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3 years ago

What is D divided by z

Mathematics

2 answers:

anastassius [24]3 years ago

4 0

Answer:

D/z

Step-by-step explanation:

Since D and z are variables, we can express the division as a fraction:

D/z

horsena [70]3 years ago

3 0

Answer:

Can you please give more context so i can help you?

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Lemons tastes sour to some people. This sourness is inherited through the TAS2R38 gene. About 65% of the population has this gen

Minchanka [31]

Answer:

We are given:

$\hat{p}=0.65$

Margin of error $E=0.02$

Confidence level = 0.99

We need to find the sample size to estimate the proportion of people who carry the TAS2R388 gene.

The formula for finding the sample size is:

$n=\hat{p} (1-\hat{p})\left (\frac{z_{\frac{0.01}{2}}}{E} \right )^{2}$

Where:

$z_{\frac{0.01}{2}}=2.576$ is the critical value at 0.01 significance level.

Therefore, the required sample size is:

$n=0.65(1-0.65) \left(\frac{2.576}{0.02} \right)^{2}$

$=0.2275 \times 16589.44$

$=3774.1 \approx 3774$

Therefore, the required sample size is 3774

3 0

3 years ago

Whats 119200/100000 in a proper fraction and is it closer to 0,1,2, or 3

sesenic [268]

119200/100000

move the decimal place to the left 5 times.

1.192 = $1 \frac{24}{125}$

It is much closer to 1

7 0

2 years ago

24 13/1,000 as a decimal number

Tems11 [23]

Answer:

24.013.

Step-by-step explanation:

Simplify: 24 13/1,000 = 24013/1000. Than just use a calculator to make it into a decimal.

4 0

3 years ago

Jack has $25 to spend at the mall. Write an inequality that expresses symbolically The amount of money, m, that jack can spend

astraxan [27]

25 >= m

m is the money that Jack can spend.
when i put ">=" i meant greater than or equal to.

3 0

2 years ago

Read 2 more answers

(D^2 +2D +1)y=e^-x log(x) solve using method variation of parameter?

Ad libitum [116K]

First you'll need the complementary solutions. The characteristic equation for this ODE is

$D^2+2D+1=(D+1)^2=0\implies D=-1$

with multiplicity 2. This means two linearly independent solutions will be $y_1=e^{-x}$ and $y_2=xe^{-x}$ .

Via variation of parameters, the particular solution will take the form

$y_p=y_1u_1+y_2u_2$

where

$u_1=-\displaystyle\int\frac{y_2e^{-x}\log x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\int\frac{y_1e^{-x}\log x}{W(y_1,y_2)}\,\mathrm dx$

The Wronskian is

$W(y_1,y_2)=\begin{vmatrix}e^{-x}&xe^{-x}\\-e^{-x}&e^{-x}(1-x)\end{vmatrix}=e^{-2x}(1-x)+xe^{-2x}=e^{-2x}$

So, you have

$u_1=-\displaystyle\int\frac{xe^{-x}e^{-x}\log x}{e^{-2x}}\,\mathrm dx$
$u_1=-\displaystyle\int x\log x\,\mathrm dx$
$u_1=\dfrac14x^2-\dfrac12x^2\log x$

and

$u_2=\displaystyle\int\frac{e^{-x}e^{-x}\log x}{e^{-2x}}\,\mathrm dx$
$u_2=\displaystyle\int\log x\,\mathrm dx$
$u_2=x(\log x-1)$

So the solution to the ODE is

$y=C_1y_1+C_2y_2+y_1u_1+y_2u_2$
$y=(C_1+C_2x)e^{-x}+\left(\dfrac14x^2-\dfrac12x^2\log x\right)e^{-x}+(\log x-1)x^2e^{-x}$
$y=(C_1+C_2x)e^{-x}+\dfrac14x^2e^{-x}(2\log x-3)$

3 0

3 years ago