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aleksandr82 [10.1K]
1 year ago
14

A man made a loss of 15% by selling an article for $595. Find the cost price of the article​

Mathematics
1 answer:
drek231 [11]1 year ago
5 0

Answer:

$700

Step-by-step explanation:

If he made a loss of 15%, then it means that selling price is 85% of the cost price.

If the cost price is x, then we have;

0.85x = 595

x = 595/0.85

x = $700

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Select all the equations on which the point (10,0) lies.
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Answer:

  B, C

Step-by-step explanation:

Since y=0, you are looking for equations such that 10 times the coefficient of x is the constant in the equation.

  a) 5·10 ≠ 15

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  c) 10 = 10 . . . . select this one

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  f) 6·10 ≠ 50

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The following linear equation is shown: y = 2% + 5
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y=8

Step-by-step explanation:

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HELP 60 PTS Write ln⁡〖(a^2 b^3)/cd〗 in expanded form. SHOW ALL WORK TO RECEIVE CREDIT.
Vinil7 [7]

Answer:

\frac{(a*a)(b*b*b)}{c*d}

Step-by-step explanation:

\frac{a^{2} * b^{3}}{cd}

A number to an exponent is the number times itself. A number to the "n"th power would be that number times itself "n" times.

This means that the fraction above could also be expressed as;

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If these parameters, "a", "b", "c", and "d" are given a value, then one can solve this equation.

6 0
2 years ago
Calculate the allele frequencies in this population of palm trees. A model consisting of a box containing fourteen, randomly arr
bogdanovich [222]

Answer:

The correct answer is

0.5 F, 0.5 f

Step-by-step explanation:

We note the following

Number of colored circles in the box = 14

Number of red circles in the box = 4

Number of purple circles in the box = 6

Number of blue circles in the box = 4

The allele frequency are as follows

Where the frequency is given as

Genotype            Frequency                         Relative frequency        

FF = Red                    4                                       4/14 (0.29≈0.3)                  

Ff = Purple                 6                                       6/14 (0.43≈0.4)

ff = Blue                     4                                       4/14(0.29≈0.3)

Within this population,

We however  have 4 FF = 8 F

                                6 Ff =   6 F + 6 f  and

                                4 ff = 8 f

Total allele = 8+6+6+8 = 28

Relative frequency of F = (8+6)/28 = 14/28 = 0.5

relative frequency of f = (8+6)/28 = 0.5

Therefore the allele frequencies in the palm tree population is

0.5 F, 0.5 f

When in equilibrium we have

However the FF has the product of F×F which is = F² = 0.29 so the frequency of F = √(0.29) = 0.535≈ 0.5

The frequency of Ff is Ff or fF =  0.43 since there is equal number of each allele in Ff we have fF or Ff = Ff = 0.43

Which hives 0.43/2 = F =f ≈ 0.2

To  

and ff = 0.29 so that f = 0.535 ≈ 0.5

Therefore f = F  = 0.5 + 0.2 = 0.7

5 0
3 years ago
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