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3 years ago

I will give u 30 points if u help me

Mathematics

1 answer:

nalin [4]3 years ago

5 0

Hope it helps :)
Have a good day :))

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A large container holds 5 gallons of water. It begins leaking at a constant rate. After 10 minutes, the container has 3 gallons

Anna007 [38]

5 - 10 x = 3
10 x = 5 - 3
10 x = 2
x = 10 : 5
x = 0.2
The rate is : 0.2 gallons / minute
5 - t * 0.2 = 0
t * 0.2 = 5
t = 5 : 0.2
t = 25 min
Answer: After 25 minutes the container will be empty.

5 0

3 years ago

Write this decimal in a fraction in its simplest form.

sammy [17]

The simplified form of that decimal in fraction form is 3/20. Hope this helps.

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3 years ago

Certain tubes manufactured by a company have a mean lifetime of 800 hours and a standard deviation of 60 hours. Find the probabi

Anna71 [15]

Answer:

Step-by-step explanation:

given that certain tubes manufactured by a company have a mean lifetime of 800 hours and a standard deviation of 60 hours.

Sample size n =16

Std error of sample mean = $\frac{\sigma}{\sqrt{n} } \\= 15$

x bar follows N(800, 15)

the probability that a random sample of 16 tubes taken from the group will have a mean lifetime

(a) between 790 and 810 hours,

= $P(790$

(b) less than 785 hours

$=P(X$

, (c) more than 820 hours,

$=P(X>820)\\=p(Z>1.333)\\= 0.0913$

(d) between 770 and 830 hours

= $P(|Z|$

4 0

3 years ago

Eli invested $ 330 $330 in an account in the year 1999, and the value has been growing exponentially at a constant rate. The val

Cloud [144]

Answer:

The value of the account in the year 2009 will be $682.

Step-by-step explanation:

The acount's balance, in t years after 1999, can be modeled by the following equation.

$A(t) = Pe^{rt}$

In which A(t) is the amount after t years, P is the initial money deposited, and r is the rate of interest.

$330 in an account in the year 1999

This means that $P = 330$

$590 in the year 2007

2007 is 8 years after 1999, so P(8) = 590.

We use this to find r.

$A(t) = Pe^{rt}$

$590 = 330e^{8r}$

$e^{8r} = \frac{590}{330}$

$e^{8r} = 1.79$

Applying ln to both sides:

$\ln{e^{8r}} = \ln{1.79}$

$8r = \ln{1.79}$

$r = \frac{\ln{1.79}}{8}$

$r = 0.0726$

Determine the value of the account, to the nearest dollar, in the year 2009.

2009 is 10 years after 1999, so this is A(10).

$A(t) = 330e^{0.0726t}$

$A(10) = 330e^{0.0726*10} = 682$

The value of the account in the year 2009 will be $682.

4 0

3 years ago

Please help ill give brianleist answer!!!!

GaryK [48]

Answer:

i think its option ccc ydi aeg

because angle y is congruent to angle a

angle i is congruent to angle d

3 0

3 years ago

Read 2 more answers