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Svetradugi [14.3K]

2 years ago

12

The monthly salary of a married civil servant is RS 25700 and 10% of his salary is deducted as his provident fund .How much inco

me tax should he pay in a year when the tax rate is 1% .

1 answer:

OverLord2011 [107]2 years ago

7 0

Step-by-step explanation:

you deduct the 10 percent provident funds first .

Then u calculate for the 1 percent of d deduction .

ur answer should be 27756

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Can someone please help!!!!!

Gekata [30.6K]

Option two N is greater than three

6 0

2 years ago

Use the confidence level and sample data to find a confidence interval for estimating the population mu. Round your answer to th

inysia [295]

Answer:

c. 9.5 lb < mu < 11.1 lb.

Step-by-step explanation:

Confidence interval can be stated as M±ME where

M is the sample mean (10.3)
ME is the margin of error

margin of error (ME) around the mean can be calculated using the formula

ME= $\frac{z*s}{\sqrt{N} }$ where

z is the corresponding statistic in 95% confidence level (1.96)

s is the standard deviation of the sample (2.4)

N is the sample size (37)

Putting thesenumbers in the formula we get:

ME= $\frac{1.96*2.4}{\sqrt{37} }$ ≈ 0.7733 ≈ 0.8

Then the 95% confidence interval would be 10.3 ± 0.8

7 0

3 years ago

What is a composite number that is less than 5?

babunello [35]

I believe 4, because two times two equals four

4 0

3 years ago

Read 2 more answers

Compute the surface area of the portion of the sphere with center the origin and radius 4 that lies inside the cylinder x^2+y^2=

Tom [10]

Answer:

16π

Step-by-step explanation:

Given that:

The sphere of the radius = $x^2 + y^2 +z^2 = 4^2$

$z^2 = 16 -x^2 -y^2$

$z = \sqrt{16-x^2-y^2}$

The partial derivatives of $Z_x = \dfrac{-2x}{2 \sqrt{16-x^2 -y^2}}$

$Z_x = \dfrac{-x}{\sqrt{16-x^2 -y^2}}$

Similarly;

$Z_y = \dfrac{-y}{\sqrt{16-x^2 -y^2}}$

∴

$dS = \sqrt{1 + Z_x^2 +Z_y^2} \ \ . dA$

$=\sqrt{1 + \dfrac{x^2}{16-x^2-y^2} + \dfrac{y^2}{16-x^2-y^2} }\ \ .dA$

$=\sqrt{ \dfrac{16}{16-x^2-y^2} }\ \ .dA$

$=\dfrac{4}{\sqrt{ 16-x^2-y^2} }\ \ .dA$

Now; the region R = x² + y² = 12

Let;

x = rcosθ = x; x varies from 0 to 2π

y = rsinθ = y; y varies from 0 to $\sqrt{12}$

dA = rdrdθ

∴

The surface area $S = \int \limits _R \int \ dS$

$= \int \limits _R\int \ \dfrac{4}{\sqrt{ 16-x^2 -y^2} } \ dA$

$= \int \limits^{2 \pi}_{0} } \int^{\sqrt{12}}_{0} \dfrac{4}{\sqrt{16-r^2}} \ \ rdrd \theta$

$= 2 \pi \int^{\sqrt{12}}_{0} \ \dfrac{4r}{\sqrt{16-r^2}}\ dr$

$= 2 \pi \times 4 \Bigg [ \dfrac{\sqrt{16-r^2}}{\dfrac{1}{2}(-2)} \Bigg]^{\sqrt{12}}_{0}$

$= 8\pi ( - \sqrt{4} + \sqrt{16})$

= 8π ( -2 + 4)

= 8π(2)

= 16π

4 0

2 years ago

Not sure about this one. Just double checking.<br><br> Simplify.<br><br> −4(3−1)+2

ANEK [815]

<span>−4(3−1)+2
=</span><span>−4(2)+2
= -8 + 2
= -6</span>

4 0

2 years ago

Read 2 more answers