An integer may be a multiple of 3.
An integer may be 1 greater than a multiple of 3.
An integer may be 2 greater than a multiple of 3.
It is redundant to say an integer is 3 greater than a multiple of 3 (that's just a multiple of 3, we've got it covered). Same for 4, 5, 6, 7...
Let's consider a number which is a multiple of 3. Clearly, we can write 3+3+3+3+... until we reach the number. It can be written as only 3's.
Let's consider a number which is 2 greater than a multiple of 3. If we subtract 5 from that number, it'll be a multiple of 3. That means we can write the number as 5+3+3+3+3+... Of course, the number must be at least 8.
Let's consider a number which is 1 greater than a multiple of 3. If we subtract 5 from that number, it'll be 2 greater than a multiple of 3. If we subtract another 5, it'll be a multiple of 3. That means we can write the number as 5+5+3+3+3+3+... Of course, the number must be at least 13.
That's it. We considered all the numbers. We forgot 9, 10, 11, and 12, but these are easy peasy.
Beautiful question.
Answer:
y=6x+5
Step-by-step explanation:
I am not sure if this is the answer but it's worth a try...
Answer:
QRS=38
Step-by-step explanation:
90-52=38
To be precise, the size of your sample space is <span><span>(<span>2410</span>)</span><span>(<span>2410</span>)</span></span>. This number does go on the bottom of the fraction, and what goes on top is the size of the event. Break up the event into independent events 1. choose the 2 defective bulbs, and 2. choose the remaining 8 bulbs. I don't have much choice in event 1. There is only one way to choose both of the defective balls. In other words, <span><span>(<span>22</span>)</span><span>(<span>22</span>)</span></span> (choosing 2 defective bulbs from a set of 2 defective bulbs). For event 2, there are <span><span>24−2=22</span><span>24−2=22</span></span> nondefective bulbs, and I must choose <span>88</span> of them, so that's <span><span>(<span>228</span>)</span><span>(<span>228</span>)</span></span>. Finally, since events 1 and 2 are independent, we multiply the answers for the combined event: <span><span><span>(<span>22</span>)</span><span>(<span>228</span>)</span></span><span><span>(<span>22</span>)</span><span>(<span>228</span>)</span></span></span>
<span><span>P=<span><span><span>(<span>22</span>)</span><span>(<span>228</span>)</span></span><span>(<span>2410</span>)</span></span></span><span>P=<span><span><span>(<span>22</span>)</span><span>(<span>228</span>)</span></span><span>(<span>2410</span>)</span></span></span></span>
Or, since <span><span><span>(<span>22</span>)</span>=1</span><span><span>(<span>22</span>)</span>=1</span></span>,
<span><span>P=<span><span>(<span>228</span>)</span><span>(<span>2410</span>)</span></span></span><span>P=<span><span>(<span>228</span>)</span><span>(<span>2410</span>)</span></span></span></span>
Hope this helps!
