bag I contain 3 red and 4 black balls while another bag II contains 5 red and 6 black balls. One ball is drawn at random from one bags and it is to be red. Find the probability that it was drawn from bag II g

Answer:

The sum of three times a number and twelve is fewer than twenty translates into an inequality

3x+12 < 20

Step-by-step explanation:

it's not the perfect answer of your question but it's kind of similar to it

Answer:

a. 8.75

b. 0.72

c. 9.31

d. 23.04

e. 6.98

Step-by-step explanation:

a.

in this number 8.7495980

the digit in the hundredth place is 4

the digit in the thousandth  place is 9

Since 9 ≥ 5 ,then 8.7495980 rounded to nearest hundredth is :

8,75

b.

in this number 0.72091

the digit in the hundredth place is 2

the digit in the thousandth  place is 0

Since 0 < 5 ,then  0.72091 rounded to nearest hundredth is :

0.72

c.

in this number 9.3113

the digit in the hundredth place is 1

the digit in the thousandth  place is 1

Since 1 < 5 ,then  9.3113 rounded to nearest hundredth is :

9.31

d.

in this number 23.0368

the digit in the hundredth place is 3

the digit in the thousandth  place is 6

Since 6 ≥ 5 ,then 23.0368 rounded to nearest hundredth is :

23.04

e.

in this number 6.9788

the digit in the hundredth place is 7

the digit in the thousandth  place is 8

Since 8 ≥ 5 ,then 6.9788 rounded to nearest hundredth is :

6.98

Step-by-step explanation:

here,

let MP be x then,

SP=x-20%of X

=x - 20x/100

=x- 0.2x

=0.8x

CP= x- 30% of x

= x- 30x/100

=x- 0.3x

=0.7x

now,profit (P) = SP -CP

= 0.8x-0.7x

=0.1x

so,

P%=( P\CP ) ×100%

= (0.1x/0.7x) ×100%

= 14.28 %

Answer:

[tex]all \: real \: numbers \: except \: \\ x + 3 < 0 \\ \: \: \: \: x < - 3[/tex]

Answer:

X is the domain and sqr root of X+3 is the range.

Answer:

m = (6-5)/(4-1)

m = 1/3

.............

Answer:

Step-by-step explanation:

[tex]\\ \rm\longmapsto Length(Breadth)=Area[/tex]

[tex]\\ \rm\longmapsto Length(7)=672[/tex]

[tex]\\ \rm\longmapsto Length=\dfrac{672}{7}[/tex]

[tex]\\ \rm\longmapsto Length=96m[/tex]

Answer:

Step-by-step explanation:

The fifth root of -243 = - 3

If you now square  - 3, you get (-3)(-3) = 9

Answer: x < 7  

Step-by-step explanation:

Answer:

y = -30

Step-by-step explanation:

Set the ratio: [tex]\frac{x_1}{y_1} = \frac{x_2}{y_2}[/tex]

Let:

[tex](x_1 , y_1) = (-4 , 10)\\(x_2 , y_2) = (12 , y)\\[/tex]

Plug in the corresponding terms to the corresponding variables:

[tex]\frac{-4}{10} = \frac{12}{y}[/tex]

First, cross multiply.

[tex](-4 * y) = (10 * 12)\\-4y = 10 * 12\\-4y = 120[/tex]

Next, isolate the variable, y. Divide -4 from both sides of the equation:

[tex]\frac{-4y}{-4} = \frac{120}{-4}\\y = \frac{120}{-4}\\y = -30[/tex]

-30 is your answer for y.

Answer:

16. 42

21. 18

Step-by-step explanation:

Answer:

you can check the picture, i tried to make it as simple as possible. so I answered by the format.

Answer:

x=0

Step-by-step explanation:

0+4=4

4x2=8

it's pretty simple once you get the hang of it

Answer:

It is 36 cm

Step-by-step explanation:

In equilateral triangle, all sides are equal.

[tex]{ \sf{perimeter = side + side + side}}[/tex]

but side = 12 cm

[tex] { \sf{p = (12 + 12 + 12)}} \\ { \sf{p = 36 \: cm}}[/tex]

Answer: Perimeter = 36 cm

Concept:

Here, we need to know how to find the perimeter of an equilateral triangle.

An equilateral triangle is a triangle in which all three sides have the same length.

