The smallest prime number of p for which p^3 + 4p^2 + 4p has exactly 30 positive divisors is 43.
<h3>What is the smallest prime number of p for which p must have exactly 30 positive divisors?</h3>
The smallest number of p in the polynomial equation p^3 + 4p^2 + 4p for which p must have exactly 30 divisors can be determined by factoring the polynomial expression, then equating it to the value of 30.
i.e.
By factorization, we have:
Now, to get exactly 30 divisor.
- (p+2)² requires to give us 15 factors.
Therefore, we can have an equation p + 2 = p₁ × p₂²
where:
- p₁ and p₂ relate to different values of odd prime numbers.
So, for the least values of p + 2, Let us assume that:
p + 2 = 5 × 3²
p + 2 = 5 × 9
p + 2 = 45
p = 45 - 2
p = 43
Therefore, we can conclude that the smallest prime number p such that
p^3 + 4p^2 + 4p has exactly 30 positive divisors is 43.
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Answer:
68% of the sample can be expected to fall between 28 and 32 cm
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 30
Standard deviation = 2
What proportion of the sample can be expected to fall between 28 and 32 cm
28 = 30-2
28 is one standard deviation below the mean
32 = 30 + 2
32 is one standard deviation above the mean
By the Empirical Rule, 68% of the sample can be expected to fall between 28 and 32 cm
Answer:
i really dont know im in 7th grade im so sorry
Step-by-step explanation:
Answer:
its 1
Step-by-step explanation: 341/5= 68 r1
Answer:
f[g(1)] = 3
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Algebra I</u>
Step-by-step explanation:
<u>Step 1: Define</u>
f(x) = 2x + 1
g(x) = 3x - 2
<u>Step 2: Find g(1)</u>
- Substitute in <em>x</em>: g(1) = 3(1) - 2
- Multiply: g(1) = 3 - 2
- Subtract: g(1) = 1
<u>Step 3: Find f[g(1)]</u>
- Substitute in g(1): f[g(1)] = 2(1) + 1
- Multiply: f[g(1)] = 2 + 1
- Add: f[g(1)] = 3
