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3 years ago

What is A decreased by 20% and then increased by 20% of that? Please show your work.

Mathematics

1 answer:

Eduardwww [97]3 years ago

3 0

Answer:

A - (20% of A) = 80% of A + (20% of (80% of A) = 96% of A

So if A is 100:

100 - (.2 * 100) = 80 + (.2 * 80) = 96

Step-by-step explanation:

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If i have 480 grams of pasta for 4 people how many kilograms will i need for 14 people

Afina-wow [57]

$\bf \begin{array}{ccll} \stackrel{pasta}{grams}&people\\ \text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\ 480&4\\ g&14 \end{array}\implies \cfrac{480}{g}=\cfrac{4}{14}\implies \cfrac{480}{g}=\cfrac{2}{7} \\\\\\ \cfrac{480\cdot 7}{2}=g\implies 1680=g$

now, there are 1000 grams in one Kilogram, so in 1680 grams, there are 1680/1000 Kilograms, namely 1.680.

7 0

3 years ago

Out of 450 applicants for a job, 206 are male and 62 are male and have a graduate degree.

xxMikexx [17]

Answer:

0.3009 is the probability that the applicant has graduate degree given he is a male.

Step-by-step explanation:

We are given he following in the question:

M: Applicant is male.

G: Applicant have a graduate degree

Total number of applicants = 450

Number of male applicants = 206

$n(M) = 206$

Number of applicants that are male and have a graduate degree = 62

$n(M\cap G) = 62$

$\text{Probability} = \displaystyle\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

$P(M) = \dfrac{206}{450} = 0.4578$

$P(M\cap G) = \dfrac{n(M\cap G)}{n} = \dfrac{62}{450} = 0.1378$

We have to find the probability that the applicant has graduate degree given he is a male.

$P(G|M) = \dfrac{P(G\cap M)}{P(M)} = \dfrac{\frac{62}{450}}{\frac{206}{450}} = \dfrac{62}{206} = 0.3009$

Thus, 0.3009 is the probability that the applicant has graduate degree given he is a male.

5 0

3 years ago

PLSS HELPPPP MEE ITS DUE TMRW

jolli1 [7]

Answer:

the turning point is (-2,-1)

i dont know the turning point but i hope this helps

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

A sphere of radius r is cut by a plane h units above the equator, where

Anika [276]

Consider the top half of a sphere centered at the origin with radius $r$ , which can be described by the equation

$z=\sqrt{r^2-x^2-y^2}$

and consider a plane

$z=h$

with $0$ . Call the region between the two surfaces $R$ . The volume of $R$ is given by the triple integral

$\displaystyle\iiint_R\mathrm dV=\int_{-\sqrt{r^2-h^2}}^{\sqrt{r^2-h^2}}\int_{-\sqrt{r^2-h^2-x^2}}^{\sqrt{r^2-h^2-x^2}}\int_h^{\sqrt{r^2-x^2-y^2}}\mathrm dz\,\mathrm dy\,\mathrm dx$

Converting to polar coordinates will help make this computation easier. Set

$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\var\phi\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

Now, the volume can be computed with the integral

$\displaystyle\iiint_R\mathrm dV=\int_0^{2\pi}\int_0^{\arctan\frac{\sqrt{r^2-h^2}}h}\int_{h\sec\varphi}^r\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta$

You should get

$\dfrac{2\pi}3\left(r^3\arctan\dfrac{\sqrt{r^2-h^2}}h-\dfrac{h^3}2\left(\dfrac{r\sqrt{r^2-h^2}}{h^2}+\ln\dfrac{r+\sqrt{r^2-h^2}}h\right)\right)$