Perimeter (equilateral triangle) = 3s

s = side length

Solve:

Given information

Side length = 12 cm

Given expression

Perimeter = 3s

Substitute values into the expression

Perimeter = 3 (12)

Simplify by multiplication

Perimeter= [tex]\boxed{36cm}[/tex]

Hope this helps!! :)

Please let me know if you have any questions

Answer:

B. hexagon

Step-by-step explanation:

Answer:

i think this might be it? but most likely :) <3

Step-by-step explanation:

Answer:

183 centimeters

Step-by-step explanation:

2 multiply the length value by 91.44

Answer:

8.5 × 10^5

Step-by-step explanation:

Move the decimal so there is one non-zero digit to the left of the decimal point. The number of decimal places you move will be the exponent on the 10. If the decimal is being moved to the right, the exponent will be negative. If the decimal is being moved to the left, the exponent will be positive.

[tex]\\ \rm\Rrightarrow (2.5\times 10^3)^3\times 4\times 10^2[/tex]

[tex]\\ \rm\Rrightarrow 2.5\times 3.4 10^3 \times 10^2[/tex]

[tex]\\ \rm\Rrightarrow 8.5\times 10^{3+2}[/tex]

[tex]\\ \rm\Rrightarrow 8.5\times 10^5[/tex]

Answer:

$12.5 dollars

Step-by-step explanation:

cc

Answer:

x=60

Step-by-step explanation:

x=180-120

Step-by-step explanation:

According to the quotient rule of differentiation, if [tex]y(x) = u(x)/v(x),[/tex] then its derivative is given by

[tex]\dfrac{dy}{dx} = \dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^2}\:\:\:\:\:\:\:\:\:(1)[/tex]

We can see that

[tex]u(x) = \ln (4x^2 + 1)[/tex]

[tex]\dfrac{du}{dx} = \dfrac{8x}{4x^2 + 1}[/tex]

[tex]v(x) = 2x - 3[/tex]

[tex]\dfrac{dv}{dx} = 2[/tex]

Plugging the above expressions into Eqn(1), we find that the derivative is

[tex]\dfrac{dy}{dx} = \dfrac{\dfrac{8x(2x - 3)}{(4x^2 + 1)} - 2\ln (4x^2 + 1)}{(2x - 3)^2}[/tex]

[tex]\:\:\:\:\:\:\:= \dfrac{8x}{(4x^2 + 1)(2x - 3)} - \dfrac{2\ln (4x^2 + 1)}{(2x - 3)^2}[/tex]

please mark this answer as brainlist

Check the pictures attached for the answers.

9514 1404 393

Answer:

  next: ←, ↑, →, ↓, ←

Step-by-step explanation:

It appears you want to identify a rule describing the sequence, then provide several additional elements in the sequence.

It looks to me like each successive element has been rotated 90° CW from the previous one.

Continuing that pattern, the arrows will point left, up, right, down, left.

Answer:

Virchow

Step-by-step explanation:

He was the one who discovered that all cells come from pre-existing cells, the third part of the cell theory.

Probabilities are used to determine the chance of an event. The following are the summary of the solution.

Given that:

[tex]n = 8[/tex]

[tex]G = 6[/tex] --- Green

[tex]Y = 2[/tex] --- Yellow

(a) Probability that the two cellphones are green (without replacement).

Since the cellphone is not replaced, the probability is calculated as follows:

[tex]Pr = \frac Gn \times \frac{G - 1}{n-1}[/tex]

So, we have:

[tex]Pr = \frac 68 \times \frac{6 - 1}{8-1}[/tex]

[tex]Pr = \frac 68 \times \frac 57[/tex]

[tex]Pr = \frac{30}{56}[/tex]

[tex]Pr = \frac{15}{28}[/tex]

Hence, the probability that the two cellphones are green (without replacement) is 15/28

(b) Probability that one green and one yellow is selected (without replacement).

Since the cellphone is not replaced, the probability is calculated as follows:

[tex]Pr = \frac Gn \times \frac{Y}{n-1} + \frac Yn \times \frac{G}{n-1}[/tex]  ---- The subtraction means the cellphones are not replaced

This gives

[tex]Pr = \frac 68 \times \frac{2}{8-1} + \frac 28 \times \frac{6}{8-1}[/tex]

[tex]Pr = \frac 34 \times \frac{2}{7} + \frac 14 \times \frac{6}{7}[/tex]

[tex]Pr = \frac 32 \times \frac17 + \frac 12 \times \frac 37[/tex]

[tex]Pr = \frac{3}{14} + \frac{3}{14}[/tex]

Take LCM

[tex]Pr = \frac{3+3}{14}[/tex]

[tex]Pr = \frac{6}{14}[/tex]

[tex]Pr = \frac{3}{7}[/tex]

Hence, the probability that one green and one yellow is selected (without replacement) is 3/7

(c) Probability that the all three cellphones are yellow (with replacement).

