Ask question

Ask question

All categories

3 years ago

14

On a vacation in Puerto Rico rose jumped off a cliff into a river in El Yunque go forest reserve her height as a function of tim

e could be modeled by the function h(t) = -16tequals -16t ^2 + 16t + 480, where T is the time after jumping in seconds and H is the height in feet how many seconds did it take Rose to hit the water

1 answer:

gayaneshka [121]3 years ago

6 0

Answer: t = 6 s

Step-by-step explanation:

When Rose hit the water, the height of the jump is 0, so: h(t) = 0.

Resolving it, we have:

-16t² + 16t + 480 = 0

To simplify the second degree equation, divided it by 16:

-t² + t + 30 = 0

Now, solve the equation using Bhaskara:

Δ = 1² - 4. (-1) . 30

Δ = 1 + 120

Δ = 121

t = $\frac{-1+\sqrt{121} }{-2}$

t₁ = $\frac{-1 + 11}{-2}$ = - 5

t₂ = $\frac{-11-11}{-2}$ = 6

Since time can be negative, it took t = 6s for Rose to hit the water.

You might be interested in

Simplify the expression x8 y-26/x14 y-5 X x-39 y-21.

KonstantinChe [14]

$\frac{x^8y^{-26}}{x^{14}y^{-5}}[x^{-36}y^{-21}]$

We will apply these following power rule

$x^m*x^n = x^{m+n}$
$x^m/x^n=x^{m-n}$

$[ \frac{x^8}{x^{14}}][ \frac{y^{-26}}{y^{-5}}][x^{-39}y^{-21}]$
$[x^{8-14}][y^{-26-5}][x^{-39}y^{-21}]$
$x^{-6}y^{-21}[x^{-39}y^{-21}]$
$[x^{-6-39}][y^{-21-21}]$
$x^{-45}y^{-42}$

8 0

3 years ago

45.15 ÷ 21 pleases solve it for me

sergiy2304 [10]

Answer:

2.15

Step-by-step explanation:

5 0

3 years ago

A consumer products company is formulating a new shampoo and is interested in foam height (in mm). Foam height is approximately

Genrish500 [490]

Answer:

a) 0.057

b) 0.5234

c) 0.4766

Step-by-step explanation:

a)

To find the p-value if the sample average is 185, we first compute the z-score associated to this value, we use the formula

$z=\frac{\bar x-\mu}{\sigma/\sqrt N}$

where

$\bar x=mean\; of\;the \;sample$

$\mu=mean\; established\; in\; H_0$

$\sigma=standard \; deviation$

N = size of the sample.

So,

$z=\frac{185-175}{20/\sqrt {10}}=1.5811$

$\boxed {z=1.5811}$

As the sample suggests that the real mean could be greater than the established in the null hypothesis, then we are interested in the area under the normal curve to the right of 1.5811 and this would be your p-value.

We compute the area of the normal curve for values to the right of 1.5811 either with a table or with a computer and find that this area is equal to 0.0569 = 0.057 rounded to 3 decimals.

So the p-value is

$\boxed {p=0.057}$

b)

Since the z-score associated to an α value of 0.05 is 1.64 and the z-score of the alternative hypothesis is 1.5811 which is less than 1.64 (z critical), we cannot reject the null, so we are making a Type II error since 175 is not the true mean.

We can compute the probability of such an error following the next steps:

<u>Step 1 </u>

Compute $\bar x_{critical}$

$1.64=z_{critical}=\frac{\bar x_{critical}-\mu_0}{\sigma/\sqrt{n}}$

$\frac{\bar x_{critical}-\mu_0}{\sigma/\sqrt{n}}=\frac{\bar x_{critical}-175}{6.3245}=1.64\Rightarrow \bar x_{critical}=185.3721$

So <em>we would make a Type II error if our sample mean is less than 185.3721</em>.

<u>Step 2</u>

Compute the probability that your sample mean is less than 185.3711

$P(\bar x < 185.3711)=P(z< \frac{185.3711-185}{6.3245})=P(z$

So, <em>the probability of making a Type II error is 0.5234 = 52.34% </em>

c)

<em>The power of a hypothesis test is 1 minus the probability of a Type II error</em>. So, the power of the test is

1 - 0.5234 = 0.4766

3 0

3 years ago

Solve screen shot attachment EASY

Leona [35]

Answer:

$x = \frac{-3d-9}{k-d}$

Step-by-step explanation:

Step 1: Write out equation

d(-3 + x) = kx + 9

Step 2: Distribute <em>d</em>

-3d + dx = kx + 9

Step 3: Subtract 9 on both sides

-3d + dx - 9 = kx

Step 4: Subtract dx on both sides

-3d - 9 = kx - dx

Step 5: Factor out <em>x</em>

-3d - 9 = x(k - d)

Step 6: Divide both sides by k - d

$x = \frac{-3d-9}{k-d}$

8 0

3 years ago

I don’t get this explain pls

Ludmilka [50]

Im not 100% sure but RT and US are the same distance so theyd be the same...

6 0

3 years ago