So 9 students = 25% (100%-75%=25%)
100 : 25 = 4
4 * 9 =
36 students
Answer:
Evaluating Polynomials:
a. f(1) = -10
b. f(-3) = -239
c. f(2)² = 3125
Factoring Polynomials:
a. (x + 1)(x² - 5x + 6) = (x + 1)(x - 3)(x - 2)
b. (x² - x - 6)(x² + 6x + 9) = (x - 3)(x + 2)(x + 3)(x + 3)
c. x³ + 3x² - 4x - 12 = (x + 3)(x - 2)(x + 2)(x - 2)(x + 2)
Answer:
(-5,9+√8)
Step-by-step explanation:
There are two ways in which you can find a point that lies in the circle. One of them is to do y the subject of the formula, and another one is to determine the center of the circumference, and with the information of the radius, you can sum this value upward or downward.
the general equation of a circle is:
with center at (h,k)
you have the following equation:
Then, the center is (-5,9)
if you sum the value of the radius in one of the fourth directions (up, down, left, right), for example upward you have
Then, one point that lies in the circle is (-5,9+√8)
Answer:
75% confidence interval is 91.8±16.66. That is between 75.1 and 108.5 pounds.
Step-by-step explanation:
The question is missing. It is as follows:
Adult wild mountain lions (18 months or older) captured and released for the first time in the San Andres Mountains had the following weights (pounds): 69 104 125 129 60 64
Assume that the population of x values has an approximately normal distribution.
Find a 75% confidence interval for the population average weight μ of all adult mountain lions in the specified region. (Round your answers to one decimal place.)
75% Confidence Interval can be calculated using M±ME where
- M is the sample mean weight of the wild mountain lions ()
- ME is the margin of error of the mean
And margin of error (ME) of the mean can be calculated using the formula
ME= where
- t is the corresponding statistic in the 75% confidence level and 5 degrees of freedom (1.30)
- s is the standard deviation of the sample(31.4)
Thus, ME= ≈16.66
Then 75% confidence interval is 91.8±16.66. That is between 75.1 and 108.5
Answer:
1070 cm.
Step-by-step explanation:
Its correct for Khan Academy:)