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vladimir1956 [14]
4 years ago
9

A shop sells candles in boxes of 3 for $12.60 each or 1 candle for $5.50. How many candles can Julia buy with $20?

Mathematics
1 answer:
kherson [118]4 years ago
4 0

Answer: the number of scented candles ... x

the number of unscented candles ... y

x + y = 28

$16 * x + $10 * y = $400  ... 16 * x + 10 * y = 400

Step-by-step explanation: This is the equation

16 * (28 - y) + 10 * y = 40016 * 28 - 16 * y + 10 * y = 400448 - 400 = 16 * y - 10 * y48 = 6 * y    /6y = 48 / 6y = 8x = 28 - y = 28 - 8 = 20Result: The candle shop sold 20 scented candles and 8 unscented candles today.

Please give me Brainliest

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In 1898 L. J. Bortkiewicz published a book entitled The Law of Small Numbers. He used data collected over 20 years to show that
attashe74 [19]

Answer:

(a) The probability of more than one death in a corps in a year is 0.1252.

(b) The probability of no deaths in a corps over 7 years is 0.0130.

Step-by-step explanation:

Let <em>X</em> = number of soldiers killed by horse kicks in 1 year.

The random variable X\sim Poisson(\lambda = 0.62).

The probability function of a Poisson distribution is:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,...

(a)

Compute the probability of more than one death in a corps in a year as follows:

P (X > 1) = 1 - P (X ≤ 1)

             = 1 - P (X = 0) - P (X = 1)

             =1-\frac{e^{-0.62}(0.62)^{0}}{0!}-\frac{e^{-0.62}(0.62)^{1}}{1!}\\=1-0.54335-0.33144\\=0.12521\\\approx0.1252

Thus, the probability of more than one death in a corps in a year is 0.1252.

(b)

The average deaths over 7 year period is: \lambda=7\times0.62=4.34

Compute the probability of no deaths in a corps over 7 years as follows:

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Thus, the probability of no deaths in a corps over 7 years is 0.0130.

6 0
3 years ago
4 raised to power two minus 8u minus 9=0​
coldgirl [10]

Answer:

x can be 3, -3, i, or -i.

Step-by-step explanation:

If you can't find the factoring by looking at this, simplify the equation.

let  

x

2

=

y

to make things easier to see

Now we have  

y

2

−

8

y

−

9

=

0

See it now?

We can factor into  

(

y

−

9

)

(

y

+

1

)

Now substitute back in  

x

2

for  

y

.

(

x

2

−

9

)

(

x

2

+

1

)

Since  

(

x

2

−

9

)

is a difference of two squares,

(

x

−

3

)

(

x

+

3

)

Now we have  

(

x

−

3

)

(

x

+

3

)

(

x

2

+

1

)

x can be 3, -3 for the first two parts

x

2

+

1

=

0

can become  

x

2

=

−

1

Taking the positive and negative root means  

x

=

±

√

−

1

Thus  

x

=

±

i

in addition to 3 and -3.

4 0
3 years ago
Read 2 more answers
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