Suppose that the handedness of the last 15 U. S. presidents is as follows: (i) 40% were left-handed (L) (ii) 47% were democrats (D) (iii) If a president is left-handed, there is a 13% chance that the president is a Democrat. What is the probability that a randomly chosen U. S. president is left-handed and a democrat?

0.0520

Step-by-step explanation:

Given the data above of 15 past presidents, let (L) denote left handedness and (D) denote democrat

Therefore, 40% = L = 0.4

47% = D = 0.47

13% chance that a left handed president is a democrat = 0.13

probability that a randomly chosen U.S. president is left-handed and a democrat

1. 0.4×0.47 = 0.1880

2. 0.4×0.13 = 0.0520

3. 0.47×0.13 = 0.0611

4. 0.4/0.47 = 0.8510

5. 0.4/0.13 = 3.0769

6. 0.47/0.13 = 3.6154

But, we are trying to find P(L and D) = P(L)×P(D|L)

= 0.40 × 013

= 0.0520

ce pula mea vrei ma

step-by-step explanation:

gayule

answer: solve 5\cdot 2^x=2405⋅2  

x

=2405, dot, 2, start superscript, x, end superscript, equals, 240.

solution

to solve for xxx, we must first isolate the exponential part. to do this, divide both sides by 555 as shown below. we do not multiply the 555 and the 222 as this goes against the order of operations!

\begin{aligned} 5\cdot 2^x& =240 2^x& =48 \\ \end{aligned}  

5⋅2  

x

2  

x

​    

=240

=48

​

step-by-step explanation: