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ehidna [41]
3 years ago
12

Does this graph represent a function?

Mathematics
1 answer:
kupik [55]3 years ago
5 0
Yes and the correct one is A
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What is the product of 2x
Irina18 [472]

Answer:

2x

Step-by-step explanation:


2x if you don't know the value of x.

5 0
4 years ago
Read 2 more answers
What is 26 divided by 43040 equal
g100num [7]
6.040892193 E-4
which equals to 
0.000604<span>0892193

move your decimal place 4 units to the left
because it's a negative (E-4), which makes your number smaller

OR

multiply that number by 10^-4</span>
6 0
3 years ago
Reid and Maria both play soccer. This season, Reid scored 4 less than twice the number of goals that Maria scored. The differenc
kap26 [50]

Answer:

I think its D

Step-by-step explanation:

Because its asking for 4 less then twice the number 6 so it would be 10 and 16 btw im so sorry if im wrong

7 0
3 years ago
Please help! Problem is attached
Darya [45]

Answer:

  (a)  -3

Step-by-step explanation:

The value of f(x) approaches 3cos(π) = -3 as x approaches π from either direction. The limit is -3.

__

<em>Additional comment</em>

The fact that the function is defined to have a different value at x=π simply means there is a jump discontinuity there. The function is <em>not continuous</em>, because the limit and the function value differ. That doesn't detract from the fact that there is a limit, and it is -3.

6 0
3 years ago
If a variable has a distribution that is​ bell-shaped with mean 28 and standard deviation 5​, then according to the Empirical​ R
mars1129 [50]

Answer:

99.7% data lies between 13 to 43

Step-by-step explanation:

Mean:  \mu = 28

Standard deviation : \sigma = 5

68% of data lies between \mu - \sigma to \mu +\sigma

95% of the data lies between \mu - 2\sigma to \mu +2\sigma

99.7% data lies between  \mu - 3\sigma to \mu +3\sigma

We are supposed to find  99.7 of the data will lie between which​ values

99.7% data lies between  28 - 3(5) to 28 +3(5)

99.7% data lies between  13 to 43

Hence 99.7% data lies between 13 to 43

8 0
3 years ago
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