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8_murik_8 [283]

3 years ago

12

7a to the 2 power - a to the 2 power + 16 = what

1 answer:

soldi70 [24.7K]3 years ago

8 0

Answer:

48a^2 + 16

Step-by-step explanation:

7a to the 2 power - a to the 2 power + 16 = (7a)^2 - a^2 + 16.

This latter result can be rewritten as 49a^2 - a^2 + 16, or 48a^2 + 16

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Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th

grigory [225]

Answer:

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

The sample space is:

$S = \{1,1,2,2,2,3\}$

And the probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{6}$

$P(1) = \frac{1}{3}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{6}$

$P(2) = \frac{1}{2}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{6}$

P(Odd Number) is then calculated as:

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{6}$

$P(Odd\ Number) = \frac{3}{6}$

$P(Odd\ Number) = \frac{1}{2}$

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

$S = \{1,1,2,2,2,3,3\}$

The probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{7}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{7}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{7}$

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{7}$

$P(Odd\ Number) = \frac{3}{7}$

Comparing P(Odd Number) before and after

$P(Odd\ Number) = \frac{1}{2}$ --- Before

$P(Odd\ Number) = \frac{3}{7}$ --- After

<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>

7 0

2 years ago

Help me .......................................

SpyIntel [72]

Answer:

X^3 AS THE NUMERATOR 8 AS THE DENOMINATOR

Step-by-step explanation:

6 0

3 years ago

By how much is the product of 9/5 and 8 1/4 greater than 5

Oliga [24]

Given:

The product of $\dfrac{9}{5}$ and $8\dfrac{1}{4}$ greater than 5.

To find:

By how much the product of $\dfrac{9}{5}$ and $8\dfrac{1}{4}$ greater than 5.

Solution:

First we need to find the product of $\dfrac{9}{5}$ and $8\dfrac{1}{4}$ .

$\text{Product}=\dfrac{9}{5}\times 8\dfrac{1}{4}$

$\text{Product}=\dfrac{9}{5}\times \dfrac{8(4)+1}{4}$

$\text{Product}=\dfrac{9}{5}\times \dfrac{33}{4}$

$\text{Product}=14.85$

Now subtract 5 from the product.

$14.85-5=9.85$

Therefore, the product of $\dfrac{9}{5}$ and $8\dfrac{1}{4}$ is 9.85 greater than 5.

3 0

2 years ago

Find the mode in 18 7 4 19 3 18 19 12

scoundrel [369]

Hello there,

We need to remember that Mode means: the number that appears more than others.

18 will be your correct answer because the number 18 appears 2 times. And there is no other number that appears 2 times except 18.

18 is your answer.

~Jurgen

7 0

3 years ago

Please help me!! How do you factor this???<br> 16x^5+12xy-9y^5

Aleksandr [31]

First simplifying step by step.

<span>16x^5+12xy-9y^5

Answer is = </span><span><span><span>16<span>x^5</span></span>+<span><span>12x</span>y</span></span>+</span>−<span>9<span>y<span>5

It will help you.</span></span></span>

7 0

2 years ago