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2 years ago

What is the 185th digit in the following pattern 12345678910111213141516

Mathematics

1 answer:

Delicious77 [7]2 years ago

5 0

Answer:

The 185th digit following that pattern of '12345678910111213141516' would be the number '5'

Step-by-step explanation:

The reason the 185th digit is '5' is because the pattern increases it's standard number by the following number (ex, 1,2,3,4,5, etc), and eventually it reached 185 as one of it's numbers, '5' would be the digit because the '18' part of 185 would only be the 184th digit, hence you are needed to include the '5' to complete the sequence of that number to reach the 185th digit.

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The length of life of an instrument produced by a machine has a normal ditribution with a mean of 12 months and standard deviati

Zielflug [23.3K]

Answer:

a) $P(X$

And we can find this probability using the normal standard distirbution or excel and we got:

$P(z$

b) $P(7$

And we can find this probability with this difference:

$P(-2.5$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

$P(-2.5$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

a) less than 7 months.

Let X the random variable that represent the length of life of an instrument of a population, and for this case we know the distribution for X is given by:

$X \sim N(12,2)$

Where $\mu=12$ and $\sigma=2$

We are interested on this probability

$P(X$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X$

And we can find this probability using the normal standard distirbution or excel and we got:

$P(z$

b) between 7 and 12 months.

$P(7$

And we can find this probability with this difference:

$P(-2.5$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

$P(-2.5$

8 0

1 year ago

1. Yejin plans to retire at age 60. She wants to create an annuity fund, which will pay her a monthly allowance of $4000 during

dolphi86 [110]

9514 1404 393

Answer:

$3891.10

Step-by-step explanation:

This question is a bit unusual in that the interest is compounded annually, but payments and withdrawals are made monthly. The effective monthly rate is ...

1.05^(1/12) -1 = 0.407412378% = i

We assume that payments to the annuity are made at the end of the month, and that withdrawals are made at the beginning of the month. (The last payment and the first withdrawal are made on the same day.)

The amount of money required in the fund is ...

A = $4000(1 -(1.00407^-360))/(1 -1.00407^-1) = $757,712.88

The amount of money needed each month to be put into the fund is P, where ...

$757,712.88 = P(1 -1.00407)^(-12(60-28))/(1 -1.00407^-1) = 194.7297P

P = $757,712.88/194.7297 = $3891.10

Yejin needs to save $3891.10 each month to meet her retirement goal.

_____

<em>Sanity check</em>

Yejin wants payments for 30 years from the fund to which she has contributed for 32 years. The similarity of the time periods means that Yejin's monthly contribution will need to be very similar to the amount she plans to withdraw.

The only ways to reduce the required contribution are to earn a higher interest on deposits, or to adjust the relative time periods (retire later).

3 0

2 years ago

A poll found that 5% of teenagers (ages 13 to 17) suffer from arachnophobia and are extremely afraid of spiders. At a summer cam

brilliants [131]

Answer:

a) Calculate the probability that at least one of them suffers from arachnophobia.

x = number of students suffering from arachnophobia

= P(x ≥ 1)

= 1 - P(x = 0)

= 1 - [0.05⁰ x (1 - 0.05)¹¹⁻⁰ ]

= 1 - (0.95)¹¹

= 0.4311999 = 0.4312

b) Calculate the probability that exactly 2 of them suffer from arachnophobia? 0.08666

= P(x = 2)

= (¹¹₂) x (0.05)² x (0.95)⁹

where ¹¹₂ = 11! / (2!9!) = (11 x 10) / (2 x 1) = 55

= 55 x 0.0025 x 0.630249409 = 0.086659293 = 0.0867

c) Calculate the probability that at most 1 of them suffers from arachnophobia?

P(x ≤ 1)

= P(x = 0) + P(x = 1)

= [(¹¹₀) x 0.05⁰ x 0.95¹¹] + [(¹¹₁) x 0.05¹ x 0.95¹⁰]

= (1 x 1 x 0.5688) + (11 x 0.05 x 0.598736939) = 0.5688 + 0.3293 = 0.8981

4 0

2 years ago

Julia wrote a linear equation in standard form as –6x – 3y = 10. Arthur noticed that the equation was not written in standard fo

aleksklad [387]

Answer:

Option A - $6x +3y = -10$ is the standard form of the equation.

Step-by-step explanation:

Given : Linear equation - $-6x -3y = 10$

To convert into the standard form.

Standard form of the equation is in form Ax+By=C

So to convert this equation into the standard form we have to multiply both side of the equation by -1, so that the value for A is positive.

After multiplying by -1 the form is

$-6x\times (-1) -3y\times(-1)= 10\times(-1)$

$6x +3y = -10$ is the standard form of the equation.

Therefore, Option A is correct.

8 0

2 years ago

Read 2 more answers

Five light bulbs that are either on or off are lined up in a row. A certain code uses different combinations of lights turned on

Paraphin [41]

Answer:

Step-by-step explanation:

The number of different ways that 2 lights can be chosen from 5 is ₅C₂. Use a calculator or find the value manually:

₅C₂ = 5! / (2! (5−2)!)

₅C₂ = 5! / (2! 3!)

₅C₂ = 5×4×3×2×1 / (2×1 × 3×2×1)

₅C₂ = 5×4 / 2

₅C₂ = 10

7 0

2 years ago