Dwayne Wade made 439 field goals out of 937 attempts. What was Dwayne Wade's ratio of successful field goals to attempt?

2 answers:

8 0

Let's set this ratio up as a fraction: (successful field goals)/(field goal attempts)
There were 439 successful field goals and 937 attempts: 439/937
The fraction cannot be reduced, so the ratio is 439/937.

Scilla [17]2 years ago

8 0

Answer:

$\frac{439}{937}$

Step-by-step explanation:

Given: Dwayne Wade made 439 field goals out of 937 attempts

To Find: Dwayne Wade's ratio of successful field goals to attempt

Solution:

Total field goals made by Dwayne Wade= $439$

Total attempt Dwayne Wade made = $937$

Now,

ratio of successful field goals to attempt made = $\frac{\text{Total successful goals fielded by Dwayne Wade}}{\text{Total attempts}}$

on putting values,

ratio of successful field goals to attempt made = $\frac{439}{937}$

As the fraction can not be simplified more,

Dwayne Wade's ratio of successful field goals to attempt is $\frac{439}{937}$

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Which of the following functions gives the length of the base edge, a(v), of a right square pyramid that is 8 inches tall as a f

Reil [10]

Answer:

$\large\boxed{a(V)=\sqrt{\dfrac{3V}{8}}}$

Step-by-step explanation:

The formula of a volume of a square pyramid:

$V=\dfrac{1}{3}a^2h$

a - base edge

h - height of a pyramid

We have H = 8in.

Substitute and solve for a:

$\dfrac{1}{3}a^2(8)=V\\\\\dfrac{8}{3}a^2=V\qquad\text{multiply both sides by}\ \dfrac{3}{8}\\\\\dfrac{3\!\!\!\!\diagup^1}{8\!\!\!\!\diagup_1}\cdot\dfrac{8\!\!\!\!\diagup^1}{3\!\!\!\!\diagup_1}a^2=\dfrac{3}{8}V\\\\a^2=\dfrac{3V}{8}\Rightarrow a=\sqrt{\dfrac{3V}{8}}$