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Gwar [14]
3 years ago
14

the cost of an annual tuition at a university increased from 10,500 to 11,300 what is the percent increase in tuition to the nea

rest tenth of a percent
Mathematics
1 answer:
Maurinko [17]3 years ago
5 0
Percent increase
find increase first
10500 to 11300
11300-10500=800
so
percent increase
change/original
origianal=10500
change=800
800/10500=8/105=0.0761
percent means parts out of 100
0.0761/1 times 100/100=7.61/100=7.61%

rond 7.61% to tenth or to 7.6%

7.6%

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A sine function has an amplitude of 3, a period of pi, and a phase shift of pi/2. What is the y-intercept of the function?
jarptica [38.1K]

Based on the calculations, we can logically deduce that the y-intercept of this sine function is equal to: A. 3.

<u>Given the following data:</u>

  • Amplitude = 3
  • Period = π
  • Phase shift = π/2

<h3>How to determine the y-intercept of this function?</h3>

Mathematically, a sine function is modeled by this equation:

y = Asin(ωt + ø)

<u>Where:</u>

  • A represents the amplitude.
  • ω represents angular velocity.
  • t represents the period.
  • ø represents the phase shift.

Also, the period of a sine wave is given by:

t = 2π/ω

2 = 2π/ω

ω = 2

Substituting the given parameters into the equation, we have;

y = 3sin(2t + π/2)

At t = 0, we have:

y = 3sin(2(0) + π/2)

y = 3sin(π/2)

y = 3sin(90)

y = 3 × 1

y = 3.

In conclusion, we can logically deduce that the y-intercept of this sine function is equal to 3.

Read more on phase shift here: brainly.com/question/27692212

#SPJ1

5 0
2 years ago
3/5 + 1/8 = ??<br><br> Simplify answer fully
mylen [45]
Hope this helps you...!!!

8 0
3 years ago
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Ken grew 4/5 of an inch last year. Sang grew 3/8 of an inch. Who grew more and by how much?
bonufazy [111]
Ken grew .425 of an inch more than sang because 4/5 is .8 and 3/8 is .375 what you do is you take those two numbers and subtract the smaller one from the bigger one and you get .425 and ken had the bigger growth. plz rate me brainliest
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3 years ago
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Can someone help pls​
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4 0
3 years ago
Each of 16 students measured the circumference of a tennis ball by four different methods, which were: A: Estimate the circumfer
almond37 [142]

Answer:

Following are the solution to the given equation:

Step-by-step explanation:

Please find the complete question in the attachment file.

In point a:

\to \mu=\frac{\sum xi}{n}

       =22.8

\to \sigma=\sqrt{\frac{\sum (xi-\mu)^2}{n-1}}

       =\sqrt{\frac{119.18}{16-1}}\\\\ =\sqrt{\frac{119.18}{15}}\\\\ = \sqrt{7.94533333}\\\\=2.8187

In point b:

\to \mu=\frac{\sum xi}{n}

       =20.6875  

\to \sigma=\sqrt{\frac{\sum (xi-\mu)^2}{n-1}}

       =\sqrt{\frac{26.3375}{16-1}}\\\\=\sqrt{\frac{26.3375}{15}}\\\\ =\sqrt{1.75583333}\\\\ =1.3251

In point c:

 \to \mu=\frac{\sum xi}{n}

         =21  

\to \sigma=\sqrt{\frac{\sum (xi-\mu)^2}{n-1}}

       =\sqrt{\frac{2.62}{16-1}}\\\\ =\sqrt{\frac{2.62}{15}} \\\\= \sqrt{0.174666667}\\\\=0.4179

In point d:

\to \mu=\frac{\sum xi}{n}

       =20.8375  

\to \sigma=\sqrt{\frac{\sum (xi-\mu)^2}{n-1}}

       =\sqrt{\frac{8.2975}{16-1}}\\\\ =\sqrt{\frac{8.2975}{15}} \\\\  =\sqrt{0.553166667} \\\\ =0.7438

6 0
3 years ago
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