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Brilliant_brown [7]
3 years ago
11

Please I need answers for this one or two are fine

Mathematics
2 answers:
Mandarinka [93]3 years ago
7 0

Answer: A= 1100

B= 11000

C= 110000

Step-by-step explanation:

Ivanshal [37]3 years ago
3 0

Answer:

a:1100....b:11000....c:110000

Step-by-step explanation:

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Help!!Need help please help me
irina [24]
 8x-6y=-96  add to this -4 times the second equation...
-8x-12y=-48
___________

-18y=-144

y=8, this makes 8x-6y=-96 become:

8x-48=-96

8x=-48

x=-6

so the solution to the system of equations is the point:

(-6,8)
3 0
3 years ago
Between 0 degrees Celsius and 30 degrees Celsius, the volume V (in cubic centimeters) of 1 kg of water at a temperature T is giv
Ira Lisetskai [31]

Answer:

T = 3.967 C

Step-by-step explanation:

Density = mass / volume

Use the mass = 1kg and volume as the equation given V, we will come up with the following equation

D = 1 / 999.87−0.06426T+0.0085043T^2−0.0000679T^3

   = (999.87−0.06426T+0.0085043T^2−0.0000679T^3)^-1

Find the first derivative of D with respect to temperature T

dD/dT = \dfrac{70000000\left(291T^2-24298T+91800\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^2}

Let dD/dT = 0 to find the critical value we will get

\dfrac{70000000\left(291T^2-24298T+91800\right)}  = 0

Using formula of quadratic, we get the roots:

T =  79.53 and T = 3.967

Since the temperature is only between 0 and 30, pick T = 3.967

Find 2nd derivative to check whether the equation will have maximum value:

-\dfrac{140000000\left(395178T^4-65993368T^3+3286558821T^2+2886200857800T-121415215620000\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^3}

Substituting the value with T=3.967,

d2D/dT2 = -1.54 x 10^(-8)    a negative value. Hence It is a maximum value

Substitute T =3.967 into equation V, we get V = 0.001 i.e. the volume when the the density is the highest is at 0.001 m3 with density of

D = 1/0.001 = 1000 kg/m3

Therefore T = 3.967 C

3 0
4 years ago
Pls help me ! i need help
Blizzard [7]

Answer:

what do you need help with?

7 0
4 years ago
Read 2 more answers
The median of a continuous random variable having distribution function F is that value m such that F(m) = 1/2 . That is, a rand
neonofarm [45]

Answer:

Step-by-step explanation:

To find median and mode for

a) In a uniform distribution median would be

(a+b)/2 and mode = any value

b) X is N

we know that in a normal bell shaped curve, mean = median = mode

Hence mode = median = \mu

c) Exponential with parameter lambda

Median = \frac{ln2}{\lambda }

Mode =0

7 0
3 years ago
HELP PLEASE!!!<br> A:645,473<br> B:2,151,575<br> C:8,175,987<br> D:10,542,719
marysya [2.9K]
38% of 21,515,714 is 8175971.32 which can be rounded to 8,175,971.
Hope this helps :)<span>
</span>
6 0
3 years ago
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