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3 years ago

10

The table below shows the results of a survey of 1000 American college students and their age breakdown. Use the table to answer

questions

2 answers:

Yakvenalex [24]3 years ago

7 0

The answer for this is c

mel-nik [20]3 years ago

3 0

Answer:

c

Step-by-step explanation:

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onsider two populations for which μ1 = 38, σ1 = 2, μ2 = 23, and σ2 = 4. Suppose that two independent random samples of sizes n1

levacccp [35]

Answer:

Center = 15

Spread = 0.599

Shape = Normal

Step-by-step explanation:

The provided information is:

$\mu_1=38 \ \ \ \ \ \ \ \ \mu_2 = 23\\\sigma_1 =2 \ \ \ \ \ \ \ \ \ \sigma_2=4\\n_1=45 \ \ \ \ \ \ \ \ n_2 =59$

Thus the center(mean) of the distribution is:

$\begin{aligned}Mean &= \mu_1-\mu_2\\&=38-23\\&=15\end{aligned}$

The spread (standard deviation) of the distribution is:

$\begin{aligned}\textrm{Standard deviation}&=\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}\\&=\sqrt{\frac{4}{45}+\frac{16}{59}}\\&=0.599\end{aligned}$

The shape of the distribution is also normally distributed.

3 0

3 years ago

Find an equation for these word problems giving brainliest

ch4aika [34]

25=.75x or 25 divided by .75= x
7 + 4= x so 7+4 = 11

5 0

3 years ago

Sin 2x + tan 2x = 4tan/ 1- tan^4 prove that

nalin [4]

$\dfrac{4\tan x}{1-\tan^4x}=\dfrac{4\tan x}{(1-\tan^2x)(1+\tan^2x)}$
$=\dfrac{4\tan x}{(1-(\sec^2x-1)\sec^2x}$
$=\dfrac{4\tan x\cos^2x}{2-\sec^2x}$
$=\dfrac{4\sin x\cos x}{2-\sec^2x}$
$=\dfrac{4\sin x\cos^3x}{2\cos^2x-1}$
$=\dfrac{2\sin2x\cos^2x}{\cos2x}$
$=2\tan2x\cos^2x$
$=2\tan2x\left(\dfrac{1+\cos2x}2\right)$
$=\tan2x(1+\cos2x)$
$=\tan2x+\sin2x$

7 0

3 years ago

Factor: 128 + 2d^3 <br><br> cubes lol

BARSIC [14]

The factorization if the given expression 128 + 2d³ is 2(64 + d³).

<h3>Factorization</h3>

128 + 2d³

There are two parameters in the expression;

128
2d³

The common factors between 128 and 2d³ is 2

So,

128 + 2d³

= 2(64 + d³)

Learn more about factorization:

brainly.com/question/25829061

#SPJ1

4 0

2 years ago

Given ABCD is a parallelogram ABE equals 180° CBE is congruent to CEB prove DAE is congruent to CEA

katrin2010 [14]

Answer:

The proved is given below.

Step-by-step explanation:

Given ABCD is a parallelogram and also ∠ABE equals 180°, ∠CBE ≅ ∠CEB

We have to prove that ∠DAE ≅ ∠CEA

As ∠CEA=∠CBE → (1)

and given ABCD is a parallelogram implies AD||CB

∴ ∠DAE=∠CBE → (2) as these are corresponding angles.

From equation (1) and (2), we get

∠DAE=∠CEA

Hence, ∠DAE is congruent to ∠CEA

Hence Proved

6 0

3 years ago