Which expression is equivalent to −1/4x+1/2 ?

Maxis taking a cross-country road trip. Gas prices vary as the friends travel across the US from $4 dollars per gallon on the ea

Alex787 [66]

Answer:

The quantity of fuel bought at $4 is 40.84 gallons

The quantity of fuel bought at $3 is 116.699 gallons

The quantity of fuel bought at $5 is 17.505 gallons

Step-by-step explanation:

Here we have

Gas used on East coast = X gallons

Gas consumed in the mid-US = Y gallons

Gas consumed in the West coast = Z gallons

Where Y = 2×(X + Z)..........(1)

4·X + 3·Y + 5·Z = 515.........(2)

X + Y + Z = 150..................(3)

Therefore substituting for Y in equation (3) we have

X + 2×(X + Z) + Z = 150

3·X + 3·Z = 150

X = (150 - 3·Z)/3 = 50 - Z

Therefore, equation (2) becomes;

4·(50 - Z) + 3·(2×((50 - Z) + Z)) + 5·Z = 515

200 - 4·Z + 300 -6·Z + 6·Z + 5·Z = 515

500 + 2·Z = 515

Z = 15

X = 50 - Z = 50 - 15 = 35

Y = 2×(X + Z) = 2×(35 + 15) = 2 × 50 = 100

(b) Whereby the bought 174 gallons based on the same ratio as in part (a), we have;

Ratio in part (a) is given as follows;

Proportion of Z gallons = 15/150 = 1/10

Proportion of X gallons = 35/150 = 7/30

Proportion of Y gallons = 100/150 = 2/3

Therefore the proportion of 174 gallons that was consumed in the East coast, X gallons = (7/30)×174 = 203/5 gallons = 40.6 gallons

The proportion of 174 gallons that was consumed in the mid-US, Y gallons = (2/3)×174 = 116 gallons

The proportion of 174 gallons that was consumed in the West coast, Z gallons = (1/10)×174 = 87/5 gallons = 17.4 gallons

Based on price of per gallon, we have

The proportion of $515 that was consumed in the East coast on X gallons = $4 ×35 = $140

Proportion = 140/515= 28/103

The proportion of $515 that was consumed in the mid-US on Y gallons = $3 ×100= $300

Proportion = 300/515= 60/103

The proportion of $515 that was consumed in the West coast on Z gallons = $5 ×15= $75

Proportion = 75/515= 15/103

Therefore, the quantity of fuel bought at $4 = ((28/103) × 601)/4 = 40.84 gallons

The quantity of fuel bought at $3 = ((60/103) × 601)/3 = 116.699 gallons

The quantity of fuel bought at $5 = ((15/103) × 601)/5 = 17.505 gallons.

5 0

3 years ago

Combine like terms to create an equivalent expression. Enter coefficients as simplified proper or improper fractions or integers

Art [367]

Answer:

Step-by-step explanation:

The first (and only step) is to remove the brackets.

-1/2 * - 3y = 3/2y Two minus make a plus

-1/2 * 10 = - 5

Answer: 3/2 y - 5

8 0

3 years ago

A certain railway company claims that its trains run late 5 minutes on the average. The actual times (minutes) that 10 randomly

iragen [17]

Answer:

Option C) We would reject the claim that the trains run late an average of 5 minutes and conclude that they run late an average of more than 5 minutes since the 99% confidence interval exceeds 5.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 5 minutes

Sample mean, $\bar{x}$ = 9.130 minutes

Sample size, n = 10

Alpha, α = 0.01

Sample standard deviation, s = 1.4 minutes

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 5\text{ minutes}\\H_A: \mu \neq 5\text{ minutes}$

We have to construct 99% confidence interval.

99% Confidence interval:

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$

Putting the values, we get,

$t_{critical}\text{ at degree of freedom 9 and}~\alpha_{0.01} = \pm 3.249$

$9.130 \pm 3.249(\dfrac{1.4}{\sqrt{10}} ) = 9.130 \pm 1.4383 = (7.6917,10.5683)$

99% Confidence interval: (7.6917,10.5683)

Since the population mean does not lie in he calculated confidence interval, thus, we fail to accept the null hypothesis and accept the alternate hypothesis.

Option C) We would reject the claim that the trains run late an average of 5 minutes and conclude that they run late an average of more than 5 minutes since the 99% confidence interval exceeds 5.

6 0

3 years ago

Find |−5|. A. −5 B. 5 C. one-fifth D. –one-fifth HELP!!!

Volgvan

recall that when it comes to absolute value expressions, once we remove the bars, it turns always to positive, so | -5 |, the bars go poof, and tada!! 5.

4 0

3 years ago

WILL GIVE A BRAINLIEST IF UR RIGHT PLEASE HELP!!

kvasek [131]

$\frac{x^{3}y^{5}}{x^{5}y^{2}}$
$\frac{y^{3}}{x^{2}}$
$\frac{6^{3}}{3^{2}}$
$\frac{216}{9}$
$24$

8 0

3 years ago