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3 years ago

This is science but pls help

Mathematics

1 answer:

Finger [1]3 years ago

8 0

2 is the answer!! stream dynamite and back door :)

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A school band found they could arrange themselves in rows of 6, 7, or 8 with no one left over. What is the minimum number of stu

mylen [45]

Answer:

168 is the answer if i m not wrong.I took the LCM.

6 0

3 years ago

For the pair of functions, find the indicated sum, difference, product, or quotient. f(x) = 16 - x2; g(x) = 4 - x Find (f + g)(x

Rudik [331]

(16-x²)+(4-x) or (-x²+16)+(-x+4)

Combine like terms

(-x²+20-x) or (-x²-x+20)

7 0

4 years ago

Read 2 more answers

List the single-digit divisors of 2100. I really need help

liberstina [14]

Answer:

1, 2, 3, 4, 5, 6, 7

Step-by-step explanation:

Prime factors of 2100 are:

2100 = 2*2*3*5*5*7

The single-digit divisors are:

1, 2, 3, 2*2= 4, 5, 2*3= 6, 7

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Another solution is you divide 2100 by all numbers 1 through 9 and list those divisible:

2100/1 = 2100, yes
2100/2 = 1050, yes
2100/3 = 700, yes
2100/4 = 525, yes
2100/5 = 420, yes
2100/6 = 350, yes
2100/7 = 300, yes
2100/8 = 262.5, no
2100/9 = 233.33, no

So all the numbers from 1 to 7

5 0

3 years ago

Past records indicate that the probability of online retail orders that turn out to be fraudulent is 0.08. Suppose that, on a gi

Sunny_sXe [5.5K]

Answer:

The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.

Step-by-step explanation:

We can model this with a binomial random variable, with sample size n=20 and probability of success p=0.08.

The probability of k online retail orders that turn out to be fraudulent in the sample is:

$P(x=k)=\dbinom{n}{k}p^k(1-p)^{n-k}=\dbinom{20}{k}\cdot0.08^k\cdot0.92^{20-k}$

We have to calculate the probability that 2 or more online retail orders that turn out to be fraudulent. This can be calculated as:

$P(x\geq2)=1-[P(x=0)+P(x=1)]\\\\\\P(x=0)=\dbinom{20}{0}\cdot0.08^{0}\cdot0.92^{20}=1\cdot1\cdot0.189=0.189\\\\\\P(x=1)=\dbinom{20}{1}\cdot0.08^{1}\cdot0.92^{19}=20\cdot0.08\cdot0.205=0.328\\\\\\\\P(x\geq2)=1-[0.189+0.328]\\\\P(x\geq2)=1-0.517=0.483$

The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.

5 0

4 years ago

Solve the equation: -3p + 10 = 1 for p

Delicious77 [7]

Answer:

Step-by-step explanation

because i added and subtrcted and minus

3 0

3 years ago

Read 2 more answers