PLEASE HELP, WILL GIVE BRAINLIEST!

The USS Enterprise was 1,123 feet in length.

Kitty [74]

Answer:

The scale for a model that is 30 inches long is 449.2 inches.

The scale for a model that is 2 feet long is 561.5 feet.

Step-by-step explanation:

Consider the provided information.

The USS Enterprise was 1,123 feet in length.

Part (A) What is the scale for a model that is 30 inches long?

1 feet = 12 inches.

First convert the length of USS enterprise in inches.

1×1,123 feet = 12×1,123 inches

1,123 feet = 13476 inches

Divide original length with the length of model.

The scale for the model is:

$\frac{13476}{30}=449.2\ inches$

Hence, the scale for a model that is 30 inches long is 449.2 inches.

Part (B) What is the scale for a model that is 2 feet long?

Since, both the units are in feet, so simply divide original length with the length of model.

The scale for the model is:

$\frac{1123}{2}=561.5\ feet$

Hence, the scale for a model that is 2 feet long is 561.5 feet.

6 0

3 years ago

Ellie earns money by raking leaves. She earns $2.25 for each bag she fills with leaves. This week, she earns $24,75. How many ba

Pepsi [2]

Answer:

she raked 11 bags of leaves

Step-by-step explanation:

24.75 divided by 2.25= 11

8 0

2 years ago

ecn 221 The sodium content of a popular sports drink is listed as 206 mg in a 32-oz bottle. Analysis of 14 bottles indicates a s

spayn [35]

Answer:

$H_{0}: \mu = 206\text{ mg}\\H_A: \mu \neq 206\text{ mg}$

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 206 mg

Sample mean, $\bar{x}$ = 217.5 mg

Sample size, n = 14

Sample standard deviation, s = 14.9 mg

Claim:

The mean sodium content for the sports drink is not 206 mg. It is different than 206 mg.

Thus, we design the null and the alternate hypothesis

$H_{0}: \mu = 206\text{ mg}\\H_A: \mu \neq 206\text{ mg}$

We use two-tailed t test to perform this hypothesis.

3 0

2 years ago

Let v1 =(-6,4) and v2=(-3,6) compute the following what i sthe angle between v1 and v2

Agata [3.3K]

$\bf ~~~~~~~~~~~~\textit{angle between two vectors }
\\\\
cos(\theta)=\cfrac{\stackrel{\textit{dot product}}{u \cdot v}}{\stackrel{\textit{magnitude product}}{||u||~||v||}} \implies 
\measuredangle \theta = cos^{-1}\left(\cfrac{u \cdot v}{||u||~||v||}\right)\\\\
-------------------------------$

$\bf \begin{cases}
v1=\ \textless \ -6,4\ \textgreater \ \\
v2=\ \textless \ -3,6\ \textgreater \ \\
------------\\
v1\cdot v2=(-6\cdot -3)+(4\cdot 6)\\
\qquad \qquad 42\\
||v1||=\sqrt{(-6)^2+4^2}\\
\qquad \sqrt{52}\\
||v2||=\sqrt{(-3)^2+6^2}\\
\qquad \sqrt{45}
\end{cases}\implies \measuredangle \theta =cos^{-1}\left( \cfrac{42}{\sqrt{52}\cdot \sqrt{45}} \right)
\\\\\\
\measuredangle \theta =cos^{-1}\left( \cfrac{42}{\sqrt{2340}} \right)\implies \measuredangle \theta \approx 29.74488129694^o$

8 0

3 years ago

Y+1=-2x-3 in standard form

nignag [31]

Hope this answer is helpful for you

7 0

3 years ago