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2 years ago

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molly is designing a new drone for the air force. the drone is supposed to be 30 feet long and 4 feet tall. she is building a pr

ototype for testing that will be 12 feet long. how tall should the prototype be?

2 answers:

nata0808 [166]2 years ago

8 0

First you set up your ratio: Length to height.
The Drone is 30/4 which can be equivalent to 7.5.
The prototype is 12/x which can be equivalent to 7.5 as well. Re-write your equation to solve for x.
x*7.5 = 12 ---> x=12/7.5. = 1.6. Now plug in 1.6 as x to test your equation so you get an answer of 7.5: 12/1.6 = 7.5.
So, the prototype for testing should be 12 feet long and 1.6 feet (19.2 inches) high.

Sholpan [36]2 years ago

7 0

19.2 i think tell me if i'm right gl

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Answer:

$y=32000(1+0.08)^x$

Step-by-step explanation:

Exponential growth function is $y=a(1+r)^x$

Where 'a' is the initial population

r is the rate of growth and x is the time period in years

a steady population of 32,000. So initial population is 32,000

an increase of 8% per year. the rate of increase is 8% that is 0.08

a= 32000 and r= 0.08

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- At 2 hours, the traveler is 725 miles from home.

- At 3 hours, the distance is constant, at 880 miles. --> TRUE.

- The total distance from home after 6 hours is 1,062.5 miles --> TRUE.

Step-by-step explanation:

The function D(t) is defined as follows:

$D(t) = 300t+125$ for $t< 2.5$

$D(t) = 880$ for $2.5 \leq 3.5$

$D(t) = 75t+612.5$ for $t\leq 6$

Where

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Now let's analyze the different statements:

- The starting distance, at 0 hours, is 300 miles. --> FALSE. In fact, if we substitute t = 0 into the 1st equation, we get

$D(0) = (300)(0)+125 = 125$

So, the distance at t = 0 is 125 miles.

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$D(2) = (300)(2)+125 = 725$

- At 2.5 hours, the traveler is 875 miles from home. --> FALSE. In fact, for t=2.5 we have to use the 2nd equation, which states that the distance is:

D(t) = 880

So, not 875 miles.

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D(t) = 880

- The total distance from home after 6 hours is 1,062.5 miles --> TRUE. In fact, if we replace t = 6 into the last equation,

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Answer: $\int\limits^a_E {\int\limits^a_E {\int\limits^a_E {xy} } \, dV$ = 1087.5

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An equation of a plane is found with a point and a normal vector. <u>Normal</u> <u>vector</u> is a perpendicular vector on the plane.

Given the points, determine the vectors:

P = (5,0,0); Q = (0,9,0); R = (0,0,4)

vector PQ = (5,0,0) - (0,9,0) = (5,-9,0)

vector QR = (0,9,0) - (0,0,4) = (0,9,-4)

Knowing that cross product of two vectors will be perpendicular to these vectors, you can use the cross product as normal vector:

n = PQ × QR = $\left[\begin{array}{ccc}i&j&k\\5&-9&0\\0&9&-4\end{array}\right]$ $\left[\begin{array}{ccc}i&j\\5&-9\\0&9\end{array}\right]$

n = 36i + 0j + 45k - (0k + 0i - 20j)

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Then, replacing with point P and normal vector n:

$36(x-5) + 20(y-0) + 45(z-0) = 0$

The equation is: 36x + 20y + 45z - 180 = 0

Second, in evaluating the triple integral, set limits:

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$z = \frac{180-36x-20y}{45}$

When z = 0:

$y = 9 + \frac{-9x}{5}$

When z=0 and y=0:

x = 5

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$\int\limits^5_0 {\int\limits {\int\ {xy} \, dz } \, dy } \, dx$

Calculating:

$\int\limits^5_0 {\int\limits {\int\ {xyz} \, dy } \, dx$

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$\frac{1}{45} \int\limits^5_0 {90xy^{2}-18x^{2}y^{2}-\frac{20}{3} xy^{3} } \, dx$

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$\frac{1}{45} [30375-60750+118462.5-39150]$

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2 years ago

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Nadusha1986 [10]

Answer:

Step-by-step explanation:

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For the null hypothesis,

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Since the population standard deviation is not given, the distribution is a student's t.

Since n = 80,

Degrees of freedom, df = n - 1 = 80 - 1 = 79

t = (x - µ)/(s/√n)

Where

x = sample mean = 15.6

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s = samples standard deviation = 4.5

t = (15.6 - 17)/(4.5/√80) = - 2.78

We would determine the p value using the t test calculator. It becomes

p = 0.0034

Since alpha, 0.05 > than the p value, 0.0043, then we would reject the null hypothesis.

The data supports the professor’s claim. The average number of hours per week spent studying for students at her college is less than 17 hours per week.

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2 years ago