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artcher [175]
3 years ago
12

77 is written as the product of two prime numbers 7 × 11. Can 702 be written as the product of two prime numbers? If not, why? W

hat about 703?​
Mathematics
1 answer:
kari74 [83]3 years ago
6 0

Answer:

702 can't be written as the product of 2 prime numbers because if you divide 702 by 2 then you get 351 and that is not a prime number. Therefore, 702 is a prime number.

The prime factorization of 703 is 19 × 37. Since it has a total of 2 prime factors, 703 is a composite number.

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The Census Bureau says that the 10 most common names in the United States are (in order) Smith, Johnson, Williams, Jones, Brown,
scZoUnD [109]
Using binomial distribution where success is the appearing of any of the top 10 most common names, thus probability of success (p) is 9.6% = 0.096 and the probability of failure = 1 - 0.096 = 0.904. Number of trials is 11.
Binomial distribution probability is given by P(x) = nCx (p)^x (q)^(n - x)
Probability that none of the top 10 most common names appears is P(0) = 11C0 (0.096)^0 (0.904)^(11 - 0) = (0.904)^11 = 0.3295
Thus, the probability that at least one of the 10 most common names appear is 1 - 0.3295 = 0.6705

Therefore, I will be supprised that none of the names of the authors were among the 10 most common names given that the probability that at least one of the names appear is 67%.
6 0
3 years ago
Solve this equation: -88=5y-13
nadya68 [22]

Answer:

y=-15

Step-by-step explanation:

4 0
3 years ago
Adam conducted an experiment wherein he monitors the ripeness of a piece of fruit over several days. Based on this information w
Tanzania [10]
The amount of days would be the independent variable and the ripeness of the fruit would be the dependent variable
5 0
3 years ago
How many pairs of consecutive natural numbers have a product of less than 40000? I am in 5th grade. This is supposed to be easy
Ugo [173]

Answer:

There are 199 pairs of consecutive natural numbers whose product is less than 40000.

Step-by-step explanation:

We notice that such statement can be translated into this inequation:

n \cdot (n+1) < 40000

Now we solve this inequation to the highest value of n that satisfy the inequation:

n^{2}+n < 40000

n^{2}+n -40000

The Quadratic Formula shows that roots are:

n_{1,2} = \frac{-1\pm\sqrt{1^{2}-4\cdot (1)\cdot (-40000)}}{2\cdot (1)}

n_{1,2} = -\frac{1}{2}\pm \frac{1}{2} \cdot \sqrt{160001}

n_{1} = -\frac{1}{2}+\frac{1}{2}\cdot \sqrt{160001}

n_{1} \approx 199.501

n_{2} = -\frac{1}{2}-\frac{1}{2}\cdot \sqrt{160001}

n_{2} \approx -200.501

Only the first root is valid source to determine the highest possible value of n, which is n_{max} = 199. Each natural number represents an element itself and each pair represents an element as a function of the lowest consecutive natural number. Hence, there are 199 pairs of consecutive natural numbers whose product is less than 40000.

6 0
2 years ago
Can anyone help meeee!
KIM [24]

y = -2/3x -3

x=0 y=-3  point (0,-3)  over 0 down 3

x=3  y=-2/3(3) -3 =-2-3 =-5   (3,-5)  go to the right 3 down 5

3 0
3 years ago
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