Ask question

Ask question

All categories

2 years ago

11

A car Travel 476 miles on 14 gallons of gas. Write an eqation relating the distance d to the nuber of gallons

1 answer:

Klio2033 [76]2 years ago

3 0

476miles per 14gallons of gas

is the same as 476 / 14 = 34

so 34miles / 1gallon of gas.

34x =578 where x is the number of gallons

578 / 34 = 17gallons

You might be interested in

At a certain university, the average attendance at basketball games has been 3125. Due to the dismal showing of the team this ye

Neporo4naja [7]

Answer:

The test statistic needed to evaluate the claim is t = -1.08.

Step-by-step explanation:

The test statistic is:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the expected value of the mean, s is the standard deviation of the sample and n is the size of the sample.

At a certain university, the average attendance at basketball games has been 3125. The athletic director claims that the attendance is the same as last year.

This means that $\mu = 3125$

Due to the dismal showing of the team this year, the attendance for the first 9 games has averaged only 2915 with a standard deviation of 585.

This means that $n = 9, X = 2915, s = 585$

What is the test statistic needed to evaluate the claim?

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{2915 - 3125}{\frac{585}{\sqrt{9}}}$

$t = -1.08$

The test statistic needed to evaluate the claim is t = -1.08.

6 0

2 years ago

Find X<br><br> Options:<br><br> 6<br> 4 square root 6<br> 6<br> square root 146

elena-14-01-66 [18.8K]

For this case we can apply the Pythagorean theorem to find "x". Taking the rectangle triangle of base 5 we have:

$x = \sqrt {11 ^ 2-5 ^ 2}\\x = \sqrt {121-25}\\x = \sqrt {96}\\\x = \sqrt {4 ^ 2 * 6}$

By definition of power properties we have:

$\sqrt [n] {a ^ m} = a ^ {\frac {m} {n}}$

So:

$x = 4 \sqrt {6}$

Answer:

$x = 4 \sqrt {6}$

8 0

2 years ago

Read 2 more answers

By using the table in the handbook, the present value of $12,000 for six years compounded at 6 percent semiannually is

denis23 [38]

Take a look at the attachment to see the solution.
A = future value
P = principal (P = 12,000)
r = interest rate (r=6)
n = time periods (n=12)

6 0

3 years ago

irakobra [83]

Answer: =296

Step-by-step explanation: −2*4=-8+20=28*3=84+48=132*2=164−2=162*4+20*3+48*2= 296

8 0

2 years ago

It has been said that newspapers are disappearing, replaced by various electronic media. In 2004, newspaper circulation (the num

shutvik [7]

Answer:

A. $y=-1.2x+59$

B. 42.2 million

Step-by-step explanation:

Part A:

Given:

Newspapers circulated in year 2004, $y_{1}=59\textrm{ million}$

Newspapers circulated in year 2014, $y_{2}=47\textrm{ million}$

Let the time $x$ start at the year 2004. So, $x_{1}=0$

For the year 2014, $x_{2}=2014-2004=10$

Therefore, linear relationship between newspapers circulated and time passed since 2004 is given as:

$y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{1})\\y-59=\frac{47-59}{10-0}(x-0)\\y-59=-\frac{12}{10}x\\y=-1.2x+59$

Therefore, the equation describing the relationship is: $y=-1.2x+59$

Part B:

For the year 2018, $x=2018-2004=14$

Plug in 14 for $x$ in the above equation and solve for $y$ . This gives,

$y=-1.2(14)+59\\y=-16.8+59\\y=42.2$

Therefore, in the year 2018, the newspaper circulation will be 42.2 million.

8 0

2 years ago