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goldenfox [79]
3 years ago
12

What are the domain and range of the function f(x)=5^x-3 +1

Mathematics
2 answers:
Ne4ueva [31]3 years ago
8 0

Answer:

\text{Domain and Range of function is all real numbers i.e } (-\infty, \infty)

Step-by-step explanation:

Given the function

f(x)=5^{x-3} +1

we have to find the domain and range of the function.

The domain is the set of all possible values of independent variable i.e of x

The range is the complete set of all possible resulting values of the dependent variable i.e of y

Here the above function is defined for every value of R i.e for all value of real numbers.

\text{Domain and Range of function is all real numbers i.e } (-\infty, \infty)

sineoko [7]3 years ago
6 0

D​omain: (-∞,∞)

Range: (-∞,∞)

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A=2 b=3 c=4<br><br> 1) ab - c =<br> 2) 6c - 2b=<br> 3) a + b - c + 5 =<br> 4) 7c - 2a =
bekas [8.4K]

Step-by-step explanation:

1)ab-c

=2(3)-4

=6-4

=2

2)6c-2b

=6(4)-2(3)

=24-6

=18

3)a+b-c+5

=2+3-4+5

=10-4

=6

5)7c-2a

=7(4)-2(2)

=28-4

=24

7 0
3 years ago
Read 2 more answers
How I do this someone ples help I need help with 25 and 27.
marta [7]

Number 15 is no solution.

Number 27 is that x has to be between -2 or 2.

This could be explained as -2≤x≤2

8 0
3 years ago
A violin student records the number of hours she spends practicing during each of nine consecutive weeks: 6.2 5.0 4.3 7.4 5.8 7.
spin [16.1K]

Answer:

A) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the first  quartile.

Correct we satisfy that 1.2 <1.7 the lower limit

Step-by-step explanation:

Assuming this complete question: A violin student records the number of hours she spends practicing during each of nine consecutive weeks:

6.2 5.0 4.3 7.4 5.8 7.2 8.4 1.2 6.3

For this case we need to sort the data first on increasing way and we got:

1.2, 4.3, 5.0, 5.8, 6.2, 6.3, 7.2, 7.4, 8.4

For this case we have 9 values the median would be on the 5 position:

Median = 6.2

The first quartile would be 5 since we analyze 1.2, 4.3, 5.0, 5.8, 6.2 and the middle point is 5.0

The third quartile would be 7.2 since we analyze 6.2, 6.3, 7.2, 7.4, 8.4 and the middle point is 7.2

Then the interquartile rnage would be:

IQR = Q_3 -Q_1= 7.2-5=2.2

And 1.5 IQR = 1.5*2.2=3.3

The lower limit on this case would be:

LL= Q_1 -1.5 IQR= 5-3.3=1.7

And our value is 1.2 is lower than the lower limit 1.7

Considering the smallest data value (1.2) and using the 1.5 × IQR rule, we would:

A) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the first  quartile.

Correct we satisfy that 1.2 <1.7 the lower limit

B) not classify the value 1.2 as an outlier, because it is not more than 1.5 × IQR below  the first quartile.

False 1.2 is more than 1.5 IQR below the first quartile

C) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the  median.

False, we analyze if is an outlier comparing with the Q1 or Q3 not with the median

D) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the  mean.

False we analyze if is an outlier comparing with the Q1 or Q3 not with the mean

5 0
3 years ago
What's the answer for this problem?
tiny-mole [99]
4%of 500 = 20

500  plus or minus 20 gives range as (480,520)
8 0
3 years ago
PLEASE HELP ME WITH THIS IT IS TIMED
Zanzabum

Answer:

7 1/2

Step-by-step explanation:

i got 7 1/2 because the one at the bottom is 7 1 /2

7 0
3 years ago
Read 2 more answers
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