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Furkat [3]
3 years ago
15

If x equals 6 what is 5x-(3+9

Mathematics
2 answers:
RSB [31]3 years ago
7 0
I think it is 18 Hope this helps. :)
aleksklad [387]3 years ago
5 0
Let's plug it in first
5*6-(3+9)
Using PEMDAS, clear the parentheses first
5*6-(12)
Now, we should use multiplication or division since there is no exponent
30-12
Finally, we can add or subtract
18
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Which is the best description of the shape of the distribution of the number of pets? Use the drop-down menu to complete the sta
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Roughly symmetric is the answer
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If one factor of 6.x2 + 5x – 6 is
sattari [20]
C. 2x + 3

You have many ways to solve this!
You could...

A) multiply all the answer choices by 3x-2 and see what matches

B) Factor the trinomial

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Which expression is equivalent to 7a2b + 10a2b2 + 14a2b3?
Llana [10]
The given expression can be simplified in many ways by grouping like terms. The simplest form is obtained by factoring out a²b which gives us the following expression.

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3 0
3 years ago
Read 2 more answers
Twenty times y is at most 100 in interval notation
DiKsa [7]

Answer:

y\in (-\infty ,5]

Step-by-step explanation:

In this problem, we need to write "Twenty times y is at most 100 in interval notation ".

20 times y means, 20y

Atmost means an inequality which is \le

ATQ,

20\times y\le 100

i.e.

20y\le 100\\\\y\le 5

We can also write it as :

y\in (-\infty ,5]

Hence, the required interval notation is y\in (-\infty ,5].

7 0
3 years ago
The reference desk of a university library receives requests for assistance. Assume that a Poisson probability distribution with
NISA [10]

Answer:

a) 0.125

b) 7

c) 0.875 hr

d) 1 hr

e) 0.875

Step-by-step explanation:l

Given:

Arrival rate, λ = 7

Service rate, μ = 8

a) probability that no requests for assistance are in the system (system is idle).

Let's first find p.

a) ρ = λ/μ

\frac{7}{8} = 0.875

Probability that the system is idle =

1 - p

= 1 - 0.875

=0.125

probability that no requests for assistance are in the system is 0.125

b) average number of requests that will be waiting for service will be given as:

λ/(μ - λ)

= \frac{7}{8 - 7}

= 7

(c) Average time in minutes before service

= λ/[μ(μ - λ)]

= \frac{7}{8(8 - 7)}

= 0.875 hour

(d) average time at the reference desk in minutes.

Average time in the system js given as: 1/(μ - λ)

= \frac{1}{(8 - 7)}

= 1 hour

(e) Probability that a new arrival has to wait for service will be:

λ/μ =

{7}{8}

= 0.875

5 0
3 years ago
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