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Travka [436]
3 years ago
15

Que tres numeros impares suman 30

Mathematics
1 answer:
emmainna [20.7K]3 years ago
6 0
7.5 + 9.5 + 13 = 30 <=== 3 odd numbers that equal 30...
<span>no te conozco otra forma , excepto para usar decimales</span>
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Help please?!?!?!?!?!?!?
nadezda [96]

Answer:

Choice D is  correct answer.

Step-by-step explanation:

We have given two function.

f(x) =2ˣ+5x  and g(x)  = 3x-5

We have to find the addition of given two function.

(f+g)(x) =  ?

The formula to find the addition, we have

(f+g)(x) = f(x) + g(x)

Putting given values in above formula, we have

(f+g)(x) =   (2ˣ+5x)+(3x-5)

(f+g)(x) =   2ˣ+5x+3x-5

Adding like terms, we have

(f+g)(x) =   2ˣ+8x-5 which is the answer.

6 0
3 years ago
1) A family consisting of three persons—A, B, and C—goes to a medical clinic that always has a doctor at each of stations 1, 2,
crimeas [40]

Answer:

Step-by-step explanation:

Let us record the station number 1, 2 or 3 for each family member A, B or C.

I am attaching a table containing total outcomes. Outcomes are presented along rows while the assigned station to each member is written along columns. For ease of understanding, 1 3 2 in the table should be interpreted as family member A being assigned to station 1, member B to station 3 and member C to station number 2, respectively.

From table it is clear that the total outcomes possible are 27.

We know that, probability can be defined as,

PROBABITILY = \frac{NUMBER\;OF\;DESIRED\;OUTCOMES}{TOTAL\;NUMBER\;OF\;OUTCOMES}

a) All Members Assigned to the Same Station.

Cases for all members being assigned to same station are as follows:

[1 1 1], [2 2 2], [3 3 3] (outcome number 1, 14 and 27 in the table).

Therefore,

PROBABILITY\;(Case\;a) = \frac{3}{27}\\\\PROBABILITY\;(Case\;a) = 0.111

b) At Most Two Members Assigned to the Same Station.

It means that maximum of 2 members can have the same station. Cases for this situation are as follows:

[1 1 2], [1 1 3], [1 2 1], [1 2 2], [1 3 1], [1 3 3], [2 1 1], [2 1 2], [2 2 1], [2 2 3], [2 3 2],

[2 3 3], [3 1 1], [3 1 3], [3 2 2], [3 2 3], [3 3 1], [3 3 2]

(outcome number 2, 3, 4, 5, 7, 9, 10, 11, 13, 15, 17, 18, 19, 21, 23, 24, 25 and 26 in the table).

Therefore,

PROBABILITY\;(Case\;b) = \frac{18}{27}\\\\PROBABILITY\;(Case\;b) = 0.666

c) All Members Assigned to a Different Station.

For this scenario, we have the following results:

[1 2 3], [1 3 2], [2 1 3], [2 3 1], [3 1 2], [3 2 1] (outcome number 6, 8, 12, 16, 20 and 22 in the table).

Therefore,

PROBABILITY\;(Case\;c) = \frac{6}{27}\\\\PROBABILITY\;(Case\;c) = 0.222

3 0
4 years ago
URGENT!!!!!
hichkok12 [17]

Answer:

36

Step-by-step explanation:

1. Cross Multiply. \frac{5}{6} =\frac{30}{x}

2. Solve. \frac{5}{6} =\frac{30}{x} =36

3. Your answer is 36


3 0
3 years ago
Read 2 more answers
For what value of the constant c is the function fcontinuous on (−[infinity], [infinity])?
Arlecino [84]

Answer:

For c=\frac{1}{7} the function f(x) is continuous on (-\infty,\infty).

Step-by-step explanation:

We have the following function

f(x) = \left\{        \begin{array}{ll}            cx^2+5x & \quad x

For the function f(x) to be continuous on (-\infty,\infty) it is sufficient to have continuity at x = 6, we need to ensure that as x approaches 6, the left and right limits match, this means that

\lim_{x \to 6^{-} } f(x)=\lim_{x \to 6^{+} } f(x)=f(x),

which holds if and only if

c\left(6\right)^2+5\left(6\right)=\left(6\right)^2-c\left(6\right)\\36c+30=36-6c\\42c=6

namely if c=\frac{1}{7}.

6 0
3 years ago
Can someone help me on problems 9, 10 and 11 please!!!
mojhsa [17]
9. i dissagree because 0.70 and 0.7 is the same thing
10. is 1.45
11. is 0.37
3 0
3 years ago
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