All categories

3 years ago

Simplify 12^3/12^7 A. 12^4 B 1/12^4 C 1/12^-4 D 12^10

Mathematics

2 answers:

mrs_skeptik [129]3 years ago

8 0

The answer is B.

Explanation:

If you divide exponents, you subtract the exponents.

Andrei [34K]3 years ago

4 0

Answer:

its b if you need an explanation can use an online calculater and it shows you the sbs

Step-by-step explanation:

You might be interested in

I need help on this how do i do this

kap26 [50]

Answer: 4311/9900 = 0.4354 4936/990 = 4.985 530/900 = 0.58 674/990 = 0.680 8099/990 = 8.180 explanation: divide the numbers and select the answers.

7 0

2 years ago

What is the approximate area of the shaded portion of the diagram? (Use 3.14 as an estimate for pi.)

Nadusha1986 [10]

Answer:

218

Step-by-step explanation:

first find the area of the circle-

$A=\pi r^{2} \\A=\pi 10^{2} \\A=100\pi \\A=314$

to find the area of the triangle, we need the base and the height. The base we know is 16, since we don't know the height yet, use the Pythagorean theorem to find the height. One leg is 16 the hypotenuse is 20 (2* the radius).

$a^{2} +b^{2} =c^{2} \\a^{2} +16^{2} =20^{2} \\a^{2} +256=400\\a^{2} =144\\a=12$

Now I know the height is 12. Find the area of the triangle

$A=\frac{1}{2} bh\\A=\frac{1}{2} (16)(20)\\A=96$

subtract the area of the triangle from the area of the circle

314-96=218

6 0

3 years ago

What is 0.6767 as a simplified fraction

vodka [1.7K]

Answer: 6767/10000

Step-by-step explanation:

Just simplify it by its nearest tenth

6 0

3 years ago

2 pounds equals how many onces

Semenov [28]

There are 16 oz in a pound, so 2 lbs is 32 oz.

4 0

3 years ago

Read 2 more answers

The probability density function of the time you arrive at a terminal (in minutes after 8:00 A.M.) is f(x) = 0.1 exp(−0.1x) for

Blababa [14]

$f_X(x)=\begin{cases}0.1e^{-0.1x}&\text{for }x>0\\0&\text{otherwise}\end{cases}$

a. 9:00 AM is the 60 minute mark:

$f_X(60)=0.1e^{-0.1\cdot60}\approx0.000248$

b. 8:15 and 8:30 AM are the 15 and 30 minute marks, respectively. The probability of arriving at some point between them is

$\displaystyle\int_{15}^{30}f_X(x)\,\mathrm dx\approx0.173$

c. The probability of arriving on any given day before 8:40 AM (the 40 minute mark) is

$\displaystyle\int_0^{40}f_X(x)\,\mathrm dx\approx0.982$

The probability of doing so for at least 2 of 5 days is

$\displaystyle\sum_{n=2}^5\binom5n(0.982)^n(1-0.982)^{5-n}\approx1$

i.e. you're virtually guaranteed to arrive within the first 40 minutes at least twice.

d. Integrate the PDF to obtain the CDF:

$F_X(x)=\displaystyle\int_{-\infty}^xf_X(t)\,\mathrm dt=\begin{cases}0&\text{for }x$

Then the desired probability is

$F_X(30)-F_X(15)\approx0.950-0.777=0.173$

7 0

3 years ago