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3 years ago

Is (ab^2) equivalent to ab^2? Explain.

Mathematics

2 answers:

olga2289 [7]3 years ago

7 0

Answer:

yes

Step-by-step explanation:

ok lets substitute a for 6 and b for 7

it will look like this ((6)(7))^2

now lets solve

6x7=42

42^2=1764

ok now for the second one (6)(7)^2

now lets solve

6x7=42

42^2=1764

i hope this helps

Zarrin [17]3 years ago

6 0

Answer: yes it is equivalent!

Step-by-step explanation:

all you do is take away the parentheses and then it is the same exact thing!

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Write 16; base 2 as a power of the given base

mixas84 [53]

2to the power of 16 is is 32 because basically you are multiplying 16 x2

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3 years ago

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netineya [11]

Answer:

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Step-by-step explanation:

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4 years ago

2 and 1/3 devided by 2/3

goldfiish [28.3K]

Answer:

3/5 , 7/2 , 3 1/2

Step-by-step explanation:

Two Plus One - Third

2 + $\frac{1}{3}$

Make Both Fractions-

$\frac{2}{1}$ + $\frac{1}{3}$

Make Same Denominator ( Bottom # Of Fraction )-

$\frac{2}{1}$ * $\frac{3}{3}$ = $\frac{6}{3}$

New Equation-

$\frac{6}{3}$ + $\frac{1}{3}$ = $\frac{7}{3}$

Seven - Thirds Divided By Two-Thirds.

$\frac{7}{3} / \frac{2}{3}$

Divide In Parts-

7 / 2 = 3.5

<h2><em>Final Answer of '2 and 1/3 divided by 2/3 is 3.5, 7/2, 3 1/2! </em><u><em>Don't Forget To Rate Brainiest!</em></u></h2><h2 /><h2><em> </em><em>-Aslina</em></h2>

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3 years ago

Customers are used to evaluate preliminary product designs. In the past, 95% of highly successful products received good reviews

Andrej [43]

Answer: The required probability is 0.615.

Step-by-step explanation:

Let A be the event that a product receives good reviews.

Let E₁ be the event that the product is highly successful.

Let E₂ be the event that the product is moderately successful.

Let E₃ be the event that the product is poor.

$P(E_1)=0.40\\\\P(E_2)=0.35\\\\P(E_3)=0.25\\\\P(A|E_1)=0.95\\\\P(A|E_2)=0.60\\\\P(A|E_3)=0.10$

(a) What is the probability that a product attains a good review.

So, it becomes,

$P(A)=P(E_1)P(A|E_1)+P(E_2)P(A|E_2)+P(E_3)P(A|E_3)\\\\P(A)=0.4\times 0.95+0.0.35\times 0.6+0.25\times 0.1\\\\P(A)=0.615$

Hence, the required probability is 0.615.

4 0

3 years ago