Step-by-step explanation:
The gender of a child which is either a boy or a girl is determined by the XX-chromosomes, or XY-chromosomes.
Since the couple plan to have 5 children, the chance of a child being a boy is equal to the chance of it being a girl - the chances are 50/50.
What we do to achieve our aim is to run a simulation that would add an X or Y to an X for all 5 children.
Doing this 125 times, we obtain the number of trials we desire.
For each trial, we get for each child, C:
C1: X + (X or Y)
C2: X + (X or Y)
C3: X + (X or Y)
C4: X + (X or Y)
C5: X + (X or Y)
Since the chance of having an X is equal to the chance of having a Y, they equal probability, which is 0.5 for each.
<span>1. multiply to -18
add to -17 = Multiplying -18 and 1 will give -18 as the result and then on
adding -18 and + 1 the result comes to -17
</span><span>2. multiply to 36 add
to -13 = Multiplying -9 and -4 will give -36 as the result and then on
adding -9 and -4 the result comes to -13
</span><span>3. multiply to -24 add to -5<span>= </span></span><span>Multiplying -8 and +3 will give -24 as the result and then on
adding -8 and +3 the result comes to -5
4. multiply to -18 add to 7= </span><span>Multiplying +9 and -2 will give -18 as the result and then on
adding +9 and -2 the result comes to 7
5. multiply to -36 add to 9= </span><span><span>Multiplying +12 and -3 will give -36 as the result and then on
adding +12 and -3 the result comes to 9
</span> 6. multiply to 24 add to 10= </span><span><span>Multiplying +6 and +4 will give 24 as the result and then on
adding +6 and +4 the result comes to 10
</span> 7. multiply to 18 add to -9= </span><span><span><span>Multiplying -6 and -3 will give -18 as the result and then on
adding -6 and -3 the result comes to -9
</span> 8. multiply to -36 add to -16</span>= </span><span>Multiplying -18 and +2 will give -36 as the result and then on
adding -18 and +2 the result comes to -16</span>
I think it’s 0.85d, but i’m not sure
Answer:
Solution : 8i
Step-by-step explanation:
We can use the trivial identities cos(π / 2) = 0, and sin(π / 2) = 1 to solve this problem. Let's substitute,
![8\left[cos\left(\frac{\pi }{2}\right)+isin\left(\frac{\pi \:}{2}\right)\right]](https://tex.z-dn.net/?f=8%5Cleft%5Bcos%5Cleft%28%5Cfrac%7B%5Cpi%20%7D%7B2%7D%5Cright%29%2Bisin%5Cleft%28%5Cfrac%7B%5Cpi%20%5C%3A%7D%7B2%7D%5Cright%29%5Cright%5D)
= 
And of course 1i = i, so we have the expression 8(0 + i ). Distributing the " 8, " 8( 0 ) = 0, and 8(i) = 8i, making the fourth answer the correct solution.
Which set of numbers correctly shows all of the factors of 28? 1, 2, 4, 7, 14, 28 1, 2, 7, 14, 28 0, 1, 7, 14, 28 1, 2, 4, 7, 8,
slega [8]
Answer:
1, 2, 4, 7, 14, 28
Step-by-step explanation:
Factors of 28 are the numbers that divide into 28 evenly.
1, 2, 4, 7, 14, 28
