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2 years ago

What is the smallest positive integer n such that n! ends in at least 2019 zeros?

Mathematics

1 answer:

Lady_Fox [76]2 years ago

3 0

Answer:

I think there's 502

Step-by-step explanation:

if this is a olunomical question that should be it I believe, because you have to write in standered form then when it equals zero, then factor it

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Which ordered pair in the form (X,Y) solution to this equation (x+y)y=10

Verizon [17]

Answer:

X=3

Y=2

Step-by-step explanation:

(3+2)2=10

If you use this, please give me brainliest

3 0

4 years ago

Find the difference -10-(-7)=

Lelu [443]

Answer:

-3

Step-by-step explanation:

Since the - in -7 and the subtraction sign - are the same, you have to group them so now it is + (positive). -10 + 7 = -3

4 0

3 years ago

The number of vibrations n n per second of a nylon guitar string varies directly with the square root of the tension T T and inv

emmasim [6.3K]

Answer: $T=40.96\ KgF$

Step-by-step explanation:

We know that:

$n$ : The number of vibrations per second of the nylon guitar string.

$T$ : Tension.

$L$ : The length of the string.

Since $n$ varies directly with the square root of the $T$ and inversely with $L$ , the equation has the following form:

$n=k*\frac{\sqrt{T} }{L}$

Where "k" is the constant of variation.

Knowing that when $n=15$ and $L=0.6$ , $T=256$ , we can find the value of "k":

$n=k*\frac{\sqrt{T} }{L}\\\\L*n=k\sqrt{T}\\\\\frac{L*n}{\sqrt{T}}=k\\\\k=\frac{(0.6)(15)}{\sqrt{256}}\\\\k=0.5625$

Finally, in order to find the tension when the length is 0.3 meters and the number of vibrations is 12, you need to substitute these values and the value of "k" into $n=k*\frac{\sqrt{T} }{L}$ and solve for $T$ :

$12=(0.5625)\frac{\sqrt{T} }{0.3}\\\\\frac{12(0.3)}{(0.5625)}=\sqrt{T}\\\\(6.4)^2=T\\\\T=40.96\ KgF$

8 0

3 years ago

A set of 5 numbers have a mode of 5, median of 6 and a mean of 7. Find the five numbers that meet these conditions.

Margaret [11]

Answer:

1. Mean = 4.57

2. Median = 5

3. Mode = 5

4. Range = 7

Five Number Summary

5. Minimum = 1

6. First Quartile = 4

7. Median = 5

8. Third Quartile = 6

9. Maximum = 8

Step-by-step explanation:

Let's arrange the numbers in order (smallest to largest):

1,2,2,4,4,4,5,5,5,5,6,6,7,8

We need to find:

1. mean

2. median

3. mode

4. range

Five Number Summary

5. Minimum

6. First Quartile

7. Median

8. Third Quartile

9. Maximum1. Mean = 4.57

2. Median = 5

3. Mode = 5

4. Range = 7

Five Number Summary

5. Minimum = 1

6. First Quartile = 4

7. Median = 5

8. Third Quartile = 6

9. Maximum = 8

Step-by-step explanation:

Let's arrange the numbers in order (smallest to largest):

1,2,2,4,4,4,5,5,5,5,6,6,7,8

We need to find:

1. mean

2. median

3. mode

4. range

Five Number Summary

5. Minimum

6. First Quartile

7. Median

8. Third Quartile

9. Maximum

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of

Aleks [24]

Answer:

CI for 90% = ( 16.34, 18.66)

Therefore at 90% confidence interval (a,b) = ( 16.34, 18.66)

And,

CI for 95% = ( 16.11, 18.89)

Therefore at 95% confidence interval (a,b) = ( 16.11, 18.89)

Step-by-step explanation:

Answer: = ( 2.64, 3.14)

Therefore at 95% confidence interval (a,b) = ( 2.64, 3.14)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = 17.5

Standard deviation r = 5.7

Number of samples n = 65

Confidence interval = 90% and 95%

z(at 90% confidence) = 1.645

z(at 95% confidence) = 1.96

Substituting the values we have; for 90%

17.5+/-1.645(5.7/√65)

17.5+/-1.645(0.707)

17.5 +/- 1.16

= ( 16.34, 18.66)

Therefore at 90% confidence interval (a,b) = ( 16.34, 18.66)

Substituting the values we have; for 95%

17.5+/-1.96(5.7/√65)

17.5+/-1.96(0.707)

17.5 +/- 1.39

= ( 16.11, 18.89)

Therefore at 95% confidence interval (a,b) = ( 16.11, 18.89)

3 0

3 years ago

What is the smallest positive integer n such that n! ends in at least 2019 zeros?​

What is the smallest positive integer n such that n! ends in at least 2019 zeros?