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hram777 [196]
3 years ago
7

the present age of the father is twice HD old is the age of its own three years hence if two years hence the age of the father w

ill be thrice as old as the age of this 11 years ago find your age before one before 5 years​
Mathematics
1 answer:
NemiM [27]3 years ago
6 0
Big sundi and more bigger with spice
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Which inequality is equivalent to this one?<br> Help with this question
Ira Lisetskai [31]

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A. the first one

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
missy bought 4 bags of chocolates. each bag had 8 pieces of chocolate in it . after her kids ate 6 of the chocolate missy packed
Novay_Z [31]

Answer:

There will be 2 numbers of individual bags of chocolates left with Missy.

Step-by-step explanation:

Missy bought 4 bags of chocolates and there are 8 pieces of chocolates in each bag.

Therefore, there was a total (4× 8) = 32 chocolates.

Now, her kids ate 6 of the chocolates.

So, there remains (32 - 6) = 26 chocolates left.

If Missy packed the remaining chocolates into individual bags with 13 chocolates in each bag, then there will be \frac{26}{13} =2 numbers of individual bags of chocolates left with Missy. (Answer)

5 0
3 years ago
An airline finds that 5% of the persons making reservations on a certain flight will not show up for the flight. If the airline
vivado [14]

Answer:

The answer to the question is;

The probability that a seat will be available for every person holding a reservation and planning to fly is 0.63307.

Step-by-step explanation:

Let the sample size =n = 100

The success probability = 5 % = 0.05

Number of tickets sold = 105 tickets

In the case where there the airline has found that 5 % will not show up, then every passenger should have  a seat, we have  

A Binomial distribution is appropriate where there is a chance for a certain number of successful outcomes from a number of independent trails

However n·p and n·q must be ≥ 5 for there to be a normal approximation of a Binomial distribution thus

n·p = 105×0.05 =  5.25 ≥ 5

and n·q = n(1 - p) = 105 (1 - 0.05) = 99.75 ≥ 5

As the requirements are met, we can proceed with the approximation of the Binomial distribution by the normal distribution

 z = \frac{x-np}{\sqrt{np(1-p)}  } = \frac{4.5 - 105*0.05}{\sqrt{105*0.05(1-0.05)} } =  - 0.3358

We therefore have P(x ≥ 5) = P( x > 4.5) = P(z > -0.34) = 1 - P(z < -0.34) = 1 -0.36693 = 0.63307

Another way to solve the question is as follows

p = 0.95 q = 0.05

μ = np = 0.95*105 = 99.75, σ = \sqrt{npq} = 2.233

P (x≤100) = P(z = P(z<0.34) = 0.63307.

6 0
3 years ago
,,,,,....,,,,,:;,,,,,:(
pshichka [43]

start at -2 go over 2 and up 5

(-2, 5)

8 0
3 years ago
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