Question<br> Find the multiplicative inverse of 0.13.

A cell doubles every 15 minutes. If there is 1 cell at noon, how many cells will there be at 3 p.m.? Show work.

Alik [6]

12:15 - 1+1 = 2
12:30 - 2 + 2 = 4
12:45 - 4 + 4 = 8
1 - 8 +8 = 16
1:15.......
.......
3 - 4096

4 0

2 years ago

The diameter of the sun is about 1.4x10^6 km. The diameter of the planet Mercury is about 5000 km. What is the approximate ratio

RUDIKE [14]

B!!! I just had this question! I made a 98 ! Hope this helps

3 0

2 years ago

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A lightbulb manufacturer ships large consignments of lightbulbs to big industrial users. When the production process is function

Mashutka [201]

Answer:

a) P[ Pf | Ptd ] = 0,428

b) P[ Pf | Nd ] = 0,692

Step-by-step explanation:

We will call:

Pf Probability of functioning correctly 60 % or 0.6

Pfn Probability of malfunctioning 40 % or 0,4

PNd Probability of nondefective bulb

D₁ Probability of defective bulb when the process is correctly working

D₂ Probability of defective bulb when the process is in malfunction condition

Ptd Total probability of defective bulbs

Then applying theorem´s Bayes have

P [ A | B ] = P(A) * P [ B | A ] / P(B)

a) Probabilty of the process is functioning correctly given that we pick up a defective bulb

The total probability of defective bulbs is equal to, probabilty of defective bulbs when the process is Ok, plus probability of defective bulbs when the process is in malfunctioning condition, therefore

Ptd = (0,6)*(0,25) + (0,40)*(0,50) = 0,15 * 0,20 = 0,35

P[ Pf | Ptd ] = Pf * P [ Pd | Pf ] / Ptd = 0,6 * 0,25 / 0,35

P[ Pf | Ptd ] = 0,428

b) P [ Pf | Nd ]

P[ Pf | Nd ] = Pf * P [Nd | Pf ] / PNd

PNd = 1 - Ptd = 1 - 0,35 = 0,65

P[ Pf | Nd ] = 0,6 * 0,75 / 0,65 = 0,692

P[ Pf | Nd ] = 0,692

6 0

2 years ago

Classifying triangles according to side length and angle measurement

Anastaziya [24]

Acute with angles at 89 or smaller, Right with angles at 90 exact, and obtuse with angles with 91 and above

4 0

2 years ago

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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

2 years ago

Question Find the multiplicative inverse of 0.13.