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guajiro [1.7K]
3 years ago
15

What is the simplified form of the following expression?

Mathematics
1 answer:
geniusboy [140]3 years ago
3 0

Answer:

242

Step-by-step explanation:

Simplify the following:

11 ((9^2 - 5^2)/2^2 + 8)

Hint: | Evaluate 2^2.

2^2 = 4:

11 ((9^2 - 5^2)/4 + 8)

Hint: | Evaluate 5^2.

5^2 = 25:

11 ((9^2 - 25)/4 + 8)

Hint: | Evaluate 9^2.

9^2 = 81:

11 ((81 - 25)/4 + 8)

Hint: | Subtract 25 from 81.

| 7 | 11

| 8 | 1

- | 2 | 5

| 5 | 6:

11 (56/4 + 8)

Hint: | Reduce 56/4 to lowest terms. Start by finding the GCD of 56 and 4.

The gcd of 56 and 4 is 4, so 56/4 = (4×14)/(4×1) = 4/4×14 = 14:

11 (14 + 8)

Hint: | Evaluate 14 + 8 using long addition.

| 1 |  

| 1 | 4

+ | | 8

| 2 | 2:

11×22

Hint: | Multiply 11 and 22 together.

| 2 | 2

× | 1 | 1

| 2 | 2

2 | 2 | 0

2 | 4 | 2:

Answer: 242

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Step-by-step explanation:

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A survey was conducted that asked 1002 people how many books they had read in the past year. results indicated that x overbar eq
Shalnov [3]
The confidence interval would be (10.44, 12.16).  This means that if we take repeated samples, the true mean lies in 90% of these intervals.

To find the confidence interval, we use:
\overline{x} \pm z*(\frac{\sigma}{\sqrt{n}})

We first find the z-value associated with this.  To do this:
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Using a z-table (http://www.z-table.com) we see that this is directly between two z-scores, 1.64 and 1.65; we will use 1.645:

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On a coordinate plane, a curved line with a minimum value of (negative 2.5, negative 12) and a maximum value of (0, negative 3)
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The points on the graph of (-4, 0), (-2.5, -12), and (0, -3), gives;

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<h3>Which method can be used to find the true statement?</h3>

From the description of the graph, we have;

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The furthest point right on the graph = (0, -3) = The maximum point

The minimum point = (-2.5, -12)

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The interval of the graph where F(x) is larger than 0 is to the left of (-4, 0), is the interval (-∞, -4)

The true statement is therefore;

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Learn more about graphs of functions here:

brainly.com/question/13936209

#SPJ1

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