Which group of college students is typically offered scholarship?

stellarik [79]

I’m not sure what you mean by this? Could u help me out and I’ll answer it

6 0

3 years ago

what should I write for literature will give brain list 3 paragraphs :You woke up this morning to 15 feet of snow and can’t open

stiv31 [10]

Answer: I would call someone who worked with that stuff and just have them do it. And if not then I would grab a shovel and start shoveling it myself, or maybe call a friend of family. Most likely I would call 911 and just have them help me!

Explanation: Hope I helped!

4 0

3 years ago

Read 2 more answers

A bat emits a sound at a frequency of 3. 00 × 104 hz as it approaches a wall. The bat detects beats such that the frequency of t

OLga [1]

Taking into account the Doppler efect, the speed of the bat is closest to 5.02 $\frac{m}{s}$ .

The Doppler effect is defined as the apparent frequency change of a wave produced by the relative movement of the source with respect to its observer. In other words, this effect is the change in the perceived frequency of any wave motion when the sender and receiver, or observer, move relative to each other.

The following expression is considered the general case of the Doppler effect:

$f1=f\frac{v+-vR}{v+-vE}$

Where:

f1, f: Frequency perceived by the receiver and frequency emitted by the issuer respectively. Its unit of measurement in the International System (S.I.) is the hertz (Hz), which is the inverse unit of the second (1 Hz = 1 s⁻¹)
v: Velocity of propagation of the wave in the medium. It is constant and depends on the characteristics of the medium. In this case, the speed of sound in air is considered to be 340 m/s
vR, vE: Speed of the receiver and the sender respectively. Its unit of measure in the S.I. is the m/s
±, ∓:

We will use the + sign:

In the numerator if the receiver approaches the sender

In the denominator if the sender moves away from the receiver

We will use the sign -:

In the numerator if the receiver moves away from the emitter

In the denominator if the sender approaches the receiver

In this case, a bat emits a sound at a frequency of 3.00×10⁴ Hz= 30,000 Hz as it approaches a wall.

The bat detects beats such that the frequency of the echo is 900 hz higher than the frequency the bat is emitting.

So, you know:

f1= 30,000 Hz + 900 Hz= 30,900 Hz
f= 30,000 Hz
v= 340 m/s
vE=vR= vbat
Receiver approaches the sender

Replacing in the expression is considered the general case of the Doppler effect:

$30900 Hz=30000 Hz\frac{340 \frac{m}{s} +vbat}{340\frac{m}{s} -vbat}$

Solving:

$\frac{30900 Hz}{30000 Hz} =\frac{340 \frac{m}{s} +vbat}{340\frac{m}{s} -vbat}$

$1.03=\frac{340 \frac{m}{s} +vbat}{340\frac{m}{s} -vbat}$

1.03×( 340 $\frac{m}{s}$ -vbat)= 340 $\frac{m}{s}$ + vbat

1.03×340 $\frac{m}{s}$ - 1.03×vbat= 340 $\frac{m}{s}$ + vbat

350.2 $\frac{m}{s}$ - 1.03×vbat= 340 $\frac{m}{s}$ + vbat

350.2 $\frac{m}{s}$ - 340 $\frac{m}{s}$ = vbat + 1.03×vbat

10.2 $\frac{m}{s}$ = 2.03×vbat

10.2 $\frac{m}{s}$ ÷ 2.03 = vbat

5.02 $\frac{m}{s}$ = vbat

In summary, the speed of the bat is closest to 5.02 $\frac{m}{s}$ .

Learn more about the Doppler effect:

6 0

2 years ago

Antique Accents tracks their daily profits and has found that the distribution of profits is approximately normal with a mean of

mezya [45]

Answer;

a) 0.434

b) 0.983

c) 0.367

Explanation:

The exact question with the given parameters wasn't obtained online, but the same question, albeit with different parameters is then obtained. Hopefully, this Helps to solve the complete question with the required parameters.

Antique Accents tracks their daily profits and has found that the distribution of profis is approximately normal with a mean of $17,700.00 and a standard deviation of about $900.00. Using this information, answer the following questions For full marks your answer should be accurate to at least three decimal places. Compute the probability that tomorrow's profit will be

a) less than $16,791 or greater than $18,231

b) greater than $15,783

c) between $17,997 and $20,130

Solution

This is a normal distribution problem with

Mean = μ = $17,700

Standard deviation = σ = $900

a) less than $16,791 or greater than $18,231. P(x < 16,791) or P(X > 18,231) = P(X < 16,791) + P(x > 18,231)

We first standardize 16,791 and 18,231

The standardized score for any value is the value minus the mean then divided by the standard deviation.

For 16791

z = (x - μ)/σ = (16791 - 17700)/900 = - 1.01

For 18231

z = (x - μ)/σ = (18231 - 17700)/900 = 0.59

To determine the required probability

P(X < 16,791) + P(x > 18,231) = P(z < -1.01) + P(z > 0.59)

We'll use data from the normal probability table for these probabilities

P(X < 16,791) + P(x > 18,231) = P(z < -1.01) + P(z > 0.59)

P(z < -1.01) = 0.15625

P(z > 0.59) = 1 - (z ≤ 0.59) = 1 - 0.7224 = 0.2776

P(X < 16,791) + P(x > 18,231) = P(z < -1.01) + P(z > 0.59) = 0.15625 + 0.2776 = 0.43385 = 0.434 to 3 d.p

b) greater than $15,783. P(x > 15783)

We standardize 15783

z = (x - μ)/σ = (15783 - 17700)/900 = -2.13

To determine the required probability

P(x > 15783) = P(z > -2.13)

We'll use data from the normal probability table for this probability

P(x > 15783) = P(z > -2.13) = 1 - P(z ≤ - 2.13)

= 1 - 0.01659 = 0.98341 = 0.983 to 3 d.p.

c) between $17,997 and $20,130.

P(17,997 < x < 20,130)

We first standardize 17,997 and 20,130

The standardized score for any value is the value minus the mean then divided by the standard deviation.

For 17,997

z = (x - μ)/σ = (17,997 - 17700)/900 = 0.33

For 20,130

z = (x - μ)/σ = (20,130 - 17700)/900 = 2.70

To determine the required probability

P(17,997 < x < 20,130) = P(0.33 < x < 2.70)

We'll use data from the normal probability table for these probabilities

P(17,997 < x < 20,130) = P(0.33 < x < 2.70)

= P(z < 2.70) - P(z < 0.33)

= 0.99653 - 0.62930

= 0.36723 = 0.367 to 3 d.p.

Hope this Helps!!!

3 0

3 years ago

16. Several members of a little league team begin wearing their baseball caps backward and soon the

77julia77 [94]

Answer:

the answer to this question is social darwinism

4 0

3 years ago