Given sec theta=-4/3 and 90

Does the absolute value graph below flip or not flip? F(x)=2|x-9|+3

deff fn [24]

Answer:

Option C, the graph does not flip because 2 is positive

4 0

3 years ago

A volcano erupted

IgorC [24]

Percentage decrease = d
h1 = 2950
h2 = 2550

d/100 = (h2-h1)/h1

d/100 = (2950-2550)/2950
d=13.56

3 0

3 years ago

Help PLEASE. I got number 3 but I can't figure out number 4.

Sergeeva-Olga [200]

Answer:

Verify the identity using trig rules.

tan ( x − π /2 ) = − cot ( x ) is an identity

Step-by-step explanation:

i think this is it, hope it helps

6 0

3 years ago

The exam scores (out of 100 points) for all students taking an introductory Statistics course are used to construct the followin

ella [17]

Answer:

The inter quartile range is 55

Step-by-step explanation:

Interquartile range:-

Definition:-

The inter quartile range is the difference between the upper (Q3) and lower(Q1) quartiles, and describes the middle 50% of values when ordered from lowest to highest.

Interquartile range = higher quartile - lower quartile

given data 50.25.80.10

write the ascending order 10 25 50 80

Step:-1

Median:- The median is obtained by first arranging the data in ascending or descending order and applying the following rule.

If the number of observations is odd, then the median is

$({\frac{n+1}{2} })^{th}$ observations

Step2:-

The median is obtained by first arranging the data in ascending or descending order and applying the following rule.

If the number of observations is even, then the median is $({\frac{n}{2} })^{th}$

$({\frac{n+1}{2} })^{th}$

In given data the number of observations is '4'(even)

If the number of observations is even, then the median is $({\frac{n}{2} })^{th}$

$({\frac{n+1}{2} })^{th}$

median is $({\frac{4}{2} })^{th}$ is 2 observations and $({\frac{4+1}{2} })^{th}$ is 2.5 observations.

so we have to select average of these two observations.

therefore the median is '3' observation

median of the given data = 50

Step:3

Interquartile range = higher quartile - lower quartile

= 80-25

=55

3 0

3 years ago

From a survey taken several years ago, the starting salaries of individuals with an undergraduate degree from Business Schools a

Klio2033 [76]

Answer: 0.8413

Step-by-step explanation:

Given: Mean : $\mu=\$40,500$

Standard deviation : $\sigma = \$4,500$

The formula to calculate z-score is given by :_

$z=\dfrac{x-\mu}{\sigma}$

For x= $36,000.00, we have

$z=\dfrac{36000-40500}{4500}=-1$

The P-value = $P(z\geq-1)=1-P(z$

Hence, the probability that a randomly selected individual with an undergraduate business degree will get a starting salary of at least $36,000.00 = 0.8413

8 0

3 years ago