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irinina [24]
4 years ago
5

Write as an expression and simplify:

Mathematics
1 answer:
Bad White [126]4 years ago
5 0

Answer:

(a) The product of the sum of a and b and their difference.

(a+b)\times{(a-b)}\\=a^2-b^2

(b) The tripled product of 6a and b².

3(6a\times{b^2})\\=3(6ab^2)\\=18ab^2

(c) The difference of the squares of 2m and 7n.

(2m)^2-(7n)^2\\=4m^2-49n^2

You might be interested in
Consider a uniform distribution from aequals4 to bequals29. ​(a) Find the probability that x lies between 7 and 27. ​(b) Find th
weeeeeb [17]

Answer:

a) 80% probability that x lies between 7 and 27.

b) 28% probability that x lies between 6 and 13.

c) 44% probability that x lies between 9 and 20.

d) 28% probability that x lies between 11 and 18.

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability that we find a value x between c and d, in which d is larger than c, is given by the following formula.

P(c \leq x \leq d) = \frac{d - c}{b - a}

Uniform distribution from a = 4 to b = 29

(a) Find the probability that x lies between 7 and 27.

So c = 7, d = 27

P(7 \leq x \leq 27) = \frac{27 - 7}{29 - 4} = 0.8

80% probability that x lies between 7 and 27.

​(b) Find the probability that x lies between 6 and 13. ​

So c = 6, d = 13

P(6 \leq x \leq 13) = \frac{13 - 6}{29 - 4} = 0.28

28% probability that x lies between 6 and 13.

(c) Find the probability that x lies between 9 and 20.

​So c = 9, d = 20

P(9 \leq x \leq 20) = \frac{20 - 9}{29 - 4} = 0.44

44% probability that x lies between 9 and 20.

(d) Find the probability that x lies between 11 and 18.

So c = 11, d = 18

P(11 \leq x \leq 18) = \frac{18 - 11}{29 - 4} = 0.28

28% probability that x lies between 11 and 18.

3 0
3 years ago
2067 Supp Q.No. 2a Find the sum of all the natural numbers between 1 and 100 which are divisible by 5. Ans: 1050 ​
Alborosie

5

Answer:

1050

Step-by-step explanation:

Natural Numbers are positive whole numbers. They aren't negative, decimals, fractions. We can just divide 5 into 100 to find how many natural numbers go up to 100 and just add them but that is just to much.

There is a easier method.

<em>E.g</em><em>:</em><em> </em><em> </em><em>Natural</em><em> </em><em>N</em><em>umbers</em><em> </em><em>that</em><em> </em><em>are</em><em> </em><em>divisible</em><em> </em><em>by</em><em> </em><em>a</em><em> </em><em>N</em><em>t</em><em>h</em><em> </em><em>Number</em><em>.</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>same</em><em> </em><em>as</em><em> </em><em>adding</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>Nth</em><em> </em><em>Numbers</em><em> </em><em> </em><em>to a</em><em> </em><em>multiple</em><em> </em><em>of</em><em> </em><em>that</em><em> </em><em>Nth</em><em> </em><em>Term</em><em>.</em><em> </em><em>For</em><em> </em><em>example</em><em>,</em><em> </em><em>let</em><em> </em><em>say</em><em> </em><em>we</em><em> </em><em>need</em><em> </em><em>to</em><em> </em><em>find</em><em> </em><em>numbers</em><em> </em><em>divisible</em><em> </em><em>by</em><em> </em><em>2</em><em>.</em><em> </em><em>We</em><em> </em><em>know</em><em> </em><em>that</em><em> </em><em>4</em><em> </em><em>is</em><em> </em><em>divisible</em><em> </em><em>by</em><em> </em><em>2</em><em> </em><em>because</em><em> </em><em>4</em><em>/</em><em>2</em><em>=</em><em>2</em><em>.</em><em> </em><em> </em><em>We</em><em> </em><em>can</em><em> </em><em>add</em><em> </em><em>the</em><em> </em><em>Nth</em><em> </em><em>numbers</em><em> </em><em>which</em><em> </em><em>is</em><em> </em><em>2</em><em> </em><em>to</em><em> </em><em>4</em><em>.</em><em> </em><em>4</em><em>+</em><em>2</em><em>=</em><em>6</em><em>.</em><em> </em><em>And</em><em> </em><em>6</em><em> </em><em>is</em><em> </em><em>divisible</em><em> </em><em>by</em><em> </em><em>2</em><em> </em><em>because</em><em> </em><em>6</em><em>/</em><em>2</em><em>=</em><em>3</em><em>.</em><em> </em><em>We</em><em> </em><em>can</em><em> </em><em>call</em><em> </em><em>this</em><em> </em><em>a</em><em> </em><em>arithmetic</em><em> </em><em>series</em><em>.</em><em> </em><em>A</em><em> </em><em>series</em><em> </em><em>which</em><em> </em><em>has</em><em> </em><em>a</em><em> </em><em>pattern</em><em> </em><em>of</em><em> </em><em>adding</em><em> </em><em>a</em><em> </em><em>common</em><em> </em><em>difference</em>

