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1 year ago

Please help !!!!!!!! Now asap !!!!!

Mathematics

1 answer:

Andrej [43]1 year ago

3 0

Answer: 4.026 x 10^0, 4.026

Step-by-step explanation:

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A research study investigated differences between male and female students. Based on the study results, we can assume the popula

garri49 [273]

Using the normal distribution and the central limit theorem, it is found that the interval that contains 99.44% of the sample means for male students is (3.4, 3.6).

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.

After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

In this problem:

The mean is of $\mu = 3.5$ .

The standard deviation is of $\sigma = 0.5$ .

Sample of 100, hence $n = 100, s = \frac{0.5}{\sqrt{100}} = 0.05$

The interval that contains 95.44% of the sample means for male students is between Z = -2 and Z = 2, as the subtraction of their p-values is 0.9544, hence:

Z = -2:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$-2 = \frac{X - 3.5}{0.05}$

$X - 3.5 = -0.1$

$X = 3.4$

Z = 2:

$Z = \frac{X - \mu}{s}$

$2 = \frac{X - 3.5}{0.05}$

$X - 3.5 = 0.1$

$X = 3.6$

The interval that contains 99.44% of the sample means for male students is (3.4, 3.6).

You can learn more about the normal distribution and the central limit theorem at brainly.com/question/24663213

7 0

2 years ago

A study was done by an online retail store to determine the rate at which users used its website. A graph of the data that was c

mariarad [96]

Answer:

The correct answer is C.

Step-by-step explanation:

First, we can eliminate A and D since it is obviously only 36 months.

Note that the question asks what can be interpreted from the range of the graph.

The range is the number of users in thousands (per month) (as can be seen from the side title).

While B is technically correct, it wrongly interprets the range. C is the best choice.

The correct answer is C.

3 0

3 years ago

Max points :)) 33 please help no explanation needed..

Xelga [282]

5(2y-4) - 3y = 1. 10y-20-3y=1. 7y-20=1. 7y=21. y =3. x = 2(3)-4. x = 2. x*y =6.

3 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%20%5C%200%7D%20%5Cfrac%7B%5Csqrt%7Bcos2x%7D-%5Csqrt%5B3%5D%7Bcos3x%7D%20%7D%7

salantis [7]

Answer:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{1}{2}$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

L'Hopital's Rule

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

We are given the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}$

When we directly plug in x = 0, we see that we would have an indeterminate form:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}$

This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}$

Plugging in x = 0 again, we would get:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \frac{0}{0}$

Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)}$

Substitute in x = 0 once more:

$\displaystyle \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)} = \frac{1}{2}$