Answer:
Step-by-step explanation:
So,it would be easier to simplify the right equation, so when I say "the right equation", I mean the one I will make right now. It is 8x+9. So the first problem is 8x+9=x+?. Well, if there are no solutions, the slopes have to be the same, and the y-int doesn't matter. So the answer would be + 7x because that will make the x an 8x. If there is one solution, then the slope has to be different. So literally anything but 7x will work. Even constants. If there are infinitely many solutions, then the equations have to be the same. Meaning, you would have to add 7x and 9 to make the left equation the same as the right.
Answer:
B
Step-by-step explanation:
The second one!!!
To get a number as a percent of another, you divide and multiply by 100, so here you would have (13/52)*100=.25*100=25, making your answer 25%.
Answer:
Looking at the first question, it's asking what best describes the probability of tossing a number less than 6 on a number cube that has 6 numbers. Impossible means that it will never land on it, for example asking what the probability of landing on 7 is. Unlikely is something that doesn't happen often. The best option that fits our scenario is option C, likely.
Looking at the second question, it's asking what the probability that the teacher chooses a girl in his class. There are 15 girls and a total of 27 students in the class so we take the probability by doing 15/27. We can narrow both the numerator and the denominator using 3 which gives us 5/9. Therefore, the best option that fits our scenario is option C, 5/9.
Finally, looking at the last question, it's asking what the theoretical probability that the coin will land on heads on the next toss. Theoretical probability doesn't consider how much times Murray tossed the coin, the only thing it cares about is what the actual probability of tossing a coin is. Therefore that makes it a 50% chance of landing on a heads and a 50% chance of landing on a tails. The best option that first our scenario is option B, 1/2.
<u><em>Hope this helps! Let me know if you have any questions</em></u>
