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Gemiola [76]
4 years ago
7

A simple random sample is drawn from a normally distributed population. The value of which of the following will not be known fo

r certain but can be inferred?
Mathematics
2 answers:
Svetllana [295]4 years ago
6 0

Answer:

Answer is A. Not sure how to put whatever the heck that symbol is but yep.

Alex777 [14]4 years ago
3 0

The value of NOT s NOR x will not be known for certain but can be inferred. The correct answer between all the choices given is the first choice or letter A. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

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Prove that.<br><br>lim Vx (Vx+ 1 - Vx) = 1/2 X&gt;00 ​
faltersainse [42]

Answer:

The idea is to transform the expression by multiplying (\sqrt{x + 1} - \sqrt{x}) with its conjugate, (\sqrt{x + 1} + \sqrt{x}).

Step-by-step explanation:

For any real number a and b, (a + b)\, (a - b) = a^{2} - b^{2}.

The factor (\sqrt{x + 1} - \sqrt{x}) is irrational. However, when multiplied with its square root conjugate (\sqrt{x + 1} + \sqrt{x}), the product would become rational:

\begin{aligned} & (\sqrt{x + 1} - \sqrt{x}) \, (\sqrt{x + 1} + \sqrt{x}) \\ &= (\sqrt{x + 1})^{2} -(\sqrt{x})^{2} \\ &= (x + 1) - (x) = 1\end{aligned}.

The idea is to multiply \sqrt{x}\, (\sqrt{x + 1} - \sqrt{x}) by \displaystyle \frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x + 1} + \sqrt{x}} so as to make it easier to take the limit.

Since \displaystyle \frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x + 1} + \sqrt{x}} = 1, multiplying the expression by this fraction would not change the value of the original expression.

\begin{aligned} & \lim\limits_{x \to \infty} \sqrt{x} \, (\sqrt{x + 1} - \sqrt{x}) \\ &= \lim\limits_{x \to \infty} \left[\sqrt{x} \, (\sqrt{x + 1} - \sqrt{x})\cdot \frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x + 1} + \sqrt{x}}\right] \\ &= \lim\limits_{x \to \infty} \frac{\sqrt{x}\, ((x + 1) - x)}{\sqrt{x + 1} + \sqrt{x}} \\ &= \lim\limits_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x + 1}+ \sqrt{x}}\end{aligned}.

The order of x in both the numerator and the denominator are now both (1/2). Hence, dividing both the numerator and the denominator by x^{(1/2)} (same as \sqrt{x}) would ensure that all but the constant terms would approach 0 under this limit:

\begin{aligned} & \lim\limits_{x \to \infty} \sqrt{x} \, (\sqrt{x + 1} - \sqrt{x}) \\ &= \cdots\\ &= \lim\limits_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x + 1}+ \sqrt{x}} \\ &= \lim\limits_{x \to \infty} \frac{\sqrt{x} / \sqrt{x}}{(\sqrt{x + 1} / \sqrt{x}) + (\sqrt{x} / \sqrt{x})} \\ &= \lim\limits_{x \to \infty}\frac{1}{\sqrt{(x / x) + (1 / x)} + 1} \\ &= \lim\limits_{x \to \infty} \frac{1}{\sqrt{1 + (1/x)} + 1}\end{aligned}.

By continuity:

\begin{aligned} & \lim\limits_{x \to \infty} \sqrt{x} \, (\sqrt{x + 1} - \sqrt{x}) \\ &= \cdots\\ &= \lim\limits_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x + 1}+ \sqrt{x}} \\ &= \cdots \\ &= \lim\limits_{x \to \infty} \frac{1}{\sqrt{1 + (1/x)} + 1} \\ &= \frac{1}{\sqrt{1 + \lim\limits_{x \to \infty}(1/x)} + 1} \\ &= \frac{1}{1 + 1} \\ &= \frac{1}{2}\end{aligned}.

8 0
3 years ago
Read 2 more answers
Find a possible exponential function in y = a · bx (y=a(b)^x) form that represents the situation described below.
Rufina [12.5K]

Answer:

y = 4 · ½ˣ

Step-by-step explanation:

It passes through the point (0, 4), so:

4 = a · b⁰

a = 4

It passes through the point (2, 1), so:

1 = 4 · b²

1/4 = b²

b = ½

y = 4 · ½ˣ

7 0
3 years ago
Combine like terms to create an equivalent expression of 3.26d + 9.75d - 2.65
sleet_krkn [62]

Answer:

13.01 d -2.65

Step-by-step explanation:

3.26d + 9.75d - 2.65

Combine like terms

3.26d + 9.75d          - 2.65

13.01 d                     -2.65

4 0
3 years ago
Shelly used exactly
Liula [17]

Answer:

17/24

Step-by-step explanation:

First find a common denominator which is 24 then times the top number by the bottom number

3 0
4 years ago
1/8 of pizza cost 2 dollars. In an hour, 6 1/4 pizzas were sold. How much money was made at the food booth in that hour
Darina [25.2K]
First of all, you would make 6 1/4 into an improper fraction. This makes it 25/4, then you multiply it by 2 to get 50/8 and have common denominators. Divide that by 1/8, which would be 50. Then since each 1/8 costs 2 dollars you would multiply 50 by 2. So, the answer would be that they made 100 dollars.
4 0
3 years ago
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