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3 years ago

Is the following a function ?

Mathematics

1 answer:

Vanyuwa [196]3 years ago

6 0

Yes it is a function

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Name the plane represented by the front of the box.

andre [41]

Answer:

FBC

Step-by-step explanation:

It's the only one at the front of the box

3 0

2 years ago

Because of the rainy season, the depth in a pond increases 3% each week. Before the rainy season started, the pond was 10 feet d

Dafna11 [192]

It's not perfectly clear, but you seem to be saying that each week, the
pond becomes 3% deeper than it was THE PREVIOUS WEEK.

If I'm reading that correctly, then the pond is 10-ft deep when the rainy
season begins, and after 'x' weeks later, the depth is

   D = 10 (1.03)^x feet.

After 8 weeks, the depth is D = 10 (1.03)⁸ = 12.67 feet .

       rounded to the nearest foot, D = 13 ft.

7 0

3 years ago

Read 2 more answers

Which number line represents x2=16

Ivenika [448]

Answer:

Step-by-step explanation:

Well 2x=16 is the same as x=8 when you divide by 2. So the dot should be on 8.

6 0

2 years ago

Read 2 more answers

Suppose that, after measuring the duration of many telephone calls, a telephone company found their data was well-approximated b

Musya8 [376]

Answer:

a) 7.79%

b) 67.03%

c) Cumulative Distribution Function

$P(t) = \displaystyle\int^{\infty}_{-\infty} 0.1e^{-0.1t}~dt\\\\= \displaystyle\int^{b}_{a} 0.1e^{-0.1t}~dt, ~~a\leq t \leq b$

Step-by-step explanation:

We are given the following in the question:

$p(x) = 0.1 e^{-0.1x}$

where x is the duration of a call, in minutes.

a) P( calls last between 2 and 3 minutes)

$=\displaystyle\int^3_2 p(x)~ dx\\\\= \displaystyle\int^3_20.1e^{-0.1x}~dx\\\\=\Big[-e^{-0.1x}\Big]^3_2\\\\=-\Big[e^{-0.3}-e^{-0.2}\Big]\\\\= 0.0779\\=7.79\%$

b) P(calls last 4 minutes or more)

$=\displaystyle\int^{\infty}_4 p(x)~ dx\\\\= \displaystyle\int^{\infty}_40.1e^{-0.1x}~dx\\\\=\Big[-e^{-0.1x}\Big]^{\infty}_4\\\\=-\Big[e^{\infty}-e^{-0.4}\Big]\\\\=-(0- 0.6703)\\= 0.6703\\=67.03\%$

c) cumulative distribution function

$P(t) = \displaystyle\int^{\infty}_{-\infty} 0.1e^{-0.1t}~dt\\\\= \displaystyle\int^{b}_{a} 0.1e^{-0.1t}~dt, ~~a\leq t \leq b$

6 0

3 years ago

What is the answer to this question?

sukhopar [10]

First plug in f and g:
$3 x^{2} +1- [1-x]$
Then distribute $3 x^{2} +1-1+x$
Combine like terms
$3 x^{2} +x$
Now plug in 2 for x
$3( 2^{2} )+2$
3(4)+2=12+2=14

Hope that helps

8 0

3 years ago