Since the cellphone is replaced, the probability is calculated as follows:

[tex]Pr = \frac Yn \times \frac Yn \times \frac Yn[/tex]

So, we have:

[tex]Pr = \frac 28 \times \frac 28 \times \frac 28[/tex]

[tex]Pr = \frac 14 \times \frac 14 \times \frac 14[/tex]

[tex]Pr = \frac 1{64}[/tex]

Hence, the probability that all three cellphones are yellow (with replacement) is 1/64

(d1) Probability that the two cellphones are green (with replacement).

Since the cellphone is replaced, the probability is calculated as follows:

[tex]Pr = \frac Gn \times \frac{G}{n}[/tex]

So, we have:

[tex]Pr = \frac 68 \times \frac{6}{8}[/tex]

[tex]Pr = \frac 34 \times \frac{3}{4}[/tex]

[tex]Pr = \frac{9}{16}[/tex]

Hence, the probability that the two cellphones are green (with replacement) is 9/16

(d2) Probability that one green and one yellow is selected (with replacement).

Since the cellphone is replaced, the probability is calculated as follows:

So, we have:

[tex]Pr = \frac Gn \times \frac{Y}{n} + \frac Yn \times \frac{G}{n}[/tex]

This gives

[tex]Pr = \frac 68 \times \frac{2}{8} + \frac 28 \times \frac{6}{8}[/tex]

[tex]Pr = \frac 34 \times \frac{1}{4} + \frac 14 \times \frac{3}{4}[/tex]

[tex]Pr = \frac 3{16} + \frac{3}{16}[/tex]

Take LCM

[tex]Pr = \frac {3+3}{16}[/tex]

[tex]Pr = \frac {6}{16}[/tex]

[tex]Pr = \frac {3}{8}[/tex]

Hence, the probability that one green and one yellow is selected (with replacement) is 3/8

Read more about probabilities at:

https://brainly.com/question/795909

Answer:

3. Sanford Township, Minnesota

4. Inner Mongolia, China

5. Arunachal Pradesh, India

6.Almaty Region, Kazakhstan

7.Madhya Pradesh, India

Step-by-step explanation:

I'll only look at (37) here, since

• (38) was addressed in 24438105

• (39) was addressed in 24434477

• (40) and (41) were both addressed in 24434541

In both parts, we're considering the line integral

[tex]\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy[/tex]

and I assume C has a positive orientation in both cases

(a) It looks like the region has the curves y = x and y = x ² as its boundary***, so that the interior of C is the set D given by

[tex]D = \left\{(x,y) \mid 0\le x\le1 \text{ and }x^2\le y\le x\right\}[/tex]

• Compute the line integral directly by splitting up C into two component curves,

C₁ : x = t and y = t ² with 0 ≤ t ≤ 1

C₂ : x = 1 - t and y = 1 - t with 0 ≤ t ≤ 1

Then

[tex]\displaystyle \int_C = \int_{C_1} + \int_{C_2} \\\\ = \int_0^1 \left((3t^2-8t^4)+(4t^2-6t^3)(2t))\right)\,\mathrm dt \\+ \int_0^1 \left((-5(1-t)^2)(-1)+(4(1-t)-6(1-t)^2)(-1)\right)\,\mathrm dt \\\\ = \int_0^1 (7-18t+14t^2+8t^3-20t^4)\,\mathrm dt = \boxed{\frac23}[/tex]

*** Obviously this interpretation is incorrect if the solution is supposed to be 3/2, so make the appropriate adjustment when you work this out for yourself.

• Compute the same integral using Green's theorem:

[tex]\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy = \iint_D \frac{\partial(4y-6xy)}{\partial x} - \frac{\partial(3x^2-8y^2)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = \int_0^1\int_{x^2}^x 10y\,\mathrm dy\,\mathrm dx = \boxed{\frac23}[/tex]

(b) C is the boundary of the region

[tex]D = \left\{(x,y) \mid 0\le x\le 1\text{ and }0\le y\le1-x\right\}[/tex]

• Compute the line integral directly, splitting up C into 3 components,

C₁ : x = t and y = 0 with 0 ≤ t ≤ 1

C₂ : x = 1 - t and y = t with 0 ≤ t ≤ 1

C₃ : x = 0 and y = 1 - t with 0 ≤ t ≤ 1

Then

[tex]\displaystyle \int_C = \int_{C_1} + \int_{C_2} + \int_{C_3} \\\\ = \int_0^1 3t^2\,\mathrm dt + \int_0^1 (11t^2+4t-3)\,\mathrm dt + \int_0^1(4t-4)\,\mathrm dt \\\\ = \int_0^1 (14t^2+8t-7)\,\mathrm dt = \boxed{\frac53}[/tex]

• Using Green's theorem:

[tex]\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dx = \int_0^1\int_0^{1-x}10y\,\mathrm dy\,\mathrm dx = \boxed{\frac53}[/tex]