<span><span>1/2<span>(<span>2g−3</span>)</span></span>=<span>−<span>4<span>(g+1)</span></span></span></span>
<span><span>1/2<span>(<span>2g−3</span>)</span></span>=<span>−<span>4<span>(g+1)</span></span></span></span>
<span><span>g+<span>−3/2</span></span>=<span><span>−4g</span>−4</span></span>
<span><span><span>g+<span>−3/2</span></span>+4g</span>=<span><span><span>−4g</span>−4</span>+4g</span></span><span><span>5g+<span>−32</span></span>=−4</span>
5g+−3/2+3/2=−4+3/2
<span><span>
5g</span>=<span>−5/2</span></span>
<span><span>5g/5</span>=<span><span>−52</span>5</span></span><span>
g=<span>−1<span>2
Hoped I helped!</span></span></span>
Answer: 0.3439
Step-by-step explanation:
Given :The last four digits for telephone numbers are randomly selected (with replacement).
Here , each position can be occupied with any of the digit independently .
Total digits = 10
Total digits other than 0 = 9
For each digits , the probability that it is not 0 = 
If we select 4 digits , The probability of getting no 0 =
(By multiplication rule of independent events)
Now , the probability that for one such phone number, the last four digits include at least one 0. = 1- P(none of them is 0)
=1- 0.6561=0.3439
Hence, the probability that for one such phone number, the last four digits include at least one 0. is 0.3439 .
Answer:
2/10 (4/5)-(5/2)+ (10)=7.66