Differentiating the function
... g(x) = 5^(1+x)
we get
... g'(x) = ln(5)·5^(1+x)
Then the linear approximation near x=0 is
... y = g'(0)(x - 0) + g(0)
... y = 5·ln(5)·x + 5
With numbers filled in, this is
... y ≈ 8.047x + 5 . . . . . linear approximation to g(x)
Using this to find approximate values for 5^0.95 and 5^1.1, we can fill in x=-0.05 and x=0.1 to get
... 5^0.95 ≈ 8.047·(-0.05) +5 ≈ 4.598 . . . . approximation to 5^0.95
... 5^1.1 ≈ 8.047·0.1 +5 ≈ 5.805 . . . . approximation to 5^1.1
Answer:
Step-by-step explanation:
Start by writing these five numbers in just one format (improper fractions with denominator 10):
9/5 = 18/10, and
-2.5 = -25/10, and
-1.1 = -11/10, and
-4/5 = -8/10, and, finally
0.8 = 8/10
The smallest of these (the least) is -25/10, and
the largest is 18/10.
Arranged in ascending order, we have:
-25/10, -11/10, -8/10, 8/10, 18/10, or
-2.5, -1.1, -4/5, 0.8, 9/5
I think the correct answer is c
Answer:
4%
Step-by-step explanation:
Divide 48 by 1202
48 ÷ 1202
= 0.03993344425956738768718801996672
Round it
0.04 or 4%
The given graph shows the relationship between the social media posts and grade point average of different students in Brent's class.
From the given graph, we can say that the grade point average of students decrease with increase in number of their social media posts.
Since grade point average decreases approximately linearly with increase in use of social media posts, It is Strong negative correlation.
Hence, option D is correct i.e. Strong Negative Correlation.
