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Dima020 [189]
3 years ago
11

tina has 50 coins, all dimes and quarters. the value of the coins is $7.10. write a system of equations to determine how many di

mes and how many quarters tina has
Mathematics
1 answer:
Sliva [168]3 years ago
7 0

Answer:

A) D + Q = 50

B) .10D + .25Q = 7.10

We multiply A) by -.10

A) -.10D - .10Q = -5 then adding this to B)

B) .10D + .25Q = 7.10

.15Q = 2.10

Quarters = 14 = 3.50

Dimes = 36 = 3.60


Step-by-step explanation:


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Hello there is a solution : 

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3 years ago
Find the center and radius of the circle whose diameter has an endpoint at (-3, -4) and the origin.
Zina [86]

Given end points of diameter,

(x1,y1)=(-3,-4)

(x2,y2)=(0,0)

Now,

the equation of circle is,

(x-x1)(x-x2)+(y-y1)(y-y2)=0

or, (x+3)(x-0)+(y+4)(y-0)=0

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which is in the form of x^2 +y^2 +2gx +2fy + c=0

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Now,

radius(r) =  \sqrt{ {g}^{2}  +  {f}^{2} - c  } \\  =  \sqrt{ \frac{ {3}^{2} }{ {2}^{2} } +  {2}^{2}  - 0 }  \\  =  \sqrt{ \frac{9}{4}  + 4}  \\  =   \sqrt{ \frac{25}{4} }  \\  =  \frac{5}{2}

6 0
3 years ago
Suppose on a certain planet that a rare substance known as Raritanium can be found in some of the rocks. A raritanium-detector i
aleksandrvk [35]

Answer:

0.967 = 96.7% probability the rock sample actually contains raritanium

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Positive reading

Event B: Contains raritanium

Probability of a positive reading:

98% of 13%(positive when there is raritanium).

0.5% of 100-13 = 87%(false positive, positive when there is no raritanium). So

P(A) = 0.98*0.13 + 0.005*0.87 = 0.13175

Positive when there is raritanium:

98% of 13%

P(A) = 0.98*0.13 = 0.1274

What is the probability the rock sample actually contains raritanium?

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.1274}{0.13175} = 0.967

0.967 = 96.7% probability the rock sample actually contains raritanium

7 0
2 years ago
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asambeis [7]

Answer:

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Step-by-step explanation:

hopefully this helps :)

8 0
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Step-by-step explanation:


8 0
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