What is the quotient of 3600 and 20?

Write 625 in exponent form

tigry1 [53]

I think the answer is 25 squared.

7 0

3 years ago

Read 2 more answers

Question 1 How do metal atoms bond together?

Yanka [14]

Answer:

Step-by-step explanation:

Hope this helps :D

4 0

3 years ago

A larger number is 2 more than 4 times a smaller number. The larger number is also 8 more than 2 times the smaller number.

pogonyaev

Answer:

see the explanation

Step-by-step explanation:

Let

x ----> the smaller number

y ----> the larger number

we know that

A larger number is 2 more than 4 times a smaller number

Identify the key words

is (=)---> equal

more (+) ---> addition

times (*) ---> multiplication

That means

The larger number y is equal to multiply the smaller number x by 4 and then adds 2

so

$y=4x+2$ ----> equation A

The larger number is also 8 more than 2 times the smaller number

Identify the key words

is (=)---> equal

more (+) ---> addition

times (*) ---> multiplication

That means

The larger number y is equal to multiply the smaller number x by 2 and then adds 8

so

$y=2x+8$ ----> equation B

Solve the system of equations by substitution

equate equation A and equation B

$4x+2=2x+8$

$4x-2x=8-2\\2x=6\\x=3$

Find the value of y

substitute the value of x in equation A or equation B

$y=4(3)+2=14$

therefore

The smaller number is 3 and the larger number is 14

5 0

3 years ago

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The Center for Medicare and Medical Services reported that there were 295,000 appeals for hospitalization and other Part A Medic

Ymorist [56]

Answer:

(a) 0.00605

(b) 0.0403

(c) 0.9536

(d) 0.98809

Step-by-step explanation:

We are given that 40% of first-round appeals were successful (The Wall Street Journal, October 22, 2012) and suppose ten first-round appeals have just been received by a Medicare appeals office.

This situation can be represented through Binomial distribution as;

$P(X=r)= \binom{n}{r}p^{r}(1-p)^{n-r} ; x = 0,1,2,3,....$

where, n = number of trials (samples) taken = 10

r = number of success

p = probability of success which in our question is % of first-round

appeals that were successful, i.e.; 40%

So, here X ~ $Binom(n=10,p=0.40)$

(a) Probability that none of the appeals will be successful = P(X = 0)

P(X = 0) = $\binom{10}{0}0.40^{0}(1-0.40)^{10-0}$

= $1*0.6^{10}$ = 0.00605

(b) Probability that exactly one of the appeals will be successful = P(X = 1)

P(X = 1) = $\binom{10}{1}0.40^{1}(1-0.40)^{10-1}$

= $10*0.4^{1} *0.6^{10-1}$ = 0.0403

(c) Probability that at least two of the appeals will be successful = P(X>=2)

P(X >= 2) = 1 - P(X = 0) - P(X = 1)

= 1 - $\binom{10}{0}0.40^{0}(1-0.40)^{10-0} - \binom{10}{1}0.40^{1}(1-0.40)^{10-1}$

= 1 - 0.00605 - 0.0403 = 0.9536

(d) Probability that more than half of the appeals will be successful = P(X > 0.5)

For this probability we will convert our distribution into normal such that;

X ~ N( $\mu = n*p=4,\sigma^{2}= n*p*q = 2.4$ )

and standard normal z has distribution as;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

P(X > 0.5) = P( $\frac{X-\mu}{\sigma}$ > $\frac{0.5-4}{\sqrt{2.4} }$ ) = P(Z > -2.26) = P(Z < 2.26) = 0.98809

3 0

3 years ago

Fill in the blanks. According to estimates by the office of the Treasury Inspector General of IRS, approximately 0.07 of the tax

maks197457 [2]

Answer:

20.72; 4.39

Step-by-step explanation:

The expected number of fraudulent returns is given by the probability that a return is fraudulent multiplied by the number of returns in the sample:

$E(F) = 0.07*296=20.72$

The standard deviation of this binomial probability with p =0.07 and n = 296 is:

$\sigma =\sqrt{np(1-p)}\\\sigma =\sqrt{296*0.07(1-0.07)}\\\sigma=4.39$

Therefore, in a random sample of 296 independent returns from this year, around 20.72 returns, give or take 4.39, will be fraudulent or will contain errors that are purposely made to cheat the IRS.

5 0

4 years ago