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Vilka [71]
3 years ago
10

a purse contains 25p coins and 50 p coins. the total amount in the purse is rs.25. if the number of 50 p coins is double the num

ber of 25p coins ?how many of each are there in purse⛄
Mathematics
1 answer:
erica [24]3 years ago
4 0
Let the total no. of 25 p coins be x
50p coins = 2x 

Value of 25 p coins ( in rupees) = 0.25*x =0.25x
Value of 50p coins ( in rupees) = 0.5*2x = x

0.25x+x = 25
1.25x =25
x = 25/1.25 = 20

no. if 25p coins = 20
and 50p coins = 2*20 = 40

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Answer:

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Step-by-step explanation:

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4 0
2 years ago
A new test to detect TB has been designed. It is estimated that 88% of people taking this test have the disease. The test detect
Elodia [21]

Answer:

Correct option: (a) 0.1452

Step-by-step explanation:

The new test designed for detecting TB is being analysed.

Denote the events as follows:

<em>D</em> = a person has the disease

<em>X</em> = the test is positive.

The information provided is:

P(D)=0.88\\P(X|D)=0.97\\P(X^{c}|D^{c})=0.99

Compute the probability that a person does not have the disease as follows:

P(D^{c})=1-P(D)=1-0.88=0.12

The probability of a person not having the disease is 0.12.

Compute the probability that a randomly selected person is tested negative but does have the disease as follows:

P(X^{c}\cap D)=P(X^{c}|D)P(D)\\=[1-P(X|D)]\times P(D)\\=[1-0.97]\times 0.88\\=0.03\times 0.88\\=0.0264

Compute the probability that a randomly selected person is tested negative but does not have the disease as follows:

P(X^{c}\cap D^{c})=P(X^{c}|D^{c})P(D^{c})\\=[1-P(X|D)]\times{1- P(D)]\\=0.99\times 0.12\\=0.1188

Compute the probability that a randomly selected person is tested negative  as follows:

P(X^{c})=P(X^{c}\cap D)+P(X^{c}\cap D^{c})

           =0.0264+0.1188\\=0.1452

Thus, the probability of the test indicating that the person does not have the disease is 0.1452.

4 0
3 years ago
Solve 5x^2 − 3x + 17 = 9.
Vedmedyk [2.9K]

Answer:

The roots of the equation are x = \frac{3+\sqrt{151}i}{10} and x = \frac{3-\sqrt{151}i}{10}

and there are no real roots of the equation given above

Step-by-step explanation:

To solve:

5x² − 3x + 17 = 9

or

⇒ 5x² − 3x + 17 - 9 = 0

or

⇒ 5x² − 3x + 8 = 0

Now,

the roots of the equation in the form ax² + bx + c = 0 is given as:

x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}

in the above given equation

a = 5

b = -3

c = 8

therefore,

x = \frac{-(-3)\pm\sqrt{(-3)^2-4\times5\times8}}{2\times5}

or

x = \frac{3\pm\sqrt{9-160}}{10}

or

x = \frac{3+\sqrt{-151}}{10} and x = \frac{3-\sqrt{-151}}{10}

or

x = \frac{3+\sqrt{151}i}{10} and x = \frac{3-\sqrt{151}i}{10}

here i = √(-1)

Hence,

The roots of the equation are x = \frac{3+\sqrt{151}i}{10} and x = \frac{3-\sqrt{151}i}{10}

and there are no real roots of the equation given above

7 0
2 years ago
Help me out, please!
VMariaS [17]

Answer:

y = 2/3(x - 3)

Step-by-step explanation:

slope of the line is 2/3

we can use point (3,0)

formula: y - y = m(x - x)

y - 0 = 2/3 (x - 3)

y = 2/3(x - 3)

5 0
2 years ago
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