All categories

3 years ago

What is the solution

Mathematics

1 answer:

ladessa [460]3 years ago

3 0

Answer:

No solutions or ∅

Step-by-step explanation:

Do the algebra:

$-\frac{1}{3}+\frac{2}{3}m=\frac2 3 m - \frac 2 3 \\\\\frac 2 3 m = \frac 2 3 m - \frac 2 3\\\\0=-\frac 2 3$

Since 0 ≠ -2/3, this equation has no solutions.

Hope this helped :)

You might be interested in

** This<br> Solve each system of equations by<br> 1.<br> y= 3x +19<br> y = 5x + 33

Sveta_85 [38]

Answer:

x=-7 and y=-2

Step-by-step explanation:

hope this helps

8 0

3 years ago

What is -9 in parenthesis to the 5th power divided by -9 in parenthesis to the 7th power?

zaharov [31]

Answer:

1/81

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

A computer consulting firm presently has bids out on three projects. Let Ai = {awarded project i}, for i = 1, 2, 3, and suppose

Karo-lina-s [1.5K]

Step-by-step explanation:

The data below is what was provided in the question and it is what I solved the question with

P(A1) = 0.23

P(A2) = 0.25

P(A3) = 0.29

P(A1 n A2 ) = 0.09

P(A1 n A3) = 0.11

P(A2 n A3) = 0.07

P(A1 n A2 n A3) = 0.02

P(A2|A1) = P(A1 n A2)/P(A1)

= 0.09/0.23

= 0.3913

We have 39.13% confidence that event A2 will occur given that event A1 already occured

b.)

P(A3 n A3|A1) = P(A2 n A3)n A1)/P(A1)

= 0.02/0.23

= 0.08695

We have about 8.7% chance of events A2 and A3 occuring given that A1 already occured.

P(A2 u A3|A1)

= P(A1 n A2)u(A1 n A3)/P(A1)

= P( A1 n A2) + P(A1 n A3) - P(A1 n A2 n A3) / P(A1)

= (0.09+0.11-0.02)/0.23

= 0.18/0.23

= 0.7826

We have 78.26% chance of A2 or A3 happening given that A1 has already occured.

6 0

3 years ago

6789 divided by 89.65

Sergeeva-Olga [200]

Step-by-step explanation:

75 . 727 is your answer

hope it is helpful to you

5 0

3 years ago

Read 2 more answers

Pls help only if yk it ill give brainliest to the correct answer !

IgorC [24]

Answer:

$x=-0.3$

$x=-11.7$

Step-by-step explanation:

$4x^2+24x+13=2x^2+6\\\mathrm{Subtract\:}6\mathrm{\:from\:both\:sides}\\4x^2+24x+13-6=2x^2+6-6\\Simplify\\4x^2+24x+7=2x^2\\\mathrm{Subtract\:}2x^2\mathrm{\:from\:both\:sides}\\4x^2+24x+7-2x^2=2x^2-2x^2\\Simplify\\2x^2+24x+7=0\\\mathrm{Solve\:with\:the\:quadratic\:formula}\\Quadratic\:Equation\:Formula\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=2,\:b=24,\:c=7:\quad x_{1,\:2}=\frac{-24\pm \sqrt{24^2-4\cdot \:2\cdot \:7}}{2\cdot \:2}\\x=\frac{-24+\sqrt{24^2-4\cdot \:2\cdot \:7}}{2\cdot \:2}:\quad \frac{-12+\sqrt{130}}{2}\\x=\frac{-24-\sqrt{24^2-4\cdot \:2\cdot \:7}}{2\cdot \:2}:\quad -\frac{12+\sqrt{130}}{2}\\The\:solutions\:to\:the\:quadratic\:equation\:are:\\x=\frac{-12+\sqrt{130}}{2},\:x=-\frac{12+\sqrt{130}}{2}\\\\quad x=-0.29912\ ,\:x=-11.70087$

6 0

3 years ago