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DerKrebs [107]
3 years ago
14

A national college researcher reported that 67%of students who graduated from high school in 2012 enrolled in college. Twenty-ni

ne high school graduates are sampled What is the mean number who enroll in college in a sample of 29 high school graduates? Round the answer to two decimal places. a) What is the standard deviation of the number who enroll in college in a sample of 29 high school graduates?
Physics
1 answer:
dsp733 years ago
4 0

Answer:

Explanation:

Theorem of Binomial Distribution will apply here.

n = 29 , p = .67 , q = 0.33

mean = np = 29 x .67 = 19.43

Standard Deviation = √npq

= √29 x .67 x .33

= √6.4

= 2.53

=

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Electromagnetic transverse waves
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A brick sliding across a smooth floor has a coefficient of static friction of s=0.23 and a coefficient of kinetic friction of k=
pantera1 [17]

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0.029325

Explanation:

0.23x0.15=0.0345

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The cornea behaves as a thin lens of focal length approximately 1.80 {\rm cm}, although this varies a bit. The material of which
spayn [35]

Answer:

The height of the image will be "1.16 mm".

Explanation:

The given values are:

Object distance, u = 25 cm

Focal distance, f = 1.8 cm

On applying the lens formula, we get

⇒  \frac{1}{v} -\frac{1}{u} =\frac{1}{f}

On putting estimate values, we get

⇒  \frac{1}{v} -\frac{1}{(-25)} =\frac{1}{1.8}

⇒  \frac{1}{v} =\frac{1}{1.8} -\frac{1}{25}

⇒  v=1.94 \ cm

As a result, the image would be established mostly on right side and would be true even though v is positive.

By magnification,

m=\frac{v}{u} and m=\frac{h_{1}}{h_{0}}

⇒  \frac{v}{u} =\frac{h_{1}}{h_{0}}

⇒  \frac{1.94}{25}=\frac{{h_{1}}}{15}

⇒  {h_{1}}=1.16 \ mm

8 0
3 years ago
3. How does the mass of a proton compare to the mass of an electron?
o-na [289]

Answer:

Proton, stable subatomic particle that has a positive charge equal in magnitude to a unit of electron charge and a rest mass of 1.67262 × 10−27 kg, which is 1,836 times the mass of an electron.

7 0
3 years ago
An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.80 kW. A receiving antenna 7
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Answer:0.1759 v

Explanation:

Intensity of wave at receiver end is

I=\frac{P_{avg}}{A}

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Amplitude of electric field at receiver end

E_{max}=\sqrt{2I\mu _0c}

Amplitude of induced emf

=E_{max}d

=\sqrt{2\times 7.29\times 10-6\times 4\pi \times 3\times 10^8}\times 0.75

=17.591\times 10^{-2}=0.1759 v

7 0
3 years ago
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