Answer:
You have found 24 nickels.
Step-by-step explanation:
Let N be the number of nickles, D be the number of dimes and Q be the number of quarters. The number of coins and their face values are given by the following expressions, respectively:
As of now we have three variables and two expressions, the final expression needed to solve the linear system is given by the amount offered by the coin dealer:
Solving the linear system:
You have found 24 nickels.
The correct answer is letter D
Just believe me thats the correct answer
Let's simplify step-by-step.<span><span><span><span>52x</span>+<span><span>43<span>y92</span></span>x</span></span>+<span><span>3<span>y15</span></span>y</span></span>+<span><span><span><span><span><span><span><span><span>8<span>x6</span></span>x</span><span>y23</span></span>x</span><span>y6</span></span>x</span><span>y32</span></span>x</span>y
</span></span><span>=<span><span><span><span>52x</span>+<span><span>43x</span><span>y92</span></span></span>+<span>3<span>y16</span></span></span>+<span><span>8<span>x10</span></span><span>y62
</span></span></span></span>Answer:<span>=<span><span><span><span><span>43x</span><span>y92</span></span>+<span><span>8<span>x10</span></span><span>y62</span></span></span>+<span>3<span>y16</span></span></span>+<span>52<span>x</span></span></span></span>
Let x be a random variable representing the price of a Congo-imported black diamond. Let the higher price be p. Then,
P(x < p) = P(x < (p - mean)/sd) = P(x < (p - 60,430)/21,958.08) = P(z < 2)
Therefore,
(p - 60,430)/21,958.08 = 2
p - 60,430 = 2 x 21,958.08 = 43,916.16
p = 34,916.16 + 60,430 = 104.346.16
Therefore, The required price is $104,346.16
Note that rotation will not affect the shape and size of an object.
Rotation with respect to a point preserves the corresponding sides and the corresponding angles of the original image.
Hence, the statements
The corresponding angle measurements in each triangle between the pre-image and the image are preserved and
The corresponding lengths, from the point of rotation, between the pre-image and the image are preserved
are true.
