The equivalent value of the function if x = -3 is -4.
<h2 /><h2>Given that</h2>
Function; for the input value –3.
<h3>We have to determine</h3>
What are the x-intercepts of the function f(x)?
<h3>According to the question</h3>
To determine the value of the function following all the steps given below.
For x = -3, we will have to substitute x = -3 into the expression to have:
Then,
The value of f(x) when x is -3 is,
Hence, The equivalent value of the function if x = -3 is -4.
To know more about Function click the link given below.
brainly.com/question/26039440
Answer:
50% probability that on a randomly selected day during this period, a unit of currency B was worth more than 1.094 units of currency A.
Step-by-step explanation:
The Empirical Rule(68-95-99.7 rule) states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 1.094
Standard deviation = 0.013
a) What is the probability that on a randomly selected day during this period, a unit of currency B was worth more than 1.094 units of currency A?
The normal distribution is symmetric, which means that 50% of the units of currency B are more than 1.094 of currency A and 50% are below.
So
50% probability that on a randomly selected day during this period, a unit of currency B was worth more than 1.094 units of currency A.
Answer:
13.5 and 8.5
Step-by-step explanation:
x + (x - 5) = 22
2x - 5 = 22
2x = 27
x = 13.5; x - 5 = 8.5
In order to find this out, we just needed to find out which one has the same smallest fraction
25 / 35. Divided this by 5 and we will get 5/7 >>> smallest fraction
15/21 . Divided this by 3 and we also will get 5/7 >> smallest fraction
So the answer would be : B. 15/21
hope this helps
I'll let <em>h</em> = <em>ax</em>, so the limit is
i.e. the derivative of .
Expand the numerator to see several terms that get eliminated:
So we have
Since <em>h</em> ≠ 0 (because it is approaching 0 but never actually reaching 0), we can cancel the factor of <em>h</em> in both numerator and denominator, then plug in <em>h</em> = 0: