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Angelina_Jolie [31]

3 years ago

11

Corey was scuba diving while his friend Zoe was climbing a mountain. If Corey was at an elevation of -155 feet, and Zoe was at a

n elevation of 560 feet, what was the difference in elevation (in feet) between Corey and Zoe?

1 answer:

Volgvan3 years ago

5 0

Answer: 715 feet

| -155 - 560 | = 715ft

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Solve the equation. StartFraction dy Over dx EndFraction equals5 x Superscript 4 Baseline (1 plus y squared )Superscript three h

dangina [55]

Answer:

Step-by-step explanation:

To solve the differential equation

dy/dx = 5x^4(1 + y²)^(3/2)

First, separate the variables

dy/(1 + y²)^(3/2) = 5x^4 dx

Now, integrate both sides

To integrate dy/(1 + y²)^(3/2), use the substitution y = tan(u)

dy = (1/cos²u)du

So,

dy/(1 + y²)^(3/2) = [(1/cos²u)/(1 + tan²u)^(3/2)]du

= (1/cos²u)/(1 + (sin²u/cos²u))^(3/2)

Because cos²u + sin²u = 1 (Trigonometric identity),

The equation becomes

[1/(1/cos²u)^(3/2) × 1/cos²u] du

= cos³u/cos²u

= cosu

Integral of cosu = sinu

But y = tanu

Therefore u = arctany

We then have

cos(arctany) = y/√(1 + y²)

Now, the integral of the equation

dy/(1 + y²)^(3/2) = 5x^4 dx

Is

y/√(1 + y²) = x^5 + C

So

y - (x^5 + C)√(1 + y²) = 0

is the required implicit solution

5 0

3 years ago

(-x2y3)*(12xy3)<br>(-2xy2)3

Leokris [45]

Assuming the numbers are exponents expect the coefficient of 2, top expression, -12x^3y^6, multiply bottom
and= 72x^4y^8

4 0

2 years ago

The cell phone company decided to collect similar data but this time they collected information of texting speed according to ag

Ludmilka [50]

Step-by-step explanation:

As we know, Scatter plots resembles line graphs in that they are plotted on the x and y axis and the purpose of a scatter plot is to shows if the variables are related to each other. If yes, they can draw a line of best fit which can expose or go through most of the points in the plot.

Given the information in the question, we know:

"they collected information of texting speed according to age"

=> We have 2 variables here which are age and (texting speed)

Step 1:

Label the x-axis the input variable (age)
Label the y-axis the output variable (texting speed)

Step 2: Plot the points according to age and texting speed

Hence, from that plot, it could have two cases in the relation of age and texting speed which are "the more time, more text is written", or "the less time, more text is written".

7 0

3 years ago

Read 2 more answers

Using coupons, you spend $120 grocery shopping. This is 80% of the total retail price of the groceries. How much would you spend

mr Goodwill [35]

Answer: The correct answer is $150.

Step-by-step explanation: If you did not use the coupons while shopping you would have spent $150. The method to calculate this is to divide the amount that you paid by the percentage that you paid.

The formula that you need to use is $120/.8 = $150.

7 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