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3 years ago

Which algebraic expression is equivalent to the expression below? 2(9x+11)+3

Mathematics

2 answers:

timama [110]3 years ago

4 0

Answer:

B: 18x + 25

Step-by-step explanation:

2(9x+11)+3

Distribute the 2

2*9x + 2*11 +3

18x +22 +3

Combine like terms

18x +25

ipn [44]3 years ago

3 0

The answer is

B: 11x +25

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Brut [27]

Answer:

I believe it is C

Step-by-step explanation:

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3 years ago

Alexander says that 3x + 4y is equivalent to (3)(4) + xy because of any order, any grouping. Is he correct why or why not

snow_lady [41]

Answer:

Alexander is incorrect because the expressions are not equivalent.

Step-by-step explanation:

If the expression is evaluated for any value of x, y; the result will not be same.

For instance, let assume x = 1 and y = 2

3x + 4y = 3 + 4 = 7

(3)(4) + xy = (3)(4) + (1 * 2) = 12 + 2 = 14

So, the expressions are not the same and Alexander is incorrect.

3 0

3 years ago

What is suface area of a cone with 16in radius and 38in hight?

tresset_1 [31]

2876.74665 in

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7 0

3 years ago

A line crosses the x-axis at -7 and the y axis at 10. Is the slope of the line positive or negative?

nordsb [41]

The line is positive

The line is positive because it slants to the right.

It wouldve been negative if it slanted to the left.

3 0

2 years ago

Read 2 more answers

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

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

Part C

$\displaystyle P(X\geq 1)=1-P(X=0)\\\\P(X\geq1)=1-(1-0.042)^{53}\\\\P(X\geq1)\approx1-0.1029\\\\P(X\geq1)\approx0.8971$

6 0

2 years ago