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3 years ago

What is the value of the digit 4 in the number 524,897,123

Mathematics

2 answers:

Dimas [21]3 years ago

8 0

It's going to be ones

Evgesh-ka [11]3 years ago

5 0

The value is millions

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Explain the

Ivenika [448]

Answer:

a2=12 (the second term of the sequence is 12)

Step-by-step explanation:

a5=324

If the term to term rule is multiply by any number, we deal with geometrical sequence

The formula you should use is an= a1*r^(n-1) where n is the number of the term which we know. In our case we know

a5, so use 5 instead of n

Then you have a5=a1*r^4 where r is the number 3 (because each next term is greater than previous in 3 times)

a5=324

324= a1*3^4

324=a1*81

a1=4 (We find the first term of sequence, because having it you can easily search for every term )

Return to the formula an= a1*r^n-1

Now search for the second term using 2 instead of n in the formula

a2= a1*r^1

a2=a1*r, a1=4, r=3

a2=4*3=12

7 0

2 years ago

Select all the expressions that have a value of 180 when b = 4. o 876 – 179 + 957 - 87 O 336 12 62 + 120 0 516 - 24 ( 115 + 187

BabaBlast [244]

Answer: miata

Step-by-step explanation:

6 0

3 years ago

Challenge: x(2)+9x+c<br><br>:-D

kvv77 [185]

Answer:

×(2)+9×+c

2×+9×+c =

11x+c

3 0

3 years ago

Find the mean, variance &a standard deviation of the binomial distribution with the given values of n and p.

MrMuchimi

A random variable following a binomial distribution over $n$ trials with success probability $p$ has PMF

$f_X(x)=\dbinom nxp^x(1-p)^{n-x}$

Because it's a proper probability distribution, you know that the sum of all the probabilities over the distribution's support must be 1, i.e.

$\displaystyle\sum_xf_X(x)=\sum_{x=0}^n\binom nxp^x(1-p)^{n-x}=1$

The mean is given by the expected value of the distribution,

$\mathbb E(X)=\displaystyle\sum_xf_X(x)=\sum_{x=0}^nx\binom nxp^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle\sum_{x=1}^nx\frac{n!}{x!(n-x)!}p^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle\sum_{x=1}^n\frac{n!}{(x-1)!(n-x)!}p^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=1}^n\frac{(n-1)!}{(x-1)!((n-1)-(x-1))!}p^{x-1}(1-p)^{(n-1)-(x-1)}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^n\frac{(n-1)!}{x!((n-1)-x)!}p^x(1-p)^{(n-1)-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^n\binom{n-1}xp^x(1-p)^{(n-1)-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^{n-1}\binom{n-1}xp^x(1-p)^{(n-1)-x}$

The remaining sum has a summand which is the PMF of yet another binomial distribution with $n-1$ trials and the same success probability, so the sum is 1 and you're left with

$\mathbb E(x)=np=126\times0.27=34.02$

You can similarly derive the variance by computing $\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2$ , but I'll leave that as an exercise for you. You would find that $\mathbb V(X)=np(1-p)$ , so the variance here would be

$\mathbb V(X)=125\times0.27\times0.73=24.8346$

The standard deviation is just the square root of the variance, which is

$\sqrt{\mathbb V(X)}=\sqrt{24.3846}\approx4.9834$

7 0

3 years ago

Jackson purchased a pack of game cards that was on sale for 22% off. The sales tax in his county is 6%. Let y represent the orig

Ymorist [56]

Answer:

Final cost = 0.9328y

Step-by-step explanation:

Assume,

Original price = y

Given:

Discount = 22%

Sales tax = 6%

Computation:

Sales tax will be added on sale value

So,

Sales price = y[100%-22%]

Sales price = 0.88y

Price after sales tax :

Final cost = 0.88y[100% + 6%]

Final cost = 0.88y[106%]

Final cost = 0.9328y

8 0

2 years ago