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3 years ago

PleAsE hElP WITh mY irEaDy

Mathematics

2 answers:

lord [1]3 years ago

8 0

Answer:

A, B, D

Step-by-step explanation:

hope this helped

saw5 [17]3 years ago

5 0

Answer: 7⁴, 7 to the fourth power, 7x7x7x7

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ASAP Please! True or False? Triangle FGH is a right triangle.

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3 years ago

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(sin (x+y)) / sinxcosy = 1+ cot(x) + tan(y)

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2 years ago

Multiply and simplify: y(2x+3y)(2x+3y)

vekshin1

<span>Simplifying y(2x + 3y)(2x + 3y)

Multiply (2x + 3y) * (2x + 3y) y(2x * (2x + 3y) + 3y * (2x + 3y)) y((2x * 2x + 3y * 2x) + 3y * (2x + 3y))

Reorder the terms: y((6xy + 4x2) + 3y * (2x + 3y)) y((6xy + 4x2) + 3y * (2x + 3y)) y(6xy + 4x2 + (2x * 3y + 3y * 3y)) y(6xy + 4x2 + (6xy + 9y2))

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4 years ago

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How many ways can a president and vice-president be chosen from a committee of 6 people?

Citrus2011 [14]

Answer:

Step-by-step explanation:

There are 6 ways the president can be chosen. Assuming the vice-president must be a different person, there are then 5 possibilities for that office. The total number of possible choices is ...

6 × 5 = 30

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Each of the 6 president choices can have any of 5 different vice-president choices.

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4 years ago

The scores of individual students on the American College Testing (ACT) Program College Entrance Exam have a normal distribution

tekilochka [14]

Answer:

The probability that the average of the scores of all 400 students exceeds 19.0 is larger than the probability that a single student has a score exceeding 19.0

Step-by-step explanation:

Xi~N(18.6, 6.0), n=400, Yi~Ber(p); Z~N(0, 1);

$P(0\leq X\leq 19.0)=P(\frac{0-\mu}{\sigma} \leq \frac{X-\mu}{\sigma}\leq \frac{19-\mu}{\sigma}), Z= \frac{X-\mu}{\sigma}, \mu=18.6, \sigma=6.0$

$P(-3.1\leq Z\leq 0.0667)=\Phi(0.0667)-\Phi (-3.1)=\Phi(0.0667)-(1-\Phi (3.1))=0.52790+0.99903-1=0.52693$

P(Xi≥19.0)=0.473

$\{Yi=0, Xi< 19\\Yi=1, Xi\geq 19\}$

p=0.473

Yi~Ber(0.473)

$P(\frac{1}{n}\displaystyle\sum_{i=1}^{n}X_i\geq 19)=P(\displaystyle\sum_{i=1}^{400}X_i\geq 7600)$

Based on the Central Limit Theorem:

$\displaystyle\sum_{i=1}^{n}X_i\~{}N(n\mu, \sqrt{n}\sigma),\displaystyle\sum_{i=1}^{400}X_i\~{}N(7440, 372)$

Then:

$P(\displaystyle\sum_{i=1}^{400}X_i\geq 7600)=1-P(0$

$P(\displaystyle\sum_{i=1}^{n}Y_i=1)=P(\displaystyle\sum_{i=1}^{400}Y_i=1)$

Based on the Central Limit Theorem:

$\displaystyle\sum_{i=1}^{400}Y_i\~{}N(400\times 0.473, \sqrt{400}\times 0.499)=\displaystyle\sum_{i=1}^{400}Y_i\~{}N(189.2; 9.98)$

$P(\displaystyle\sum_{i=1}^{400}Y_i=1)\~{=}P(0.5$

Then:

the probability that the average of the scores of all 400 students exceeds 19.0 is larger than the probability that a single student has a score exceeding 19.0

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3 years ago