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3 years ago

A recent survey at a high school found that 3 out of 5 seniors played atleast one sport during high school.

Mathematics

2 answers:

stira [4]3 years ago

8 0

D would show three fifths of seniors played a sport

Lena [83]3 years ago

4 0

Answer:

D, 150 Seniors out of 250

Step-by-step explanation:

Multiply 3*150, and 5*150.

150 over 250 is the same as 3 out of 5.

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at the car wash fundraiser jenna's team washed 14 cars in 45 minutes. if they keep up at that rate, how many cars can the team w

s2008m [1.1K]

You can write it as a proportion and then solve the proportion.
(2.5 hours is 150 minutes)
14 cars x cars
--------------- = -------------------
45 minutes 150 minutes

Then solve.

x=46.6666666
So that is about 47 cars

4 0

3 years ago

T what point does the curve have maximum curvature? Y = 7ex (x, y) = what happens to the curvature as x → ∞? Κ(x) approaches as

Nookie1986 [14]

Formula for curvature for a well behaved curve y=f(x) is

K(x)= $\frac{|{y}''|}{[1+{y}'^2]^\frac{3}{2}}$

The given curve is y=7 $e^{x}$

${y}''=7e^{x}\\ {y}'=7e^{x}$

k(x)= $\frac{7e^{x}}{[{1+(7e^{x})^2}]^\frac{3}{2}}$

${k(x)}'=\frac{7(e^x)(1+49e^{2x})(49e^{2x}-\frac{1}{2})}{[1+49e^{2x}]^{3}}$

For Maxima or Minima

${k(x)}'=0$

$7(e^x)(1+49e^{2x})(98e^{2x}-1)=0$

→ $e^{x}=0∨ 1+49e^{2x}=0∨98e^{2x}-1=0$

$e^{x}=0 , ∧ 1+49e^{2x}=0$ [not possible ∵there exists no value of x satisfying these equation]

→ $98e^{2x}-1=0$

Solving this we get

x= $-\frac{1}{2}\ln{98}$

As you will evaluate ${k(x})}''$ <0 at x= $-\frac{1}{2}\ln98$

So this is the point of Maxima. we get y=7×1/√98=1/√2

(x,y)=[ $-\frac{1}{2}\ln98$ ,1/√2]

k(x)= $\lim_{x\to\infty } \frac{7e^{x}}{[{1+(7e^{x})^2}]^\frac{3}{2}}$

k(x)= $\frac{7}{\infty}$

k(x)=0

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C = π×D
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