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3 years ago

Decrease 9809821 by 60000

1 answer:

7 0

Hi!
The answer is 9749821.
I did 9809821 - 6000.
I hope this answer helps!
I would love it if you helped me with my question too!

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Kaelyn Kaitlyn start with $15 and saved $3.50 each day what expression represents the total amount of Taylor save

Vinil7 [7]

Your equation should look like this:

d = day

15 + 3.50d

Hope this helps!

6 0

3 years ago

Please answer this correctly as soon as possible I want genius or expert people to answer this correctly

Ede4ka [16]

6/54 times because there are 9 marbles and the orange marble is 1 out of those 9 marbles. If you multiply 6*9= 54 then you notice there is a 6/54 chance of picking any 1 marble out of the 54 times that you decide to pick marbles.

4 0

3 years ago

Find the surface area of this rectangular prism: Don't forget to show the unit of measure.

Ann [662]

Answer:

See Explanation

Step-by-step explanation:

The question is incomplete as the attachment of the prism is not given.

From the attachment:

$L =10cm$ -- Length

$W = 4cm$ -- Width

$H=5cm$ -- Height

Required

Determine the volume

So, the volume is:

$Volume = Length * Width (* Height$

$Volume = 10cm * 4cm * 5cm$

$Volume = 200cm^3$

8 0

3 years ago

I need help with 10,11,12

pshichka [43]

10. 50 is roughly around ~16% of 300. (16% is 50.1)

11. 35% of 150 would be 52.5

12. 80 is 200% of 40

3 0

3 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