What is the value of x2 + (2y) ÷ (2w) + 3z for w = 2, x = 5, y = 8, and z = 3? Enter your Your answer in the box.

Mathematics

2 answers:

Tatiana [17]3 years ago

6 0

Answer:

The answer is 38 if x is being squared. If x is being multiplied by 2 then the answer is 23

Step-by-step explanation:

attashe74 [19]3 years ago

4 0

Answer:

Step-by-step explanation:

To find the value of x² + (2y) ÷(2w) +3z , first we substitute the values of w, x, y and z

5² + 2(8) ÷ 2(2) + 3(3)

To simplify this further, we have to apply the rule of BODMAS

(Applying the rule of BODMAS simply means when you have an equation, if its having bracket, you have to remove the bracket first, then you move to powers then you proceed to dividing then subtraction and then addition)

5² + 2(8) ÷ 2(2) + 3(3)

=25 +16 ÷ 4 +9

=25 + 4 +9

=38

Therefore the value of x² + (2y) ÷(2w) +3z is 38

You might be interested in

The probability that a male will be colorblind is .042. Find the probabilities that in a group of 53 men, the following are true

Dvinal [7]

Answer:

See below for answers and explanations

Step-by-step explanation:

Part A

$\displaystyle P(X=x)=\binom{n}{x}p^xq^{n-x}\\\\P(X=5)=\binom{53}{5}(0.042)^5(1-0.042)^{53-5}\\\\P(X=5)=\frac{53!}{(53-5)!*5!}(0.042)^5(0.958)^{48}\\\\P(X=5)\approx0.0478$

Part B

$P(X\leq5)=P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)\\\\P(X\leq5)=\binom{53}{1}(0.042)^1(1-0.042)^{53-1}+\binom{53}{2}(0.042)^2(1-0.042)^{53-2}+\binom{53}{3}(0.042)^3(1-0.042)^{53-3}+\binom{53}{4}(0.042)^4(1-0.042)^{53-4}+\binom{53}{1}(0.042)^5(1-0.042)^{53-5}\\\\P(X\leq5)\approx0.9767$