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3 years ago

When is the difference of two decimals equal to a whole number?

Mathematics

1 answer:

trasher [3.6K]3 years ago

4 0

Answer:

The sum (or difference) of two decimals is a whole number when the sum (or difference) of the integers to the right of the decimal points is 0.

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Simplify $\frac{1+\sqrt{2}}{2+\sqrt{3}}$. Your solution can be converted to the form $A(1+\sqrt{B})-(\sqrt{C}+\sqrt{D})$, where

Zina [86]

Answer:

A+B+C+D = 13

Step-by-step explanation:

The given expression is:

$\dfrac{1+\sqrt{2}}{2+\sqrt{3}}$

We have to simply it and express it in the form of:

$A(1+\sqrt{B})-(\sqrt{C}+\sqrt{D})$

Multiply and divide the given expression with $2-\sqrt 3$ :

$\dfrac{1+\sqrt{2}}{2+\sqrt{3}} \times \dfrac{2-\sqrt 3}{2-\sqrt 3}\\\Rightarrow \dfrac{(1+\sqrt{2}) \times (2-\sqrt 3)}{(2+\sqrt{3})\times (2-\sqrt 3)}\\\Rightarrow \dfrac{2+2\sqrt2-\sqrt3-\sqrt6}{2^2-(\sqrt{3})^2}\\\Rightarrow \dfrac{2+2\sqrt2-\sqrt3-\sqrt6}{4-3}\\\Rightarrow \dfrac{2(1+\sqrt2)-(\sqrt3+\sqrt6)}{1}\\\Rightarrow 2(1+\sqrt2)-(\sqrt3+\sqrt6)$

It is the simplified form of given expression.

Formula used:

$(a+b)(a-b) = a^{2} -b^{2}$

Comparing the simplified expression with $A(1+\sqrt{B})-(\sqrt{C}+\sqrt{D})$

$2(1+\sqrt2)-(\sqrt3+\sqrt6)=A(1+\sqrt{B})-(\sqrt{C}+\sqrt{D})\\\Rightarrow A =2, B=2, C=3\ and\ D=6$

So, value of

$A+B+C+D = 2+2+3+6 = 13$

8 0

4 years ago

I need help with this question (image)

Alecsey [184]

Answer:

stop cheating

Step-by-step explanation:

nhn

8 0

3 years ago

Read 2 more answers

Miguel builds a model airplane with the scale 1 in. = 6 ft.

Zarrin [17]

6.5 in
not sure though

8 0

3 years ago

What is the value of x? Enter your answer in the box. x =

lutik1710 [3]

Answer:

x = 9

Step-by-step explanation:

Since you have a equilateral triangle all angles are 60°.

7x - 3 = 60

7x = 60+3

7x = 63

x = 63/7

x = 9

7 0

4 years ago

Read 2 more answers

A certain medical test is known to detect 73% of the people who are afflicted with the disease Y. If 10 people with the disease

Setler [38]

The probability of an event is the measurement of the chance of that event's occurrence. The probabilities of considered events are:

P(At least 8 have the disease) ≈ 0.4378
P(At most 4 have the disease) ≈ 0.0342

<h3 /><h3>How to find that a given condition can be modeled by binomial distribution?</h3>

Binomial distributions consist of n independent Bernoulli trials. Bernoulli trials are those trials that end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))

Suppose we have random variable X pertaining to a binomial distribution with parameters n and p, then it is written as

X = B(n,p)

The probability that out of n trials, there'd be x successes is given by

P(X=x) = $^nC_xp^x(1-p)^{n-x}$

Since 10 people can be either diseased or not and they be so independent of each other (assuming them to be selected randomly) , thus, we can take them being diseased or not as outputs of 10 independent Bernoulli trials.

Let we say

Success= Probability of a diseased person tagged as diseased by the clinic

Failure = Probability of a diseased person tagged as not diseased by the clinic.

Then,

P(Success) = p = 72% = 0.72 (of a single person)

P(Failure) = q = 1-p = 0.28

Let X be the number of people diagnosed diseased by the clinic out of 10 diseased people. Then we have: X ≈ B(n+10,P=0.73)

Calculating the needed probabilities, we get:

a) P(At leased 8 have disease) = P(X≥8) =P(X=8) + P(X=9) + P(X=10)

P(X≥8) = $^{10}C_8(0.73)^8(0.28)^2+^{10}C_9(0.73)^9(0.27)^1+^{10}C_{10}(0.73)^{10}(0.27)^0$

P(X≥8) ≈ 0.2548 + 0.1456 + 0.0374 ≈ 0.4378

b) P(At most 4 have the disease) = P(X≤4) = P(X=0) + P(X=1)+P(X=2)+P(X=3)+P(X=4)

P (X ≤ 4) =

$^{10}C_0(0.73)^0(0.27)^{10}+^{10}{C_1(0.73)^1(0.27)^9+^{10}{C_2(0.73)^2(0.27)^8+^{10}C_3(0.73)&^3(0.27)^7 \\$

$+^{10}C_4(0.73)^4(0.27)^6$

P (X ≤ 4) = 0.000003 + 0.000076+0.00088+0.00604+0.02719

P (X ≤ 4) = 0.0342

Thus,

The probabilities of considered events are:

P(At leased 8 have disease) = 0.4378 approx
P(At most 4 have the disease) = 0.0342 approx

Learn more about binomial distribution here:

brainly.com/question/13609688

#SPJ1

4 0

2 years ago