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3 years ago

How many intergers between 1 and 500 inclusive are divisible by both 6 and 9 but not 4

1 answer:

7 0

I think the answer is 14.
the smallest common multiple of 6 and 9 is 18. all multiples of 18 is divisible by 6 and 9.
500 ÷18=27.777777777
there are 27 numbers between 1 and 500 that are divisible by both 6 and 9, but when the quotient is an even number, the multiple of 18 is divisible by 4 as well, so only the multiples of 18 that produce an odd quotient satisfy the requirement.
from 1 to 27, there are 14 odd numbers.

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2 years ago

Wayne Gretsky scored a Poisson mean six number of points per game. sixty percent of these were goals and forty percent were assi

BartSMP [9]

Answer:

a) The mean for the total revenue he earns per game is of 13.2K while the standard deviation is of 3.63K.

b) 0.05 = 5% probability that he has four goals and two assists in one game

Step-by-step explanation:

In hockey, a point is counted for each goal or assist of the player.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval. The standard deviation is the square root of the mean.

(a) Find the mean and standard deviation for the total revenue he earns per game.

60% of six are goals, which means that 60% of the time he earned 3K.

40% of six are goals, which means that 40% of the time he earned 1K.

The mean is:

$\mu = 6*0.6*3 + 6*0.4*1 = 13.2$

The standard deviation is:

$\sigma = \sqrt{\mu} = \sqrt{13.2} = 3.63$

The mean for the total revenue he earns per game is of 13.2K while the standard deviation is of 3.63K.

(b) What is the probability that he has four goals and two assists in one game

Goals and assists are independent of each other, which means that we find the probability P(A) of scoring four goals, the probability P(B) of getting two assists, and multiply them.

Probability of four goals:

60% of 6 are goals, which means that:

$\mu = 6*0.6 = 3.6$

The probability of scoring four goals is:

$P(A) = P(X = 4) = \frac{e^{-3.6}*(3.6)^{4}}{(4)!} = 0.19122$

Probability of two assists:

40% of 2 are assists, which means that:

$\mu = 6*0.4 = 2.4$

The probability of getting two assists is:

$P(B) = P(X = 2) = \frac{e^{-2.4}*(2.4)^{2}}{(2)!} = 0.26127$

Probability of four goals and two assists:

$P(A \cap B) = P(A)*P(B) = 0.19122*0.26127 = 0.05$

0.05 = 5% probability that he has four goals and two assists in one game

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3 years ago

6. Solve the following equation for x. Show each step. 5x + 2 = 3x + 4(2x - 1)

monitta

Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers

5. The recursive algorithm given below can be used to compute gcd(a, b) where a and b are non-negative integer, not both zero.

s2008m [1.1K]

Implementating the given algorithm in python 3, the greatest common divisors of (124 and 244) and (4424 and 2111) are 4 and 1 respectively.

The program implementation is given below and the output of the sample run is attached.

def gcd(a, b):

#initialize a function named gcd which takes in two parameters

if a>b:

#checks if a is greater than b

return gcd (b, a)

#if true interchange the Parameters and Recall the function

elif a == 0:

return b

elif a == 1:

return 1

elif((a%2 == 0)and(b%2==0)):

#even numbers leave no remainder when divided by 2, checks if a and b are even

return 2 * gcd(a/2, b/2)

elif((a%2 !=0) and (b%2==0)):

#checks if a is odd and B is even

return gcd(a, b/2)

else :

return gcd(a, b-a)

#since it's a recursive function, it recalls the function with new parameters until a certain condition is satisfied

print(gcd(124, 244))

print()

#leaves a space after the first output

print(gcd(4424, 2111))

Learn more :brainly.com/question/25506437

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2 years ago