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2 years ago

Find the greatest and least number from 5,0,8,9 and 7

Mathematics

1 answer:

Dvinal [7]2 years ago

6 0

Answer:

Greatest number: 98750
Least number: 5789

Explanation:

To find the greatest number with the following values, we must arrange the numbers in descending form.

=> We can clearly tell that the numbers in descending form is 9 > 8 > 7 > 5 > 0

Hence, the greatest number with the following numbers (5,0,8,9, and 7) will be 98750.

To find the least number with the following values, we must arrange the numbers in ascending form.

=> We can clearly tell that the numbers in ascending form is 0 < 5 < 7 < 8 < 9

Hence, the least number with the following numbers (5,0,8,9, and 7) will be 5789

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BartSMP [9]

Answer:

√135

Step-by-step explanation:

3^2+b^2=12^2

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9-9+b^2=144-9

b^2=135

√135=b

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3 years ago

Give the domain and range of each relationship. Determine whether or not each relationship is a function.

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3 years ago

Clear The perimeter of a regular octagon is 88 mm. How long is each side?

Colt1911 [192]

Answer:

11 mm

Step-by-step explanation:

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3 years ago

Hey:) can you please help me posted picture of question

mr Goodwill [35]

The top row contains the product of $x^{2}$ with each term of the expressions $x^{6}+ x^{4}+ x^{2}$ and bottom row contains the product of -1 with each term of expression $x^{6}+ x^{4}+ x^{2}$ .

The final result will be the sum of terms of both rows as shown below:

$( x^{6}+ x^{4} + x^{2} )( x^{2} -1) \\ \\ 
= x^{2} ( x^{6}+ x^{4} + x^{2} )-1( x^{6}+ x^{4} + x^{2} ) \\ \\ 
= x^{8}+ x^{6}+ x^{4}- x^{6}- x^{4}- x^{2} \\ \\ 
= x^{8}- x^{2}$

So the correct answer is option D

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3 years ago

Suppose that Upper X has a discrete uniform distribution f left-parenthesis x right-parenthesis equals StartLayout left-brace1st

emmasim [6.3K]

Answer:

the probability that the sample mean is greater than 2.1 but less than 2.4 is 0.2444

Step-by-step explanation:

Given the data in the question;

x f(x) xp(x) x²p(x)

1 1/3 0.33333 0.33333

2 1/3 0.66667 1.33333

3 1/3 1.00000 3.0000

∑ 2.0000 4.6667

∑(xp(x)) = 2

∑(x²p(x)) = 4.6667

Variance σ² = ∑(x²) - ∑(x)² = 4.6667 - (2)² = 4.6667 - 4 = 0.6667

standard deviation σ = √variance = √0.6667 = 0.8165

Now since, n = 33 which is greater than 30, we can use normal approximation

for normal distribution z score ( x-μ)/σ

mean μ = 2

standard deviation = 0.817

sample size n = 33

standard of error σₓ = σ/√n = 0.817/√33 = 0.1422

so probability will be;

p( 2.1 < X < 2.4 ) = p(( 2.1-2)/0.1422) < x"-μ/σₓ < p(( 2.4-2)/0.1422)

= 0.70 < Z < 2.81

= 1 - ( 0.703 < Z < 2.812 )

FROM Z-SC0RE TABLE

= 1 - ( 0.25804 + 0.49752 )

= 1 - 0.75556

p( 2.1 < X < 2.4 ) = 0.2444

Therefore, the probability that the sample mean is greater than 2.1 but less than 2.4 is 0.2444

7 0

3 years ago

Find the greatest and least number from 5,0,8,9 and 7​

Find the greatest and least number from 5,0,8,9 and 7