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2 years ago

How many significant figures in 20340 ?

Mathematics

2 answers:

Iteru [2.4K]2 years ago

8 0

There are 4. (2,0,3,4) the last 0 doesn’t count as a sig fig

elena-14-01-66 [18.8K]2 years ago

7 0

There are 4 significant figures in 20340.

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Your company has just taken out a 1-year installment loan for $72,500 at a nominal rate of 20.0% but with equal end-of-month pay

Mariulka [41]

Answer:

100% of the 2nd monthly payment go toward the repayment of principal.

Step-by-step explanation:

The loan taken is the Principal which is mentioned as $72,500 with interest at a nominal rate of 20%. Firstly, it is important to understand that nominal rate means <em>non-compounding </em>rate. Simply put will be a "<em>one-time charged" </em>rate on the loan. Since this is given as 20% of the Principal. It is calculated thus: $\frac{20}{100}$ × $\frac{72,500}{1}$ = $14,500. So the interest on the loan is $14,500. Added to the Principal the total amount to be paid back by the company becomes: $72,500 + $14,500 = $87,000. To pay back this amount at equal end-of-month installments in 1 year (12 months), we divide the total amount by 12. i.e $\frac{87000}{12}$ = $7250. This means, the monthly payment will be $7,250. Since the monthly payment pays only 10% of the initial principal $72,500. By the second month only 20% of the Principal would have been paid. So all of the monthly payment will go towards repaying the principal

3 0

2 years ago

Half the product of two numbers a and b, added to thrice a third number c

Alexus [3.1K]

Answer:

ab/2 + 2c

Step-by-step explanation:

Half the product would be represented by “product/2”

The product of two numbers a and b can be represented by “ab”

Added to twice a third number c = + 2c

8 0

1 year ago

A case in riverhead, new york, nine different crime victims listened to voice recordings of five different men. all nine victims

Alexus [3.1K]

the complete question in the attached figure

1) Find the probability that all nine victims would select the same person. You should enter your answer as a decimal, not as a percent. Do not round, as your answer should be a terminating decimal.

P=(1/5)*(1/5)*(1/5)*(1/5)*(1/5)*(1/5)*(1/5)*(1/5)*(1/5)=0.000000512

2) what is the probability that the first victim picks SOMEONE? ANYONE?

P=(1/5)=0.20 ---------------- > 20%

3) Now let’s look at victim #2. What is the probability that they pick the same person as victim #1?

P=(1/5)*(1/5)=0.04 --------------> 4%

4) Now let’s look at victim #3. What is the probability that they pick the same person as victim #1?

P=(1/5)*(1/5)*(1/5)=0.008 --------------> 0.8%

5) Now let's look at victim #9. What is the probability that they pick the same person as victim #1?

P=(1/5)*(1/5)*(1/5)*(1/5)*(1/5)*(1/5)*(1/5)*(1/5)*(1/5)=0.000000512

6) Once you find all those probabilities, what rule would apply?

the probability is given by (1/5)^n

n=number of victims that pick the same person---------> in this problem = 9

P=(1/5)^9= 0.000000512

5 0

2 years ago

Quadrilateral ABCD is similiar to quadrilateral EFGH. The lengths of the three longest sides in quadrilateral ABCD are 24 feet,

masha68 [24]

Well the sides of these 2 quadrilaterals are proportional to each other , so you would have to find the scale factor of the 2 then see which sides have already proportional pairs , then multiply *or / to find the last one .

Answer: D. 11.7 ft

5 0

2 years ago

Read 2 more answers

Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i

jenyasd209 [6]

Answer:

If the limit that you want to find is $\lim_{x\to \infty}\dfrac{P(x)}{Q(x)}$ then you can use the following proof.

Step-by-step explanation:

Let $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$ and $Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0}$ be the given polinomials. Then

$\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}$

Observe that

$\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}$

and

$\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}$

Then

$\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}$

3 0

2 years ago