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3 years ago

What is the price of a silver chain purchased for 45.00

Mathematics

2 answers:

Trava [24]3 years ago

5 0

Answer:

$45???

Step-by-step explanation:

ArbitrLikvidat [17]3 years ago

4 0

Answer:

$45

That's all we know from the question.

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Please help, i’ll mark brainliest.

nalin [4]

Answer:

264 cu

Step-by-step explanation:

This is because you have to add it from all sides, the first prism is 144 cu and then the second one is, 120, and when you add both of those together you get 264

Thus the answer is....264

5 0

2 years ago

The length of a rectangle is represented by x, the width is represented by 6x^2-4x-2. What is the area of the rectangle?

navik [9.2K]

Answer:

$2x(3x^2-2x-1)$

Step-by-step explanation:

Area of a rectangle is simply length times width

{ $6x^2-4x-2$ } * x would be the area

That can be simplified to

$6x^3-4x^2-2x$

Which can be further simplified to

$2x(3x^2-2x-1)$

6 0

3 years ago

The ratio of 1.4/7 in the simplest form

sweet [91]

If you want to convert 1.4 into a fraction, you can do it like this:

1 equals to 1/1 and 0.4 equals to 4/10 which simplifies to 2/5.
The result is 1 2/5.

If you want to write the ratio of 1.4/7 in the simplest form, you can do this using the following steps:

1.4/7 = 1 2/5 / 7 = 7/5 / 7 = 7/5 * 1/7 = 1/5 = 0.2

The correct result would be 1/5.

3 0

3 years ago

If you’re good with permutations in math 30 please help and answer this easy question below real answers only

Len [333]

Answer: d) 2730

Step-by-step explanation:

1st movie and 2nd movie and 3rd movie

15 videos × 14 remaining videos × 13 remaining videos = 2730

6 0

3 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

7 0

3 years ago