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IrinaVladis [17]
3 years ago
10

A stretched string is 2.11 m long and has a mass of 19.5 g. When the string is oscillated at 440 Hz, which is the frequency of t

he standard A pitch that orchestras tune to, transverse waves with a wavelength of 16.1 cm travel along the string. Calculate the tension in the string.
Physics
1 answer:
Katarina [22]3 years ago
7 0

Answer:

efsfefsfsdf

Explanation:

efsfesef

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Molecules move from areas of low concentration to high concentration through a process of
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Active transport move molecules from low concentrations to high concentrations.
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In a large restaurant an average 60% customers ask for water with their meal. A random sample of 10 customers is selected. Find
Gekata [30.6K]

Answer:

a)P(6)=0.25

b)p(x

c)p(x\geq3)=0.9878

d)\sigma=\sqrt{2.4}=1.5492

Explanation:

From the question we are told that:

Population percentage p_\%=\60%

Sample size n=10

Let x =customers ask for water

Let y =customers dose not ask for water with their meal  

Generally the equation for y is mathematically given by

y=1-p_\%\\y=1-0.60\\y=0.40

Generally the equation for pmf p(x) is mathematically given by

P(x)=10C_x (0..6)^x(0.4)^{10-x}

a)

Generally the probability that exactly 6 ask for water is mathematically given by

P(x)6=10C_6 (0..6)^6(0.4)^{10-6}

P(6)=0.25

b)

Generally the probability that  less than 9 ask for water with meal  is mathematically given by

p(xg)

p(x

p(x

p(x

c)

Generally the probability that  at least 3 ask for water with meal  is mathematically given by

p(x\geq3)=1-p(x

p(x\geq3)=1-[p(0)+p(1)+p(2)]

p(x\geq3)=1-[0.00001+0.0015+0.0106]

p(x\geq3)=1-[0.0122]

p(x\geq3)=0.9878

d)

Generally the mean and standard deviation of sample size is mathematically given by

Mean

\=x=np=10(0.6)=6

Standard deviation

v(x)=npq=10(0.6)(0.4)=2.4

\sigma=\sqrt{2.4}=1.5492

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3 years ago
How to calculate sound of an echo ​
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3 years ago
A transformer has input voltage and current of 12 V and 4 A
Advocard [28]

Explanation:

It is given that,

Input voltage, V_i=12\ V

Input current, I_i=4\ A

Output current, I_o=0.8\ A

Number of turns in the secondary side of transformer, N_s=1177

We need to find the number of turns in the primary side of the transformer. The current to the number of turns in the input and output is given by :

\dfrac{N_s}{N_p}=\dfrac{I_p}{I_s}

Substituting all the above values

So,

N_p=\dfrac{N_sI_s}{I_p}\\\\N_p=\dfrac{1177\times 4}{0.8}\\\\N_p=5885

So, the number of turns in primary side of the transformer is 5885.

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