Answer:
A. Between 3.0 and 3.5 and between 4.0 and 4.5
Step-by-step explanation:
The zeroes of a function occur whenever a value of x returns zero. To predict where the zeroes lie, determine the interval(s) where the function crosses the x-axis. This occurs when either 
goes from a negative value to a positive value or vice versa.
From 
and 
, the y-values go from 4.0 (positive) to -0.2 (negative), respectively. Therefore, there must be a zero in this interval.
From 
and 
, the y-values go from -0.8 (negative) to 0.1 (positive), respectively. Therefore, there must also be a zero in this interval.
Thus, the zeros of this function occur between 3.0 and 3.5 and between 4.0 and 4.5, leading to answer choice A.
Answer:
what's the question
Step-by-step explanation:
........
The answer is D ! U just have to see the numbers the left corner
The number 12,300 is written 1.23 x 10⁴ as a scientific notation. It's coefficient is 1.23; base is 10⁴ in exponent form.
Scientific notation is a method developed by scientists to shorten the number that expresses a very large number. The scientific notation is based on powers of the base number 10.
Scientific notation has two numbers: coefficient and base. The coefficient must be greater than or equal to 1 and less than 10. The base is always 10 written in exponent form.
12,300 as coefficient in standard form in scientific notation.
1) put decimal after the first digit and drop the zeros. from 12,300 to 1.23 this is the coefficient.
2) to find the exponent, count the number of places from decimal to the end of the number.
1.2300 ; there are 4 places
So the scientific notation is 1.23 x 10⁴
Answer:
μ = 235.38
σ = 234.54
Step-by-step explanation:
Assuming the table is as follows:
![\left[\begin{array}{cc}Savings&Frequency\\\$0-\$199&339\\\$200-\$399&86\\\$400-\$599&55\\\$600-\$799&18\\\$800-\$999&11\\\$1000-\$1199&8\\\$1200-\$1399&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7DSavings%26Frequency%5C%5C%5C%240-%5C%24199%26339%5C%5C%5C%24200-%5C%24399%2686%5C%5C%5C%24400-%5C%24599%2655%5C%5C%5C%24600-%5C%24799%2618%5C%5C%5C%24800-%5C%24999%2611%5C%5C%5C%241000-%5C%241199%268%5C%5C%5C%241200-%5C%241399%263%5Cend%7Barray%7D%5Cright%5D)
This is an example of grouped data, where a range of values is given rather than a single data point. First, find the total frequency.
n = 339 + 86 + 55 + 18 + 11 + 8 + 3
n = 520
The mean is the expected value using the midpoints of each range.
μ = (339×100 + 86×300 + 55×500 + 18×700 + 11×900 + 8×1100 + 3×1300) / 520
μ = 122400 / 520
μ = 235.38
The variance is:
σ² = [(339×100² + 86×300² + 55×500² + 18×700² + 11×900² + 8×1100² + 3×1300²) − (520×235.38²)] / (520 − 1)
σ² = 55009.7
The standard deviation is:
σ = 234.54
