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jeka57 [31]
3 years ago
10

Edward is selling pencil sketches at an art festival. He has sold 8 of his large sketches and 6

Mathematics
1 answer:
oksian1 [2.3K]3 years ago
4 0

Step-by-step explanation:

147.50 becuase 8 painting equal 116$ and the smallons equal 31.50 so add them together and you get 147.50

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There are 9 students in a class: 7 boys and 2 girls.
Allisa [31]

Answer:

approximately 41.67%

Step-by-step explanation:

4 0
3 years ago
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I have to round an answer to the nearest tenth. My original answer is 15.98. Do I go to 16, or 15.9? Please answer soon.
puteri [66]

Answer:

16

Step-by-step explanation: becuase you have to round it think to urself is it closer to 15.9 or 16

3 0
3 years ago
Two containers, A and B begin with equal volumes of liquid.
Reika [66]

Answer:

volume left in A is 80ml

Step-by-step explanation:

let original volume in both be x ml

after pouring from A to B

vol in A = x - 120

vol in B = x + 120

now it is given that

vol in B = 4 times vol in A

implies

x + 120 = 4 \times (x - 120) \\ x + 120 = 4x - 480 \\ 3x = 600 \\ x = 200

hence original volume in A is 200ml. after pouring in B it is 200-120 = 80 ml

8 0
2 years ago
The shape of Oklahoma can almost be divided into 2 perfect rectangles and 1 triangle. About how many square miles does Oklahoma
Sav [38]
The answer is 63,800 mi²


hoped this helps
4 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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