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2 years ago

Can someone help me please

Mathematics

1 answer:

lozanna [386]2 years ago

3 0

Answer:

Are you in middle school ( specifically 6th grade) bc i think i had that like a week ago lol...

Step-by-step explanation:

You might be interested in

A set of 200 test scores are normally distributed with a mean of 70 and a standard deviation of 5. Approximately how many studen

spin [16.1K]

Answer: About 191 students scored between a 60 and an 80.

Step-by-step explanation:

Given : A set of 200 test scores are normally distributed with a mean of 70 and a standard deviation of 5.

i.e. $\mu=70$ and $\sigma=5$

let x be the random variable that denotes the test scores.

Then, the probability that the students scored between a 60 and an 80 :

$P(60$

The number of students scored between a 60 and an 80 = 0.9544 x 200

= 190.88 ≈ 191

Hence , about 191 students scored between a 60 and an 80.

8 0

3 years ago

Please help me with this question

Mandarinka [93]

Answer:

The answer is y=60/x

(I also had this question)

6 0

1 year 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

7 0

2 years ago

The local grocery store wants to know if the shoppers prefer Crest or Colgate toothpaste. The store gives all of its shoppers a

ch4aika [34]

i think the answer is letter C

5 0

3 years ago

Period:<br>7x-9-11=3x+4+2x

vazorg [7]

Answer:

x=12

Step-by-step explanation:

combine like terms

7 0

2 years ago

Read 2 more answers