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3 years ago

PLEASE HELP WILL MARK BRAINLIEST!!

Mathematics

1 answer:

Maru [420]3 years ago

3 0

Answer:

V ≈ 21941.33 yd.

Step-by-step explanation:

I just calculated each side with the help of my trusty calculator! Hope this helped! Comment if you are still confused!

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The lengths, in inches, of adult corn snakes are normally distributed with a population standard deviation of 6 inches and an un

liubo4ka [24]

Answer: 2.460

Step-by-step explanation:

The formula of Margin of Error for (n<30):-

$ME=t_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

Given : Sample size : n= 18

Level of confidence = 0.90

Significance level : $\alpha=1-0.90=0.1$

Using the t-distribution table ,

Critical value : $t_{n-1, \alpha/2}=t_{17,0.05}= 1.7396$

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Then, we have

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5 0

3 years ago

PLEASE HELP! WILL MARK BRAINLIEST!

makkiz [27]

Answer:

The maximum height a rider will experience is 55 feet.

Step-by-step explanation:

Let's start writing the function that defines the path of a seat on the new Ferris wheel. This function will depend of the variable ''t'' which is time.

$X(t)=(x,y)$

In which $X(t)$ are the coordinates of the seat (the x - coordinate and the y - coordinate) that depend from time.

$''x''$ and $''y''$ are functions that depend from the variable ''t''.

For this exercise :

$X(t)=[-25sin(\frac{\pi}{30}t);-25cos(\frac{\pi}{30}t)+30]$

In order to find the maximum height a rider will experience we will study the behaviour of the y - component from the function $X(t)$ .

The function to study is $y(t)=-25cos(\frac{\pi}{30}t)+30$

To find its maximum, we will derivate this function and equalize it to 0. Doing this, we will find the ''critical points'' from the function.

⇒ $y(t)=-25cos(\frac{\pi}{30}t)+30$ ⇒

$y'(t)=\frac{5}{6}\pi sin(\frac{\pi}{30}t)$

Now we equalize $y'(t)$ to 0 ⇒

$y'(t)=0$ ⇒ $\frac{5}{6}\pi sin(\frac{\pi}{30}t)=0$

In this case it is easier to look for the values of ''t'' that verify :

$sin(\frac{\pi}{30}t)=0$

Now we need to find the values of ''t''. We know that :

$sin(0)=0\\\\sin(\pi)=0\\sin(-\pi)=0$

Therefore we can write the following equivalent equations :

$\frac{\pi}{30}t=0$ (I)

$\frac{\pi}{30}t=\pi$ (II)

$\frac{\pi}{30}t=-\pi$ (III)

From (I) we obtain $t_{1}=0$

From (II) we obtain $t_{2}=30$

And finally from (III) we obtain $t_{3}=-30$

We found the three critical points of $y(t)$ . To see if they are either maximum or minimum we will use the second derivative test. Let's calculate the second derivate of $y(t)$ :