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zvonat [6]
3 years ago
14

Determine if the finite correction factor should be used. If​ so, use it in your calculations when you find the probability. In

a sample of 900 gas​ stations, the mean price for regular gasoline at the pump was $ 2.876 per gallon and the standard deviation was ​$0.009 per gallon. A random sample of size 50 is drawn from this population. What is the probability that the mean price per gallon is less than ​$2.874​?
Mathematics
1 answer:
serg [7]3 years ago
5 0

Answer:

Idk

Step-by-step explanation:

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By selling a watch on 400 and 460’. Loss and profit happened respectively.if the loss and profit are equal, find cp.
Maslowich

Answer:

430

Step-by-step explanation:

<u>Given :-</u>

  • By selling a watch on 400 and 460’. Loss and profit happened respectively.
  • The loss and profit are equal.

And we need to find out the CP .So ,

<u>Let :- </u>

  • CP be x
  • Profit and Lost be y .

<u>According to the Question :- </u>

:\implies 460 = x + y <em>( Profit)</em>

:\implies 400 = x - y <em>( Loss)</em>

<u>Adding both :- </u>

:\implies 460 + 400 = x + y + x - y

:\implies 860 = 2x

:\implies x = 860/2

:\implies x = 430

<u>Hence the Cost price of the watch is 430 .</u>

5 0
3 years ago
Pleaseee i rlly need this one rn!!
bija089 [108]

Answer:

its 20 inches

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
URGENT HOW MANY STUDENTS SPEAK SPANISH? and it isn’t 14 I tried that and it was wrong
Ray Of Light [21]

Answer:

46 students

Step-by-step explanation:

Brainliest pls? hope it helps :)

Have a good day/night

3 0
4 years ago
Tan(x)=2
nydimaria [60]
<span>Tan(x)=2
Or for x=45 tan(x)=1
then for tan(x)=2 ; x must be superior to 45 
Then the answer is </span><span>d. 63.4 degrees</span> 
8 0
3 years ago
Help me guys <br> thx so much
navik [9.2K]

Answer:

Option D, 10x^4\sqrt{6} +x^3\sqrt{30x} -10x^4\sqrt{3} -x^3\sqrt{15x}

Step-by-step explanation:

<u>Step 1:  Multiply</u>

<u />(\sqrt{10x^4} -x\sqrt{5x^2} )*(2\sqrt{15x^4} + \sqrt{3x^3})\\ (\sqrt{10 * x^2 * x^2} -x\sqrt{5 * x^2} ) * (2\sqrt{15 * x^2 * x^2} +\sqrt{3 * x^2 * x})\\(x^2\sqrt{10} -x^2\sqrt{5} )*(2x^2\sqrt{15} +x\sqrt{3x}) \\\\

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

(2x^4\sqrt{150} ) + (x^3\sqrt{30x}) + (-2x^4\sqrt{75}) + (-x^3\sqrt{15x}  )

2x^4\sqrt{5^2*6} + x^3\sqrt{30x} -2x^4\sqrt{5^2*3} -x^3\sqrt{15x}

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

Answer:  Option D, 10x^4\sqrt{6} +x^3\sqrt{30x} -10x^4\sqrt{3} -x^3\sqrt{15x}

6 0
3 years ago
Read 2 more answers
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