<em>Back</em><em> </em><em>to</em><em> </em><em>the</em><em> </em><em>problem</em><em>,</em><em> </em><em>we</em><em> </em><em>can</em><em> </em><em>use</em><em> </em><em>the</em><em> </em><em>sum</em><em> </em><em>of</em><em> </em><em>arithmetic</em><em> </em><em>series</em><em> </em><em>formula</em><em>,</em>

<em>y = x( \frac{z {}^{1}  +  {z}^{n} }{2} )</em>

<em>Where</em><em> </em><em>x</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>number</em><em> </em><em>of</em><em> </em><em>terms</em><em> </em><em>in</em><em> </em><em> </em><em>our</em><em> </em><em>sequence</em><em>.</em><em> </em><em>Z1</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>fist</em><em> </em><em>term</em><em> </em><em>of</em><em> </em><em>our</em><em> </em><em>series</em><em>.</em><em> </em><em> </em><em>ZN</em><em> </em><em>is</em><em> </em><em>our</em><em> </em><em>last</em><em> </em><em>term</em><em>.</em><em> </em><em>And</em><em> </em><em>y</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>sum</em><em> </em><em>of</em><em> </em><em>all</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>terms</em><em> </em>

<em>The</em><em> </em><em>first</em><em> </em><em>term</em><em> </em><em>is</em><em> </em><em>5</em><em>,</em><em> </em><em>the</em><em> </em><em>numbers</em><em> </em><em>of</em><em> </em><em>terms</em><em> </em><em>being</em><em> </em><em>added</em><em> </em><em>is</em><em> </em><em>2</em><em>0</em><em> </em><em>because</em><em> </em><em>1</em><em>0</em><em>0</em><em>/</em><em>5</em><em>=</em><em>2</em><em>0</em><em>.</em><em> </em><em>The</em><em> </em><em>last</em><em> </em><em>term</em><em> </em><em>is</em><em> </em><em>1</em><em>0</em><em>0</em><em>.</em>

<em>y = 20( \frac{5 + 100}{2} )</em>

<em>y = 20( \frac{105}{2} )</em>

<em>y = 1050</em>

5 0
3 years ago
Write a recursive formula for each sequence given or described below.
valentinak56 [21]

Answer:

a_{n} =a_{n-1} +3000 and a₁ =30000 for n = 2,3,4,5,6, ......

Step-by-step explanation:

Doug has joined a job with a starting salary of $30000 per year.  

Hence, if a₁ is the salary of Doug in the first year, then  

a₁ =30000

Now, each year Doug will receive a raise of $3000 in his salary.

Hence, in the 2nd year, his salary(a₂) will become ( a₁ +3000) per year.

Again, in the 3rd year, his salary(a₃) will become ( a₂ +3000) per year.

Therefor, in the similar manner the recursive formula for his salary in each year will be given as a_{n} =a_{n-1} +3000 and a₁ =30000 for n = 2,3,4,5,6, ...... {aₙ is the yearly salary of Doug in the nth year} (Answer)

7 0
3 years ago
Consider to circular swimming pools.Pool A has a radius of 12 feet and Pool B has a diameter of 7.5 meters which pool has a grea
mamaluj [8]

C = Pi * d  where C = circumference, d = diameter, r = radius

C = 2*pi *r

we need to watch the units since they are different

Pool A

C = 2 * pi * r

C = 2 * pi * 12  = 24 * pi  in feet


Pool B

C = pi * d

change meters to feet

7.5 m * 3.28 ft/ 1 m  = 24.6 ft

C = pi * 24.6  = 24.6* pi  in ft


Pool B had a greater circumference

24.6 * pi > 24 pi


6 0
3 years ago
Pls help me with this question ASAP
WINSTONCH [101]

Answer:

8

Step-by-step explanation:

(3+5)^2/8=(8)^2/8=64/8=8

5 0
3 years ago
Read 2 more answers
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