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elena-14-01-66 [18.8K]
2 years ago
15

Find a pair of numbers that provides a counterexample to show that the given statement is false. (Enter your answers as a comma-

separated list.) If the sum of two counting numbers is an even counting number, then the product of the two counting numbers is an even counting number
Mathematics
1 answer:
k0ka [10]2 years ago
4 0

Answer:

Step-by-step explanation:

We have to find a pair of numbers which give sum as even number but product is not even.

We know even numbers are multiples of 2.

When we add two even numbers we get even number and their product also would be even.

Similarly when we add two odd numbers we get answer as even number

Eg. 1+1=2\\3+3=6\\2k+1 +(2k+3) = 4k+4 i.e. even

But product need not be even.

We have

1+3=4 even\\1(3) =3 not even

Thus we have given counter example that when sum of two counting numbers is even, product need not be even.

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Jet001 [13]

Answer:

a)                      

b) number of flowers.  

c) 2830              

Step-by-step explanation:

We are given the following in the question:

The number of flowers bloomed is given by the function:

 

where s is the s is the number of seeds she planted.

The number of seeds planted per week is given by

where w represents the number of weeks.

a)  composite function that represents how many flowers Lily can expect to bloom over a certain number of weeks.

The composite function can be written as:

where f(w) gives the number of flowers that bloomed in w weeks.

b)  units of measurement for the composite function

The composite functions gives the number of flower that will bloom in w weeks. Thus, the unit of measurement is number of flowers.

c) Number of flowers in 35 weeks.

We put w = 35 in the composite function.

2830 flowers will bloom in 35 weeks.

Thanks

Step-by-step explanation:

8 0
2 years ago
Manufacture of a certain component requires three different machining operations. Machining time for each operation has a normal
Nikolay [14]

Answer:

0.0359

Step-by-step explanation:

Data provided:

mean values of three independent times are 15, 30, and 20 minutes

the standard deviations are 2, 1, and 1.6 minutes

Now,

New Mean = 15 + 30 + 25 = 65

Variance = ( standard deviation )²

or

Variance = 2² + 1² + 1.6² = 7.56

therefore,

Standard deviation = √variance

or

Standard deviation =  2.75

Thus,

Z-value = \frac{\textup{60 - 65}}{\textup{2.75}}

or

Z-value = - 1.81

from the Z-table

the Probability of Z ≤ -1.81 = 0.0359

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2 years ago
Daisy gave 1/2 of her lanzones to Nuel. Nuel gave 1/3 of the lanzones he received to Ronnie and Ronnie gave 1/4 of what he recei
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Assuming all of them had the same amount at the start

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2 years ago
at a track meet, jacob and daniel compete in the 220 m hurdles. daniel finishes in 3/4 of a minute. jacob finishes with 5/12 of
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I hope this help

Jacob  is the faster time

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2 years ago
Read 2 more answers
Trains A and B, 200 km apart on the same straight track, travel at speeds of 50 km/hr and 65 km/hr respectively. At the end of 1
Gwar [14]

Answer:

Ok, at the beginning the distance between the trains is 200km:

IPa(0s) - P(0s)I = 200km.

where Pa is the position of train A, and Pb is the position of train B.

Now, the speed of train A is:

Sa = 50km/h

Sb = 65km/h

Now, remember the relation:

Distance = Speed*time.

So, in one hour, the displacement of train A is:

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The displacement of train B is:

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Now, if at the beginning train A is 200km ahead of train B, then we have:

Pa - Pb = (Pa(0s) + 50km) - (Pb(0s) + 65km)

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Now, if train B was 200km ahead of train A, we have:

Pb - Pa = (Pb(0s) + 65km) - (Pa(0s) - 50km) =

(Pb(0s) - Pa(0s)) + (65km - 50km) = 200km + 15km = 215km

4 0
2 years ago
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