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2 years ago

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There are 1500 pupils in a school. 23% of the pupils in the school are Malays. 68% of the pupils are Chinese. The rest of the pu

pils are indians. how many more Chinese pupils than Malay pupils are there?

1 answer:

sasho [114]2 years ago

5 0

Answer:

there is 77% more chinese students.

Step-by-step explanation:

subtract the 23% of malay students from the 100% of the students in the school.

100-23=77

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A glacier in Alaska moves about 29.2 meters a day.

Leto [7]

Answer:

29,100

Step-by-step explanation:

29.1 × 10 = 291

29.1 × 100 = 2,910

29.1 × 1000 = 29,100

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2 years ago

Aaron made a $400 purchase using his credit card. The credit card has a simple interest rate of 12% each year. How much will Aar

melisa1 [442]

Answer: 448 dollars

=============================================

Work Shown:

P = 400 is the balance

r = 0.12 is the decimal form of the interest rate

t = 1 is the number of years

Use those values to get the simple interest to be

i = P*r*t

i = 400*0.12*1

i = 48

The amount of interest charged is $48

Add this onto the balance of 400 dollars and the total amount Aaron owes is P+i = 400+48 = 448 dollars.

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2 years ago

Evaluate 7^2 + (4x5) - 2^3

Phoenix [80]

So it would be 49+(20)+8

= 77

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3 years ago

Write the product form of 3 a² b ³<br>

kifflom [539]

Answer:

a³-b³= (a-b)(a²+ab+ b²)

Step-by-step explanation:

3 0

2 years ago

A professor pays 25 cents for each blackboard error made in lecture to the student who pointsout the error. In a career ofnyears

marta [7]

Answer:

(a) The probability that <em>Y</em>₂₀ exceeds 1000 is 3.91 × 10⁻⁶.

(b) <em>n</em> = 28.09

Step-by-step explanation:

The random variable <em>Y</em>ₙ is defined as the total numbers of dollars paid in <em>n</em> years.

It is provided that <em>Y</em>ₙ can be approximated by a Gaussian distribution, also known as Normal distribution.

The mean and standard deviation of <em>Y</em>ₙ are:

$\mu_{Y_{n}}=40n\\\sigma_{Y_{n}}=\sqrt{100n}$

(a)

For <em>n</em> = 20 the mean and standard deviation of <em>Y</em>₂₀ are:

$\mu_{Y_{n}}=40n=40\times20=800\\\sigma_{Y_{n}}=\sqrt{100n}=\sqrt{100\times20}=44.72\\$

Compute the probability that <em>Y</em>₂₀ exceeds 1000 as follows:

$P(Y_{n}>1000)=P(\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}>\frac{1000-800}{44.72})\\=P(Z> 4.47)\\=1-P(Z$

**Use a <em>z </em>table for probability.

Thus, the probability that <em>Y</em>₂₀ exceeds 1000 is 3.91 × 10⁻⁶.

(b)

It is provided that P (<em>Y</em>ₙ > 1000) > 0.99.

$P(Y_{n}>1000)=0.99\\1-P(Y_{n}$

The value of <em>z</em> for which P (Z < z) = 0.01 is 2.33.

Compute the value of <em>n</em> as follows:

$z=\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}\\2.33=\frac{1000-40n}{\sqrt{100n}}\\2.33=\frac{100}{\sqrt{n}}-4\sqrt{n} \\2.33=\frac{100-4n}{\sqrt{n}} \\5.4289=\frac{(100-4n)^{2}}{n}\\5.4289=\frac{10000+16n^{2}-800n}{n}\\5.4289n=10000+16n^{2}-800n\\16n^{2}-805.4289n+10000=0$

The last equation is a quadratic equation.

The roots of a quadratic equation are:

$n=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

a = 16

b = -805.4289

c = 10000

On solving the last equation the value of <em>n</em> = 28.09.

8 0

3 years ago