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3 years ago

Help if you get it right I will give you Brainliest

Mathematics

1 answer:

olga_2 [115]3 years ago

8 0

Answer:

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Assume that two marbles are drawn without replacement from a box with 1 blue, 3 white, 2 green, and 2 red marbles. find the prob

Leona [35]

The first time I believe it's 1/4 and then 1/7

8 0

3 years ago

HI I NEVER KNEW HOW TO DO THIS BUT I DO SO CAN YALL REALLY HELP ME PLEASE

Aliun [14]

Answer:

A. 4/15 and 0.26 repeating

Step-by-step explanation:

As we see in the long division, 4/15 is equal to 0.26666666666666666666666 (or. 0.26 repeating).

4/15

4 doesn't go into 15, so add a zero.

4.0/15

40 goes into 15 twice

40 - 30 is 10. 10 is less than 15 so add a 0.

4.00

100 goes into 15 6 times.

100 - 90 is 10. 10 is less than 15, so add a 0.

The 100 into 15 process repeats, so there's your answer.

8 0

3 years ago

Read 2 more answers

What is the true solution to the equation below?

Marina CMI [18]

Answer:

The solution is:

$x=4$

Step-by-step explanation:

Considering the expression

$lne^{lnx}+lne^{lnx}^{^2}=2ln8$

$\ln \left(e^{\ln \left(x\right)}\right)+\ln \left(e^{\ln \left(x\right)\cdot \:2}\right)=2\ln \left(8\right)$

$\mathrm{Apply\:log\:rule}:\quad \:log_a\left(a^b\right)=b$

$\ln \left(e^{\ln \left(x\right)}\right)=\ln \left(x\right),\:\space\ln \left(e^{\ln \left(x\right)2}\right)=\ln \left(x\right)2$

$\ln \left(x\right)+\ln \left(x\right)\cdot \:2=2\ln \left(8\right)$

$\mathrm{Add\:similar\:elements:}\:\ln \left(x\right)+2\ln \left(x\right)=3\ln \left(x\right)$

$3\ln \left(x\right)=2\ln \left(8\right)$

$\mathrm{Divide\:both\:sides\:by\:}3$

$\frac{3\ln \left(x\right)}{3}=\frac{2\ln \left(8\right)}{3}$

$\ln \left(x\right)=\frac{2\ln \left(8\right)}{3}.....A$

Solving the right side of the equation A.

$\frac{2\ln \left(8\right)}{3}$

$\ln \left(8\right):\quad 3\ln \left(2\right)$

Because

$\ln \left(8\right)$

$\mathrm{Rewrite\:}8\mathrm{\:in\:power-base\:form:}\quad 8=2^3$

⇒ $\ln \left(2^3\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _a\left(x^b\right)=b\cdot \log _a\left(x\right)$

$\ln \left(2^3\right)=3\ln \left(2\right)$

$\frac{2\ln \left(8\right)}{3}=\frac{2\cdot \:3\ln \left(2\right)}{3}$

$\mathrm{Multiply\:the\:numbers:}\:2\cdot \:3=6$

$=\frac{6\ln \left(2\right)}{3}$

$\mathrm{Divide\:the\:numbers:}\:\frac{6}{3}=2$

$=2\ln \left(2\right)$

So, equation A becomes

$\ln \left(x\right)=2\ln \left(2\right)$

$\mathrm{Apply\:log\:rule}:\quad \:a\log _c\left(b\right)=\log _c\left(b^a\right)$

$=\ln \left(2^2\right)$

$=\ln \left(4\right)$

$\ln \left(x\right)=\ln \left(4\right)$

$\mathrm{Apply\:log\:rule:\:\:If}\:\log _b\left(f\left(x\right)\right)=\log _b\left(g\left(x\right)\right)\:\mathrm{then}\:f\left(x\right)=g\left(x\right)$