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mariarad [96]
2 years ago
9

Berto has $12 to put gas in his car. If gas costs $3.75 per gallon, which ordered pair on the graph of the line relating number

of gallons of gas, x, to the total cost of the gas, y, includes the greatest amount of gas Berto can buy?
Mathematics
2 answers:
Nikolay [14]2 years ago
5 0
<span>number of gallons of gas, x
</span><span>total cost of the gas, y
</span>
3.75x <= 12
       x <= 3.2

<span>The greatest amount of gas Berto can buy is 3.2 gallons</span>
Brrunno [24]2 years ago
4 0

Answer:

(3.2, 12)

Step-by-step explanation:

Total cost for gas = $12

Cost per gallon = $3.75

Letting total cost of gas be y and number of gallons be x, maximum number of gallons of gas he can buy would be ;

Total cost of gas/ cost per gallon = 12 / 3.75 = 3.2 gallons

Therefore , the coordinate pair (x, y) = (3.2, 12)

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The blue M&amp;M was introduced in 1995. Before then, the color mix in a bag of plain M&amp;Ms was (30% Brown, 20% Yellow, 20% R
IRISSAK [1]

Answer:

The probability that the yellow M&M came from the 1994 bag is 0.07407 or 7.407%

Step-by-step explanation:

Given

Before 1995

(Br) Brown = 30%

(Y) Yellow = 20%  =0.2

(R) Red = 20%

(G) Green =10%  =0.1

(O) Orange = 10%

(T) Tan = 10%

 

After 1995

(Br) Brown = 13%

(Y) Yellow = 14%  =0.14

(R) Red = 13%

(G) Green = 20% = 0.2

(O) Orange = 16%

(Bl) Blue = 24%

Since there are two bags, let A be the bag from 1994, and B be the bag from 1996

Then let AY imply we drew a yellow M&M from the 1994 bag

AG implies we drew a green M&M from the 1994 bag

BY implies imply we drew a yellow M&M from the 1996 bag

BG implies we drew a green M&M from the 1996 bag

P(AY) =0.2

P (BY) = 0.14

P(AG) =0.1

P(BG) =0.2

Since the draws from the 1994 and 1996 bag are independent,

therefore

P(AY n BG) = 0.2 * 0.2 = 0.04  -------(1)\\P(AG n BY) =0.1 * 0.14 =0.014   --------(2)\\

The draws can happen in either of the 2 ways in (1) and (2) above

therefore total probability E is given as

E =P( AY n BG) u P(AG n BY)\\=0.04 + 0.014 =0.O54

For the yellow one to be from 1994, it implies that the event to be chosen is

P(AYnBG) = 0.2*0.2

Since the total probability is given as E=0.054

then P((AYnBG) /E) =\frac{0.04}{0.054} = 0.07407

Concluding statement: This is the condition for the Yellow one to come from 1994 and green from 1996 provided that they obey the condition from E

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2 years ago
In a​ survey, 39​% of the respondents stated that they talk to their pets on the telephone. A veterinarian believed this result
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Answer:

Step-by-step explanation:

Hypotheses:

H_0: p=0.39\\H_a:p

(Left tailed test)

p = sample proportion

n = sample size = 200

Sample proportion = p = 73/200 = 0.365

Std error = 0.03404

p difference = 0.365-0.39 = 0.025

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Find a polynomial $f(x)$ of degree $5$ such that both of these properties hold: $\bullet$ $f(x)$ is divisible by $x^3$. $\bullet
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There seems to be one character missing. But I gather that <em>f(x)</em> needs to satisfy

• x^3 divides f(x)

• (x-1)^3 divides f(x)^2

I'll also assume <em>f(x)</em> is monic, meaning the coefficient of the leading term is 1, or

f(x) = x^5 + \cdots

Since x^3 divides f(x), and

f(x) = x^3 p(x)

where p(x) is degree-2, and we can write it as

f(x)=x^3 (x^2+ax+b)

Now, we have

f(x)^2 = \left(x^3p(x)\right)^2 = x^6 p(x)^2

so if (x - 1)^3 divides f(x)^2, then p(x) is degree-2, so p(x)^2 is degree-4, and we can write

p(x)^2 = (x-1)^3 q(x)

where q(x) is degree-1.

Expanding the left side gives

p(x)^2 = x^4 + 2ax^3 + (a^2+2b)x^2 + 2abx + b^2

and dividing by (x-1)^3 leaves no remainder. If we actually compute the quotient, we wind up with

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If the remainder is supposed to be zero, then

\begin{cases}a^2+6a+2b+6 = 0 \\ 2ab-6a-8 = 0 \\ 2a+b^2+3 = 0\end{cases}

Adding these equations together and grouping terms, we get

(a^2+2ab+b^2) + (2a+2b) + (6-8+3) = 0 \\\\ (a+b)^2 + 2(a+b) + 1 = 0 \\\\ (a+b+1)^2 = 0 \implies a+b = -1

Then b=-1-a, and you can solve for <em>a</em> and <em>b</em> by substituting this into any of the three equations above. For instance,

2a+(-1-a)^2 + 3 = 0 \\\\ a^2 + 4a + 4 = 0 \\\\ (a+2)^2 = 0 \implies a=-2 \implies b=1

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p(x) = x^2 - 2x + 1 \\\\ \implies f(x) = x^3 (x^2 - 2x + 1) = \boxed{x^5-2x^4+x^3}

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