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3 years ago

How can i find the probabilities in questions like these (there are two different probabilities???)

1 answer:

Nutka1998 [239]3 years ago

4 0

Find the probability of picking a red marble. This would be 12 out of 20. The, take 1 away from the denominator because you already picked a marble. So then find the probability of picking white marble, which is 8 out of 19. The multiply them to get 84 out of 380

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Help with math pls cant figure it out

tatyana61 [14]

Answer:

$\frac{5}{4}+or-\frac{3i\sqrt{7} }{4}$

Step-by-step explanation:

You can solve this equation with the quadratic formula which is:

$m_{1,2}=\frac{-b+or-\sqrt{b^{2}-4ac } }{2a}$

All you need to do is rearrange the equation (add $-\frac{11}{2}$ to both sides and then multiply both sides by 2) to get:

$2m^2-5m+11=0$

From this new equation, we can easily see that a = 2, b = -5, and c = 11 AND you don't have to work with as many fractions. Substitute the values for a, b, and c into the quadratic formula to get:

$m_{1,2}=\frac{5+or-\sqrt{(-5)^{2}-4(2)(11) } }{2(2)}$

That simplifies to:

$\frac{5}{4}+or-\frac{\sqrt{25-88} }{4}$

Which simplifies even further to:

$\frac{5}{4}+or-\frac{\sqrt{-63} }{4}$

Which gets you:

$\frac{5}{4}+or-\frac{3i\sqrt{7} }{4}$

P.S. The plus-minus sign wasn't working for me, so I had to just write, "+or-". I hope that didn't make it too confusing for you.

P.P.S. Hope this helps :) Don't forget to mark Brainliest if it did.

7 0

3 years ago

Write the ratios for sin A and cos A. Need help.

klio [65]

Answer:

Your answer would be C, Sin A= 3/5 cos A= 4/5

Step-by-step explanation:

Remember Soh-cah-toa

Soh stands for sin= opposite/hypotenuse

Cah stands for cos= adjacent/hypotenuse

Toa stands for tan= opposite/adjacent

8 0

3 years ago

Determine the values of k for which the equation has no solution

Naddik [55]

To solve this, we would need to figure out the values of $k$ for which an absolute value is negative. From basics, you should know that an absolute value is always greater than or equal to 0.

If we rearrange this equation a little bit, we get,

$|2x+7|\leq 5+k$

So $5+k$ cannot be negative. So,

$5+k\geq 0\\k\geq -5$

For this to have a solution, k needs to be greater than or equal to $-5$ .

For no solution, k is less than -5. So, values of k which will make the equation to not have any solution is $k$ .

ANSWER: $k$

5 0

3 years ago

4. A loan made on March 12 is due September 10 of the following year. Find the exact time for the loan in a non-leap year and a

saul85 [17]

Answer:

182 days

Step-by-step explanation:

Given:

Date on which loan is made = March 12

Due date of the loan = September 10

Now,

The months in between March and September and the number of days in the particular months are

Month Number of days

April 30

May 31

June 30

July 31

August 31

------------------------------------------

Total number of days = 153

also,

Number of days left in March after 12 are (31 - 12 ) = 19 days

and,

Number of days in September before the due date = 10 days

Hence, the total time for loan = 19 + 153 + 10 = 182 days

since the included months does not have the month of February. Therefore,

the time for both leap year and non-leap year will remain same.

7 0

3 years ago

Plss answer!! MERRY Christmas!!! Critical Question. AAAAA

liberstina [14]

Answer:

Step-by-step explanation:

abc = 1

We have to prove that,

$\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}=1$

We take left hand side of the given equation and solve it,

$\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+\frac{1}{c}}+\frac{1}{1+c+\frac{1}{a}}$

Since, abc = 1,

$\frac{1}{c}=ab$ and c = $\frac{1}{ab}$

By substituting these values in the expression,

$\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+\frac{1}{c}}+\frac{1}{1+c+\frac{1}{a}}=\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+ab}+\frac{1}{1+\frac{1}{ab}+\frac{1}{a}}$

$=\frac{b}{b+ab+1}+\frac{1}{1+b+ab}+\frac{ab}{ab+1+b}$

$=\frac{1+b+ab}{1+b+ab}$

$=1$

Which equal to the right hand side of the equation.

Hence, $\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}=1$

5 0

3 years ago