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melomori [17]
3 years ago
7

How much smaller is x−3 than x+4?

Mathematics
2 answers:
Nina [5.8K]3 years ago
8 0
-3+7= POSITIVE 4 if it’s on a line then the Awnser is one
loris [4]3 years ago
4 0

Answer:

Try 1

Step-by-step explanation:

The difference between 3 and 4 is one.

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what is the probability that a card picked at random from a 52-card deck of playing cards is a club or a jack?
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|\Omega|=52\\ |A|=\underbrace{13}_{\text{clubs}}+\underbrace{4}_{\text{jacks}}-\underbrace{1}_{\text{jack of clubs}}=16\\\\ P(A)=\dfrac{16}{52}=\dfrac{4}{13}\approx31\%

7 0
3 years ago
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The graph of y= (x - 2)(x + 4) is shown. What is the y-intercept of this graph?
LenKa [72]

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The y intercept in -8

Step-by-step explanation:

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What is the domain of the relation shown<br> on the graph?
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where is the graph?........

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The line plot shows the number of points scored by Josh in his basketball games. How many games did Josh score more than six poi
andre [41]

Answer:

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Step-by-step explanation:

8 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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