = 5

<u>To find:</u>

The probability that a randomly selected students plays both = ?

<u>Solution:</u>

Total Number of students = N(A)+N(B)+N( $A \cup B$ )' =15 + 20 + 5 = 40

(As there is no student common in both the instruments we can simply add the three values to find the total number of students)

As per the venn diagram, no student plays both the instruments i.e.

$N(A\cap B) =0$

Formula for probability of an event E can be observed as:

$P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}$

$P(A\cap B) = \dfrac{N(A\cap B)}{N(U)}\\\Rightarrow \dfrac{0}{40}\\\Rightarrow P(A\cap B) = 0$

So, <em>0</em> is the probability that a randomly selected student plays both a stringed and a brass instrument.

5 0

2 years ago