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3 years ago

Square root 8 to the nearest hundredth

Mathematics

1 answer:

yawa3891 [41]3 years ago

5 0

Answer:

2.83

Step-by-step explanation:

just round to the nearest hundredth

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Find the first,fourth,and eighth term of the sequence A(n)=-3x2 ^n-1

marusya05 [52]

A(n) = 5.3^(n-1)
first term A(1) = 5.3^(1-1)= 5.3^0 = 5.1 = 5
fourth term A(4) = 5.3^(4-1) = 5.3^3 = 5x27 = 135
8th term A(8) = 5.3^(8-1) = 5.3^7 = 5.3^3.3^3.3 = 5x27x27x3= 15x729 = 10935

hope this helps! ;)

7 0

3 years ago

ALL YOU NEED TO DO IS DRAW THIS!!!!! Circle ω1 with center K of radius 4 and circle ω2 of radius 6 intersect at points W and U.

maria [59]

Answer:

WU = (14√13)/13 ≈ 6.6564

Step-by-step explanation:

Call the incenter of ∆KWU point A. Call the center of circle ω2 point B.

Then ∠KWA has half the measure of arc WA. ∠AWU is congruent to ∠KWA, so also has half the arc measure. That is, ∠KWU has the same measure as arc WA and ∠KBW.

KB is a perpendicular bisector of chord WU, so ∆KWB is a right triangle, of which WU is twice the altitude to base KB.

The length of KB can be found several ways. One of them is to use the Pythagorean theorem:

KB² = KW² +WB² = 4² +6² = 52

KB = √52 = 2√13

The area of triangle KWB is ...

area ∆KWB = (1/2)KW·WB = (1/2)(4)(6) = 12 . . . . square units

Using KB as the base in the area calculation, we have ...

area ∆KWB = (1/2)(KB)(WU/2)

12 = KB·WU/4

WU = 48/KB = 48/(2√13) = 24/√13

WU = (24√13)/13 ≈ 6.6564

7 0

3 years ago

Solve S = 2πrh for r

Elodia [21]

Divide both sides by 2πh:-

r = S / 2πh Answer

6 0

3 years ago

PLEASE HELP!!!

Klio2033 [76]

Answer:

The answer is A and B

Step-by-step explanation:

y-y=m(x-x)

y-(-3)=m(x-(-5))

y+3=0.4(x+5)

y-y=m(x-x)

y-1=0.4(x-5)

6 0

3 years ago

Let X equal the number of typos on a printed page with a mean of 4 typos per page.

timama [110]

Answer:

a) There is a 98.17% probability that a randomly selected page has at least one typo on it.

b) There is a 9.16% probability that a randomly selected page has at most one typo on it.

Step-by-step explanation:

Since we only have the mean, we can solve this problem by a Poisson distribution.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

In this problem, we have that $\mu = 4$

(a) What is the probability that a randomly selected page has at least one typo on it?

Thats is $P(X \geq 1)$ . Either a number is greater or equal than 1, or it is lesser. The sum of the probabilities must be decimal 1. So:

$P(X < 1) + P(X \geq 1) = 1$

$P(X \geq 1) = 1 - P(X < 1)$

In which

$P(X < 1) = P(X = 0)$ .

$P(X = 0) = \frac{e^{-4}*4^{0}}{(0)!} = 0.0183$

$P(X \geq 1) = 1 - P(X < 1) = 1 - 0.0183 = 0.9817$

There is a 98.17% probability that a randomly selected page has at least one typo on it.

(b) What is the probability that a randomly selected page has at most one typo on it?

This is $P = P(X = 0) + P(X = 1)$ . So:

$P(X = 0) = \frac{e^{-4}*4^{0}}{(0)!} = 0.0183$

$P(X = 1) = \frac{e^{-4}*4^{1}}{(1)!} = 0.0733$

$P = P(X = 0) + P(X = 1) = 0.0183 + 0.0733 = 0.0916$

There is a 9.16% probability that a randomly selected page has at most one typo on it.

3 0

3 years ago