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defon
3 years ago
7

Im giving extra points and brainliest

Mathematics
2 answers:
mrs_skeptik [129]3 years ago
8 0
The answer would be c I think
Veseljchak [2.6K]3 years ago
5 0

Answer: the c

Step-by-step explanation:

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I need to find the answer to 3 questions. Which measure of central tendency is most representative of the data 6, 8, 3, 5, 32, 6
Dominik [7]

Step-by-step explanation:

the first one is 2 that appears most while the second is 20 and 3

3 0
3 years ago
Help don’t understand
marusya05 [52]

Answer:

f(-14) = -6

f(-4) = 6

f(12) = 6

f(0) = -3

negative

Step-by-step explanation:

f(-14) = -6

This is because when x is -14, y is -6, as seen in the graph

f(-4) = 6

This is because when x is -4, y is 6, as seen in the graph

f(12) = 6

This is because when x is 12, y is 6, as seen in the graph

f(0) = -3

This is because when x is 0, y is -3, as seen in the graph

is f(4) positive or negative?
negative

This is because when x is 4, y is -6, as seen in the graph

6 0
2 years ago
What is the x-intercept of the graph of the function f(x)=x2-16x+64?
Katena32 [7]
X=8 

Explanation: 
f(x)=x^2-16x+64+(x-8)^2

You see there is a double x- intercept by the 8. So the parabola is tangent to the x axis. So x=8 f(x)=0 

hope this helps <span />
3 0
3 years ago
Read 2 more answers
Help! 3-4 quick please!!
Inga [223]
3: independent
4: dependent
4 0
3 years ago
The probability density function of the time you arrive at a terminal (in minutes after 8:00 A.M.) is f(x) = 0.1 exp(−0.1x) for
Blababa [14]

f_X(x)=\begin{cases}0.1e^{-0.1x}&\text{for }x>0\\0&\text{otherwise}\end{cases}

a. 9:00 AM is the 60 minute mark:

f_X(60)=0.1e^{-0.1\cdot60}\approx0.000248

b. 8:15 and 8:30 AM are the 15 and 30 minute marks, respectively. The probability of arriving at some point between them is

\displaystyle\int_{15}^{30}f_X(x)\,\mathrm dx\approx0.173

c. The probability of arriving on any given day before 8:40 AM (the 40 minute mark) is

\displaystyle\int_0^{40}f_X(x)\,\mathrm dx\approx0.982

The probability of doing so for at least 2 of 5 days is

\displaystyle\sum_{n=2}^5\binom5n(0.982)^n(1-0.982)^{5-n}\approx1

i.e. you're virtually guaranteed to arrive within the first 40 minutes at least twice.

d. Integrate the PDF to obtain the CDF:

F_X(x)=\displaystyle\int_{-\infty}^xf_X(t)\,\mathrm dt=\begin{cases}0&\text{for }x

Then the desired probability is

F_X(30)-F_X(15)\approx0.950-0.777=0.173

7 0
3 years ago
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