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4 years ago

Consider the quadratic function f(x) = x2 – 8x – 4. What is the value of the leading coefficient?

Mathematics

1 answer:

pantera1 [17]4 years ago

7 0

Answer:

x = 4 + 2 sqrt(5) or x = 4 - 2 sqrt(5)

thus :

Step-by-step explanation:

Solve for x over the real numbers:

x^2 - 8 x - 4 = 0

Hint: | Solve the quadratic equation by completing the square.

Add 4 to both sides:

x^2 - 8 x = 4

Hint: | Take one half of the coefficient of x and square it, then add it to both sides.

Add 16 to both sides:

x^2 - 8 x + 16 = 20

Hint: | Factor the left hand side.

Write the left hand side as a square:

(x - 4)^2 = 20

Hint: | Eliminate the exponent on the left hand side.

Take the square root of both sides:

x - 4 = 2 sqrt(5) or x - 4 = -2 sqrt(5)

Hint: | Look at the first equation: Solve for x.

Add 4 to both sides:

x = 4 + 2 sqrt(5) or x - 4 = -2 sqrt(5)

Hint: | Look at the second equation: Solve for x.

Add 4 to both sides:

Answer: x = 4 + 2 sqrt(5) or x = 4 - 2 sqrt(5)

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3 years ago

A student records the number of hours that they have studied each of the last 23 days. They compute a sample mean of 2.3 hours a

natita [175]

Answer:

the standard deviation increases

Step-by-step explanation:

Let x₁ , x₂, . . . , x₂₃ be the actual data observed by the student

The sample means = x₁ + x₂ + . . . , x₂₃ / 23

$= \frac{x_1 +x_2 +...x_2_3}{23}$

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⇒ $\sum xi =2.3 \times 23 = 52.9hrs$

let x₁ , x₂, . . . , x₂₃ arranged in ascending order

Then x₂₃ was 10 and has been changed to 14

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Sample mean $= \frac{x_1 +x_2 +...x_2_3}{23}$

$\frac{52.9hrs + 4}{23} \\\\= \frac{56.9}{23} \\\\= 2.47$

therefore, the new sample mean is 2.47

2) For the old data set

the median is $x_1_2(th)$ values

$[\frac{n +1}{2} ]^t^h value$

when we use the new data set only x₂₃ is changed to 14

i.e the rest all observation remain unchanged

Hence, sample median = $[{x_1_2]^t^h value$ remain unchange

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The Standard deviation of old data set is calculated

$=\sqrt{\frac{1}{n-1} \sum (xi - \bar x_{old})^2 } \\\\=\sqrt{\frac{1}{22}\sum ( xi - 2.3)^2 }---(1)$

The new sample standard sample deviation is calculated as

$= \sqrt{\frac{1}{n-1} \sum (xi-2.47)^2} ---(2)$

Now, when we compare (1) and (2) the square distance between each observation xi and old mean is less than the squared distance between each observation xi and the new mean.

Since,

(xi - 2.3)² ∑ (xi - 2.47)²

Therefore , the standard deviation increases

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Answer:

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