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Ilya [14]
2 years ago
14

PlEASE HELP WITH THIS QUESTION Lisa bought 5 CDs from a store. The price of each CD was the same. After she bought the CDs, her

account balance showed a change of −$64.95. What would have been the change to Lisa's account balance had she only bought 1 CD from the store?
−$59.95
−$12.99
$12.99
$59.95
Mathematics
1 answer:
slava [35]2 years ago
8 0

Answer:

-12.99

Step-by-step explanation:

-64.95 / 5 = -12.99

The change would be -12.99

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A dealer bought 3 gross buttons at $.38 a dozen. He sold the buttons at 5 cents each. How much was his profit per gross?(12 doze
777dan777 [17]

Answer:

$2.64

Step-by-step explanation:

Selling them at 5 cents each ($0.05), he could sell 1 dozen buttons for

12 * $0.05 = $0.60

As he bought them at $0.38 per dozen, the profit per dozen would be

$0.60 - $0.38 = $0.22

As 12 dozen is 1 gross, the profit per gross would be

12 * $0.22 = $2.64

7 0
1 year ago
Read 2 more answers
The heights of the trees for sale at two nurseries are shown below. Heights of trees at Yard Works in feet : 7, 9, 7, 12, 5 Heig
Troyanec [42]

Answer:

The mean of tree heights at Yard Works is 8 feet and at Grow Station is 9 feet.The mean absolute deviation of the tree heights at both yards is 2.

Step-by-step explanation:

1). Height of the trees at Yard Works are = 7,9,7,12,5 feet

So mean height of the trees = (7+9+7+12+5)÷5

                                               = 40÷5 =8 feet

Standard deviation of the trees at Yard works = ∑(║(height of the tree-mean height of the tree))║/(number of trees)

(height of the tree-mean height of the tree)= ║(7-8)║+║(9-8)║+║(7-8)║+║(12-8)║+║(5-8)║ = (1)+1+(1)+4+(3)= 10

Therefore standard deviation = (10)/(5) =2

2). In the same way mean height of the trees at Grow Station=(9+11+6+12+7)/5= 45/5 = 9

Now we will calculate the mean deviation of the tress at Grow Station

= ∑║(height of the tree-mean height of the tree)║/(number of trees)

= ║(9-9)║+║(11-9)║+║(6-9)║+║(12-9)║+║(7-9)║/(5)

= (0+2+3+3+2)/5

= 10/5 =2

Therefore The mean of tree heights at Yard Works is 8 feet and at Grow Station is 9 feet.The mean absolute deviation of the tree heights at both yards is 2.

                                                             

7 0
2 years ago
Read 2 more answers
Suppose that 90% of all dialysis patients will survive for at least 5 years. In a simple random sample of 100 new dialysis patie
Keith_Richards [23]

Answer:

The  probability   P(\^ p  >  0.80) =  0.99957

Step-by-step explanation:

From the question we are told that

     The population proportion is  p  =  0.90

     The sample size is  n  =  30

  Generally mean of the sampling distribution is  \mu_{\= x } =  p =  0.90

Generally the standard deviation is mathematically represented as

      \sigma  = \sqrt{\frac{p( -p )}{n} }

=>    \sigma  = \sqrt{\frac{0.90( 1 -0.90 )}{100} }

=>    \sigma  = 0.03

Generally the he probability that the proportion surviving for at least five years will exceed 80%, rounded to 5 decimal places is mathematically represented as

       P(\^ p  >  0.80) =  P(\frac{ \^ p - p }{ \sigma }  >  \frac{0.80 - 0.90}{0.03} )

Generally \frac{\^p - p}{\sigma}  =  Z(The standardized \  value  \  of  \^  p)

So  

    P(\^ p  >  0.80) =  P(Z  >  -3.33 )

From the z-table P(Z  >  -3.33 ) = 0.99957

So

    P(\^ p  >  0.80) =  0.99957

5 0
2 years ago
Which pair of arrows doesn't belong?
Vika [28.1K]

Answer:

I don't understand wat u said

4 0
1 year ago
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What is the following simplified product? Assume x greater-than-or-equal-to 0 (StartRoot 10 x Superscript 4 Baseline EndRoot min
9966 [12]

Answer:

\bold{10x^4\sqrt{6}+x^3\sqrt{30x}-10x^4\sqrt{3}-x^3\sqrt{15x}}

Step-by-step explanation:

To find:

Simplified product of:

(\sqrt{10x^4}-x\sqrt{5x^2})(2\sqrt{15x^4}+\sqrt{3x^3})

Solution:

First of all, let us have a look at some of the formula:

1. (a+b) (c+d) = ac+bc+ad+bd

2. a^b\times a^c =a^{b+c }

3. \sqrt{a^{2b}} = \sqrt{a^b.a^b}=a^b

4. \sqrt a  \times \sqrt b = \sqrt{a\times b}

Now, let us apply the above formula to solve the given expression.

(\sqrt{10x^4}-x\sqrt{5x^2})(2\sqrt{15x^4}+\sqrt{3x^3})\\\\\Rightarrow(\sqrt{10x^4})(2\sqrt{15x^4})+(\sqrt{10x^4})(\sqrt{3x^3})-(x\sqrt{5x^2})(2\sqrt{15x^4})-(x\sqrt{5x^2})(\sqrt{3x^3})\\\\\Rightarrow2\sqrt{150x^8}+\sqrt{30x^7}-2x\sqrt{75x^6}-x\sqrt{15x^5}\\\\\Rightarrow\bold{10x^4\sqrt{6}+x^3\sqrt{30x}-10x^4\sqrt{3}-x^3\sqrt{15x}}

The answer is:

\bold{10x^4\sqrt{6}+x^3\sqrt{30x}-10x^4\sqrt{3}-x^3\sqrt{15x}}

8 0
2 years ago
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