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DaniilM [7]
3 years ago
5

Can you write an equivalent fraction for 5/9 and 1/10 using the least common denominator?​

Mathematics
1 answer:
krek1111 [17]3 years ago
6 0

Answer: Yes

Step-by-step explanation:  

Rewriting input as fractions if necessary:

5/9, 1/10

For the denominators (9, 10) the least common multiple (LCM) is 90.

LCM(9, 10)

Therefore, the least common denominator (LCD) is 90.

Calculations to rewrite the original inputs as equivalent fractions with the LCD:

5/9 = 5/9 × 10/10 = 50/90

1/10 = 1/10 × 9/9 = 9/90

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13x^-5 y^0/5^-3 z^-10
VMariaS [17]

Answer:

\frac{1625z^{10}}{x^{5}}

Step-by-step explanation:

We want to simplify:

\frac{13x^{-5}y^0}{5^{-3}z^{-10}}

We rewrite as positive index to get:

\frac{13y^05^{3}z^{10}}{x^{5}}

This simplifies to

\frac{13*1*125z^{10}}{x^{5}}

This will finally give us:

\frac{1625z^{10}}{x^{5}}

We cannot simplify further.

Hence the simplest form is \frac{1625z^{10}}{x^{5}}

7 0
3 years ago
Suppose small aircraft arrive at a certain airport according to a Poisson process with rate α = 8 per hour, so that the number o
Ksenya-84 [330]

Answer:

Step-by-step explanation:

Step1:

We have Suppose small aircraft arrive at a certain airport according to a Poisson process with rate α =8 per hour, so that the number of arrivals during a time period of t hours is a Poisson rv with parameter μ = 8t

Step2:

Let “X” the number of small aircraft that arrive during time t and it follows poisson distribution parameter “”

The probability mass function of poisson distribution is given by

P(X) = , x = 0,1,2,3,...,n.

Where, μ(mean of the poisson distribution)

a).

Given that time period t = 1hr.

Then,μ = 8t

             = 8(1)

             = 8

Now,

The probability that exactly 6 small aircraft arrive during a 1-hour period is given by

P(exactly 6 small aircraft arrive during a 1-hour period) = P(X = 6)

Consider,

P(X = 6) =  

              =  

              =  

              = 0.1219.

Therefore,The probability that exactly 6 small aircraft arrive during a 1-hour period is 0.1219.

1).P(At least 6) = P(X 6)

Consider,

P(X 6) = 1 - P(X5)

                = 1 - {+++++}

                = 1 - (){+++++}

                = 1 - (0.000335){+++++}

                = 1 - (0.000335){1+8+32+85.34+170.67+273.07}

                = 1 - (0.000335){570.08}

                = 1 - 0.1909

                = 0.8090.

Therefore, the probability that at least 6 small aircraft arrive during a 1-hour period is 0.8090.

2).P(At least 10) = P(X 10)

Consider,

P(X 10) = 1 - P(X9)

                 = 1 - {+++++

5 0
3 years ago
Daniel's basic cell phone rate each month is $29.95. Add to that $5.95 for voice mail and $2.95 for text messaging. This past mo
Vlad [161]
5.95 + 2.95 = 8.9
8.9 + 29.95 = 38.85
62.35 - 38.85 = 23.5 
So Daniel spent $23.50 on long distance calling 
8 0
4 years ago
Read 2 more answers
A sewing machine can sew 500 stitches in 2/3 min.<br><br> Complete the unit rate.
Nesterboy [21]
500 stiches/2/3 min = 750 stiches per minute
7 0
3 years ago
Read 2 more answers
at 4:00 am the temperature was -9 degrees f. at noon the temperature was 18 degrees f. what was the change in temperature
Nutka1998 [239]
There was a change of 27 degrees F.
7 0
3 years ago
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