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professor190 [17]
3 years ago
6

In a surey of 375 dog and cat owners there were 215 dog owners and 193 cat owners. how many in the survey own a dog

Mathematics
2 answers:
VARVARA [1.3K]3 years ago
4 0
First, draw venn diagram and determine how many people own both cat and dog. See image attached.
x represents the number of people own both cat and dog
dog only owner + cat only owner + both cat and dog owner = 375
215 - x + 193 - x + x = 375
408 - x = 375
x = 408 - 375
x = 33
The number of people own both cat and dog is 33

Second, find the number of people own dog only
dog only = 215 - 33
dog only = 182
The number of people own dog only is 182

spin [16.1K]3 years ago
3 0
Percent wise, 57.3% own a dog
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Zapisz w jak najprostszej postaci:<br> a) -x-4(1-x)<br> b) 3(5x-2)-5(3-4x)<br> c) √2(3-x)-2(1-x√2)
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Answer:

Step-by-step explanation:

a) -x-4(1-x)  =

− 2 x −4 ( 1 − x ) .

2 x − 4

b) 3(5x-2)-5(3-4x)  =

3 ⋅ (5 x − 2 ) − 5 ⋅ ( 5 − 4 x )

35 x  −  31

c) √2(3-x)-2(1-x√2) =

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2 x √ 2 + 3 √ 2 x -√ 2 x x − 2

4 0
3 years ago
Savannah loves to talk on her cell phone . she has $100 in her budget to pay this month's cell phone bill. her cell phone plan i
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3 years ago
I am having trouble with this relative minimum of this equation.<br>​
Norma-Jean [14]

Answer:

So the approximate relative minimum is (0.4,-58.5).

Step-by-step explanation:

Ok this is a calculus approach.  You have to let me know if you want this done another way.

Here are some rules I'm going to use:

(f+g)'=f'+g'       (Sum rule)

(cf)'=c(f)'          (Constant multiple rule)

(x^n)'=nx^{n-1} (Power rule)

(c)'=0               (Constant rule)

(x)'=1                (Slope of y=x is 1)

y=4x^3+13x^2-12x-56

y'=(4x^3+13x^2-12x-56)'

y'=(4x^3)'+(13x^2)'-(12x)'-(56)'

y'=4(x^3)'+13(x^2)'-12(x)'-0

y'=4(3x^2)+13(2x^1)-12(1)

y'=12x^2+26x-12

Now we set y' equal to 0 and solve for the critical numbers.

12x^2+26x-12=0

Divide both sides by 2:

6x^2+13x-6=0

Compaer 6x^2+13x-6=0 to ax^2+bx+c=0 to determine the values for a=6,b=13,c=-6.

a=6

b=13

c=-6

We are going to use the quadratic formula to solve for our critical numbers, x.

x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}

x=\frac{-13 \pm \sqrt{13^2-4(6)(-6)}}{2(6)}

x=\frac{-13 \pm \sqrt{169+144}}{12}

x=\frac{-13 \pm \sqrt{313}}{12}

Let's separate the choices:

x=\frac{-13+\sqrt{313}}{12} \text{ or } \frac{-13-\sqrt{313}}{12}

Let's approximate both of these:

x=0.3909838 \text{ or } -2.5576505.

This is a cubic function with leading coefficient 4 and 4 is positive so we know the left and right behavior of the function. The left hand side goes to negative infinity while the right hand side goes to positive infinity. So the maximum is going to occur at the earlier x while the minimum will occur at the later x.

The relative maximum is at approximately -2.5576505.

So the relative minimum is at approximate 0.3909838.

We could also verify this with more calculus of course.

Let's find the second derivative.

f(x)=4x^3+13x^2-12x-56

f'(x)=12x^2+26x-12

f''(x)=24x+26

So if f''(a) is positive then we have a minimum at x=a.

If f''(a) is negative then we have a maximum at x=a.

Rounding to nearest tenths here:  x=-2.6 and x=.4

Let's see what f'' gives us at both of these x's.

24(-2.6)+25

-37.5  

So we have a maximum at x=-2.6.

24(.4)+25

9.6+25

34.6

So we have a minimum at x=.4.

Now let's find the corresponding y-value for our relative minimum point since that would complete your question.

We are going to use the equation that relates x and y.

I'm going to use 0.3909838 instead of .4 just so we can be closer to the correct y value.

y=4(0.3909838)^3+13(0.3909838)^2-12(0.3909838)-56

I'm shoving this into a calculator:

y=-58.4654411

So the approximate relative minimum is (0.4,-58.5).

If you graph y=4x^3+13x^2-12x-56 you should see the graph taking a dip at this point.

3 0
3 years ago
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