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Leona [35]
2 years ago
14

The mean of the weights of a group of 100 men and women

Mathematics
1 answer:
Arte-miy333 [17]2 years ago
7 0

Answer:

Step-by-step explanation:

160lb.

The mean of the weights of a group of 100 men and women is 160lb. If the number of men in the group is 60 and the mean weight of the men is 180lb, what is the mean weight of the women? For a set of data, the lower quartile is 19, the median is 31, and the upper quartile is 48.

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Write the standard form of the equation of the line passing through (0, 1) and parallel to x = 4.
Savatey [412]

Answer:

x=0

Step-by-step explanation:

The line x=4 is a vertical line through the x-axis at 4. A line parallel to it will also be a line through the x-axis. It has the form x=a where a is the x-coordinates of any points on the line. Since the line crosses through (0,1), the equation is x=0.

5 0
3 years ago
Find the linear function with the following properties. <br><br> f(0)=−9 Slope of f=−8
Ratling [72]

Answer:

  • f(x) = -8x - 9

Step-by-step explanation:

<h3>Given </h3>

<u>Function f, with </u>

  • f(0) = -9
  • Slope of f = -8
<h3>To find</h3>
  • Linear function
<h3>Solution</h3>

<u>Slope-intercept formula</u>

  • f(x) = mx + b
  • f(0) = -9 means y-intercept b = -9, and the slope is m = - 8

<u>So the function is</u>

  • f(x) = -8x - 9
6 0
3 years ago
9-c=-13<br> What’s the answer
trapecia [35]
First you do -13+9. That equals -4.
Then bring down -c=-4 because there is a 1 in front of the c so -1c.
Next divide both sides by -1. The -c side cancels out so then your left with -4/-1.
That equals positive 4.
C=4
That’s ur answer.
7 0
3 years ago
Which function is an example of exponential growth? y = 2(0.3)x y = 1.5(0.4)x y = 3(8)x y = 2(0.7)x
4vir4ik [10]
Exponential equation is
y=a(b)ˣ

when b>1, it is growth
when 0>b>1, it is decay

so

we want the 2nd number to be more than 1
that would be the 3rd one, y=3(8)ˣ
7 0
3 years ago
Find dy/dx if y =x^3+5x+2/x²-1
stiks02 [169]

<u>Differentiate using the Quotient Rule</u> –

\qquad\pink{\twoheadrightarrow \sf \dfrac{d}{dx} \bigg[\dfrac{f(x)}{g(x)} \bigg]= \dfrac{ g(x)\:\dfrac{d}{dx}\bigg[f(x)\bigg] -f(x)\dfrac{d}{dx}\:\bigg[g(x)\bigg]}{g(x)^2}}\\

According to the given question, we have –

  • f(x) = x^3+5x+2
  • g(x) = x^2-1

Let's solve it!

\qquad\green{\twoheadrightarrow \bf \dfrac{d}{dx}\bigg[ \dfrac{x^3+5x+2 }{x^2-1}\bigg]} \\

\qquad\twoheadrightarrow \sf \dfrac{(x^2-1) \dfrac{d}{dx}(x^3+5x+2) - ( x^3+5x+2)  \dfrac{d}{dx}(x^2-1)}{(x^2-1)^2 }\\

\qquad\twoheadrightarrow \sf \dfrac{(x^2-1)(3x^2+5)  -  ( x^3+5x+2) 2x}{(x^2-1)^2 }\\

\qquad\pink{\sf \because \dfrac{d}{dx} x^n = nx^{n-1} }\\

\qquad\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-(2x^4+10x^2+4x)}{(x^2-1)^2 }\\

\qquad\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-2x^4-10x^2-4x}{(x^2-1)^2 }\\

\qquad\green{\twoheadrightarrow \bf \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}\\

\qquad\pink{\therefore  \bf{\green{\underline{\underline{\dfrac{d}{dx} \dfrac{x^3+5x+2 }{x^2-1}}  =  \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}}}}\\\\

7 0
2 years ago
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