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3 years ago

What is the square root of 378904

Mathematics

1 answer:

zvonat [6]3 years ago

6 0

Simplify the following:

sqrt(378904)

sqrt(378904) = sqrt(8×47363) = sqrt(2^3×47363):

sqrt(2^3 47363)

sqrt(2^3 47363) = sqrt(2^3) sqrt(47363) = 2 sqrt(2) sqrt(47363):

2 sqrt(2) sqrt(47363)

sqrt(2) sqrt(47363) = sqrt(2×47363):

2 sqrt(2×47363)

2×47363 = 94726:

Answer: 2 sqrt(94726)

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Tonys bill at the restaurant was $9.52. If he wants to leave a 20% tip, how much is that?

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Answer:

11.42 dollars

Step-by-step explanation: since you are adding a percentage turn 20% into a decimal. if you do this you will get .2. Next, multiple .2 by 9.52 and you will get 20% of 9.52. Now add this value to 9.52 and you will get 11.42 dollars.

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A black line is shorter than a white line.the white line is shorter than a gray line. Is the black line longer or shorter than t

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3 years ago

Read 2 more answers

A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic flaws and must be classified as

Kisachek [45]

Answer:

(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

Step-by-step explanation:

Let X = number of seconds in the batch.

The probability of the random variable X is, p = 0.31.

The random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of X is:

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

(a)

Compute the probability that only one goblet is a second among six randomly selected goblets as follows:

$P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888$

Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b)

Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:

P (X ≥ 2) = 1 - P (X < 2)

$=1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776$

Thus, the probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c)

If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.

P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)

$={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453$

Thus, the probability that at most five must be selected to find four that are not seconds is 0.9453.

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3 years ago

Please Help Me!

melisa1 [442]

Answer: B or 52

Step-by-step explanation: I will explain later

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3 years ago

New streetlights will be installed along a section of the highway. The posts for the streetlights will be 7.5 m tall and made of

SIZIF [17.4K]

First, we determine the volumes of the posts may it be cylindrical in shape or rectangular prism.

(A) cylindrical:
( π(26.7/100)² - π(24.2/100)²)*(7.5) = 0.3 m³
(B) rectangular prism:
(40/100)²(7.5) - (35/100)²(7.5) = 0.28125 m³

Then, we calculate for the amount of substance
(A) cylindrical: (0.3 m³)(2700 kg/m³) = 810 kg
(B) rectangular prism : (0.28125 m³)(2700 kg/m³) = 759.375 kg

Then, calculate for the costs
(A) (810 kg)($0.38/kg) = $307.8
(B) (759.375 kg)($0.38/kg) = $288.56

Thus, the answer for A is rectangular post
B. About $19.24 can be saved.

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3 years ago