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Alik [6]
3 years ago
14

WILL MARK BRAINLIEST!!!! BRAINLIEST! Write 206 in base 7

Mathematics
2 answers:
mario62 [17]3 years ago
6 0

Answer:

appreciate it man

Step-by-step explanation:

best of luck

tensa zangetsu [6.8K]3 years ago
5 0

Answer:

104

Step-by-step explanation:

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Can someone list the coordinates of jk?
Ganezh [65]

Answer:

j= (-2,2)

k=(2,1)

Step-by-step explanation:

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3 years ago
A tile factory earns money by charging a flat fee for delivery and a sales price of $0.25 per tile. One customer paid a
ddd [48]

Answer: R(x) = 0.25x + 500

Flat fee is computed by:

The sales price of each tile is 0.25 and the customer only bought 10,000 tiles.

So, $0.25 x 10,000 = $2500

So the total sales price per tile sold was $2,500.

The buyer paid $3,000, so the flat fee was included there.

So, $3,000 - $2,500 = $500

So the flat fee was $500.

~

The revenue function is the total income from producing the units. And it has a equation of: R(x) = price per unit x number of units sold plus any fee that is included

So the function describing the revenue of the tile from this sale is:

R(x) = 0.25x + 500

8 0
3 years ago
What value for x makes the sentence true?<br> 4X-4 = 20
skad [1K]
Take the minise out the ecation so we can
4 0
3 years ago
Read 2 more answers
Ahhh help I have no idea
Monica [59]

Answer:

Solve 3-6x >= 0

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Domain: All Real Numbers <= 1/2

7 0
3 years ago
Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65
erma4kov [3.2K]

Answer:

Probability that the sample average is at most 3.00 = 0.98030

Probability that the sample average is between 2.65 and 3.00 = 0.4803

Step-by-step explanation:

We are given that the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65 and standard deviation 0.85.

Also, a random sample of 25 specimens is selected.

Let X bar = Sample average sediment density

The z score probability distribution for sample average is given by;

               Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean = 2.65

           \sigma  = standard deviation = 0.85

            n = sample size = 25

(a) Probability that the sample average sediment density is at most 3.00 is given by = P( X bar <= 3.00)

    P(X bar <= 3) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{3-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z <= 2.06) = 0.98030

(b) Probability that sample average sediment density is between 2.65 and 3.00 is given by = P(2.65 < X bar < 3.00) = P(X bar < 3) - P(X bar <= 2.65)

P(X bar < 3) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{3-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z < 2.06) = 0.98030

 P(X bar <= 2.65) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{2.65-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z <= 0) = 0.5

Therefore, P(2.65 < X bar < 3)  = 0.98030 - 0.5 = 0.4803 .

                                                                             

8 0
3 years ago
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