Answer:
y^16
Explanation:
who need to add the exponents only
7 + 9 = 16
therefore, the answer is y^16
Answer:
B. The same on the moon.
Explanation:
The density of an object is the ratio of the mass contained by the object to the volume occupied by that mass.

When the object is taken from the earth to anywhere in the universe, its mass remains constant. The dimensions of the object and hence its volume also remains constant anywhere in the universe.
Therefore, the density of the object will also remain the same as it depends upon the mass and the volume of the object.
So, the correct option is:
<u>B. The same on the moon.</u>
The Sun <span>produces energy by forming "Helium" in its core by the process of "Nuclear fusion"
Hope this helps!</span>
Answer:
(a) 
(b) 
Explanation:
Represent losing with L and winning with W.
So:
--- Given

Probability of winning would be:



The question illustrates binomial probability and will be solved using the following binomial expansion;

So:
Solving (a): Winning at least 1
We look at the above and we list out the terms where the powers of W is at least 1; i.e., 1,2,3 and 4
So, we have:

Substitute value for W and L


<em>Hence, the probability of her winning at least one is 0.7599</em>
Solving (a): Wining exactly 2
We look at the above and we list out the terms where the powers of W is exactly 2
So, we have:

Substitute value for W and L


<em>Hence, the probability of her winning exactly two is 0.2646</em>