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3 years ago

14

In November 2010 the average price of 5 gallons of regular unleaded gasoline in the United states was $14.46. what was the price

for 16 gallons of gasoline?

1 answer:

Lapatulllka [165]3 years ago

3 0

Propecia:

5 gallons ====== $ 14.46.

16 gallons ======= x dollars

X=16*14.46: 5=46.27

Answer: $ 46.27.

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The radius of circle A is three feet less than twice the diameter of circle B. If the sun mmm of the diameters of both circles i

yanalaym [24]

Answer:

Area of Circle A = 1133.54

Circumference of Circle A= 119.32

Step-by-step explanation:

Given: Sum of both circle diameters= 49 feet.

Circle A radius= Three feet less than twice the diameter of circle B.

Lets assume diameter of Circle B is "x"

∴ Circle A radius (r)= $(2x-3)$

We know, the diameter of circle is $2r$

Diameter of Circle A is $2\times (2x-3)$

Next as given sum of both circle diameter is 49 feet.

∴ $2(2x-3)+x= 49\ feet$

distributing 2 with 2x and -3

⇒ $4x-6+x= 49$

⇒ $5x-6= 49$

Adding both side by 6

⇒ $5x=55$

Cross multiplying both side.

⇒ $x=\frac{55}{5} = 11$

∴ Diameter of circle B is 11 feet.

Next, subtituting the value of x to get the value of radius for Circle A.

Radius of circle A= $(2x-3)$

⇒ Radius of circle A= $(2\times 11-3)= 22-3$

∴ Radius of circle A= 19 feet.

Area of circle= $\pi r^{2}$

Taking π= 3.14

Area of circle A= $3.14\times 19^{2} = 3.14\times 361$

∴ Area of circle A= 1133.54

Circumference of circle= 2πr

Circumference of Circle A= $2\times 3.14\times 19= 119.32$

∴ Circumference of Circle A= 119.32

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3 years ago

Read 2 more answers

Charlie wants to order lunch for his friends. He’ll order 6 sandwich’s and a $3 kids meal for his little brother Charlie had $27

Vera_Pavlovna [14]

The answer is 4, first you the $27 and subtract the $3, then u take 24 and divide it by 6

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2 years ago

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An urn contains n white balls andm black balls. (m and n are both positive numbers.) (a) If two balls are drawn without replacem

Genrish500 [490]

DISCLAIMER: Please let me rename b and w the number of black and white balls, for the sake of readability. You can switch the variable names at any time and the ideas won't change a bit!

<h2>(a)</h2>

Case 1: both balls are white.

At the beginning we have $b+w$ balls. We want to pick a white one, so we have a probability of $\frac{w}{b+w}$ of picking a white one.

If this happens, we're left with $w-1$ white balls and still $b$ black balls, for a total of $b+w-1$ balls. So, now, the probability of picking a white ball is

$\dfrac{w-1}{b+w-1}$

The probability of the two events happening one after the other is the product of the probabilities, so you pick two whites with probability

$\dfrac{w}{b+w}\cdot \dfrac{w-1}{b+w-1}=\dfrac{w(w-1)}{(b+w)(b+w-1)}$

Case 2: both balls are black

The exact same logic leads to a probability of

$\dfrac{b}{b+w}\cdot \dfrac{b-1}{b+w-1}=\dfrac{b(b-1)}{(b+w)(b+w-1)}$

These two events are mutually exclusive (we either pick two whites or two blacks!), so the total probability of picking two balls of the same colour is

$\dfrac{w(w-1)}{(b+w)(b+w-1)}+\dfrac{b(b-1)}{(b+w)(b+w-1)}=\dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}$

<h2>(b)</h2>

Case 1: both balls are white.

In this case, nothing changes between the two picks. So, you have a probability of $\frac{w}{b+w}$ of picking a white ball with the first pick, and the same probability of picking a white ball with the second pick. Similarly, you have a probability $\frac{b}{b+w}$ of picking a black ball with both picks.

This leads to an overall probability of

$\left(\dfrac{w}{b+w}\right)^2+\left(\dfrac{b}{b+w}\right)^2 = \dfrac{w^2+b^2}{(b+w)^2}$

Of picking two balls of the same colour.

<h2>(c)</h2>

We want to prove that

$\dfrac{w^2+b^2}{(b+w)^2}\geq \dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}$

Expading all squares and products, this translates to

$\dfrac{w^2+b^2}{b^2+2bw+w^2}\geq \dfrac{w^2+b^2-b-w}{b^2+2bw+w^2-b-w}$

As you can see, this inequality comes in the form

$\dfrac{x}{y}\geq \dfrac{x-k}{y-k}$

With x and y greater than k. This inequality is true whenever the numerator is smaller than the denominator:

$\dfrac{x}{y}\geq \dfrac{x-k}{y-k} \iff xy-kx \geq xy-ky \iff -kx\geq -ky \iff x\leq y$

And this is our case, because in our case we have

$x=b^2+w^2$
$y=b^2+w^2+2bw$ so, y has an extra piece and it is larger
$k=b+w$ which ensures that k<x (and thus k<y), because b and w are integers, and so b<b^2 and w<w^2

4 0

3 years ago

Haruka hiked several kilometers in the morning. She hiked only 666 kilometers in the afternoon, which was 25\%25%25, percent les

katen-ka-za [31]

Answer:

She hiked 1,554 kilometers in all.

Step-by-step explanation:

Let the hiking kilometers for morning be x

Given that:

Hiking kilometers in afternoon = 666

According to given condition that afternoon kilometers are 25% less than morning:

x = 666 + 25% of x

By simplifying:

x = 666 + 0.25x

x - 0.25x = 666

0.75x = 666

Dividing both sides by 0.75 we get:

x = 666/0.75

x = 888 kilometers

She hiked 888 kilometers in morning

And

Total = 888 + 666 =1554 km

i hope it will help you!

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2 years ago

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Use compatible numbers to estimate the quotient 2,545÷4

victus00 [196]

Answer:

600

Step-by-step explanation:

round 2,545 to 2,400 and keep 4 the same because 24 goes into 4 six times so 2400/4 equals 600

24/4 is 6 and you add the two zeros

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3 years ago