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2 years ago

Correctly identify the following anatomical features of the olfactory receptors.

1 answer:

7 0

Olfactory receptors are structurally defined to form columns of cells in the olfactory epithelium of the nose.

<h3>What are olfactory receptors?</h3>

Olfactory receptors are nerve cells found in the nose that receive stimuli (smells) from the surrounding environment.

Olfactory receptors are elongated-shaped cells that form columns on the olfactory epithelium in the nose.

In conclusion, olfactory receptors are structurally defined to form the olfactory epithelium in the nose.

Learn more about olfactory receptors here:

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PLSSS help!!!! What is the measure of angle x??

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The graph of the function f is the graph of the function g compressed horizontally by a factor of 2 and shifted to the right by

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Answer: A) f(x) = g(2x - 1)

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<u>Original function = g(x)</u>

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Benjamin writes an expression for the sum 1 cubed, 2 cubed, 3 cubed what is the value of this expression

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At a large corporation, the distribution of years of employment for the employees has mean 20.6 years and standard deviation 5.3

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Explanation:

The z score is a score used in statistics to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:

$z=\frac{x-\mu}{\sigma}\\\\where\ \mu=mean.\sigma=standard\ deviation, x=raw\ score.\\\\For\ a \ sample\ size\ n:\\\\z=\frac{x-\mu}{\sigma/\sqrt{n} }$

For a normal distribution with a mean of μ and standard deviation σ, For any sample size (n) drawn from this population, the sample mean is normally distributed, with mean $\mu_x=\mu$ and standard deviation $\sigma_x=\frac{\sigma}{\sqrt{n } } \\\\$ .

Hence given that a mean of μ = 20.6 years and standard deviation σ = 5.3 years, for a sample size of 100:

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3 years ago

Use series to approximate the definite integral i to within the indicated accuracy. i = 1/2 x3 arctan(x) dx 0 (four decimal plac

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The expression $\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx$ is an illustration of definite integrals

The approximated value of the definite integral is 0.0059

<h3>How to evaluate the definite integral?</h3>

The definite integral is given as:

$\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx$

For arctan(x), we have the following series equation:

$\arctan(x) = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 1}}{2n + 1}}$

Multiply both sides of the equation by x^3.

So, we have:

$x^3 * \arctan(x) = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 1}}{2n + 1}} * x^3$

Apply the law of indices

$x^3 * \arctan(x) = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 1 + 3}}{2n + 1}}$

$x^3 * \arctan(x) = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 4}}{2n + 1}}$

Evaluate the product

$x^3 \arctan(x) = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 4}}{2n + 1}}$

Introduce the integral sign to the equation

$\int\limits^{1/2}_{0} x^3 \arctan(x)\ dx =\int\limits^{1/2}_{0} \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 4}}{2n + 1}}$

Integrate the right hand side

$\int\limits^{1/2}_{0} x^3 \arctan(x)\ dx =[ \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 4}}{2n + 1}} ]\limits^{1/2}_{0}$

Expand the equation by substituting 1/2 and 0 for x

$\int\limits^{1/2}_{0} x^3 \arctan(x)\ dx =[ \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{(1/2)^{2n + 4}}{2n + 1}} ] - [ \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{0^{2n + 4}}{2n + 1}} ]$

Evaluate the power

$\int\limits^{1/2}_{0} x^3 \arctan(x)\ dx =[ \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{(1/2)^{2n + 4}}{2n + 1}} ] - 0$

$\int\limits^{1/2}_{0} x^3 \arctan(x)\ dx = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{(1/2)^{2n + 4}}{2n + 1}}$

The nth term of the series is then represented as:

$T_n = \frac{(-1)^n}{2^{2n + 5} * (2n + 4)(2n + 1)}$

Solve the series by setting n = 0, 1, 2, 3 ..........

$T_0 = \frac{(-1)^0}{2^{2(0) + 5} * (2(0) + 4)(2(0) + 1)} = \frac{1}{2^5 * 4 * 1} = 0.00625$

$T_1 = \frac{(-1)^1}{2^{2(1) + 5} * (2(1) + 4)(2(1) + 1)} = \frac{-1}{2^7 * 6 * 3} = -0.0003720238$

$T_2 = \frac{(-1)^2}{2^{2(2) + 5} * (2(2) + 4)(2(2) + 1)} = \frac{1}{2^9 * 8 * 5} = 0.00004340277$

$T_3 = \frac{(-1)^3}{2^{2(3) + 5} * (2(3) + 4)(2(3) + 1)} = \frac{-1}{2^{11} * 10 * 7} = -0.00000634131$

..............

At n = 2, we can see that the value of the series has 4 zeros before the first non-zero digit

This means that we add the terms before n = 2

This means that the value of $\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx$ to 4 decimal points is

$\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx = 0.00625 - 0.0003720238$

Evaluate the difference

$\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx = 0.0058779762$

Approximate to four decimal places

$\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx = 0.0059$

Hence, the approximated value of the definite integral is 0.0059

Read more about definite integrals at:

brainly.com/question/15127807

5 0

2 years ago