Range of scores making up the interquartile ranger

1 answer:

4 0

For a standard normally distributed random variable Z (with mean 0 and standard deviation 1), we get a probability of 0.0625 for a z-score of Z ≈ 1.53, since

P(Z ≥ 1.53) ≈ 0.9375

You can transform any normally distributed variable Y to Z using the relation

Z = (Y - µ) / σ

where µ and σ are the mean and standard deviation of Y, respectively.

So if s is the standard deviation of X, then

(250 - 234) / s ≈ 1.53

Solve for s :

16/s ≈ 1.53

s ≈ 10.43

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What is the factorization of 729^15+1000?

igomit [66]

Answer:

The factorization of $729x^{15} +1000$ is $(9x^{5} +10)(81x^{10} -90x^{5} +100)$

Step-by-step explanation:

This is a case of factorization by sum and difference of cubes, this type of factorization applies only in binomials of the form $(a^{3} +b^{3} )$ or $(a^{3} -b^{3})$ . It is easy to recognize because the coefficients of the terms are perfect cube numbers (which means numbers that have exact cubic root, such as 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, etc.) and the exponents of the letters a and b are multiples of three (such as 3, 6, 9, 12, 15, 18, etc.).

Let's solve the factorization of $729x^{15} +1000$ by using the sum and difference of cubes factorization.

1.) We calculate the cubic root of each term in the equation $729x^{15} +1000$ , and the exponent of the letter x is divided by 3.

$\sqrt[3]{729x^{15}} =9x^{5}$

$1000=10^{3}$ then $\sqrt[3]{10^{3}} =10$

So, we got that

$729x^{15} +1000=(9x^{5})^{3} + (10)^{3}$ which has the form of $(a^{3} +b^{3} )$ which means is a sum of cubes.

Sum of cubes

$(a^{3} +b^{3} )=(a+b)(a^{2} -ab+b^{2})$

with $a= 9x^{5}$ y $b=10$

2.) Solving the sum of cubes.

$(9x^{5})^{3} + (10)^{3}=(9x^{5} +10)((9x^{5})^{2}-(9x^{5})(10)+10^{2} )$

$(9x^{5})^{3} + (10)^{3}=(9x^{5} +10)(81x^{10}-90x^{5}+100)$

8 0

2 years ago

AYÚDENME!!!! T.T PORFA

Vitek1552 [10]

Answer:

Por el teorema de 30 grados en un triangulo rectangulo la linea roja es $\sqrt{3}$ veces la longitud del cateto opuesto al angulo de de 30. Por tanto la linea roja será: