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1 year ago

A car trip is 1/2h shorter that a bus trip. How long is the car trip if the bus trip is 3/4h?

Mathematics

1 answer:

olchik [2.2K]1 year ago

7 0

Answer:

1/4h

Step-by-step explanation:

If the time for the car trip is x, then given that a car trip is 1/2h shorter than a bus trip, it means that adding 1/2h to the time for a car trip gives the time for the bus trip.

Hence, the time for the bus trip

= (1/2 + x)h

since the bus trip is 3/4h, it means that

1/2 + x = 3/4

collecting like terms,

x = 3/4 - 1/2

= 3/4 - 2/4

= 1/4

Hence the car trip is 1/4h.

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A control chart is developed to monitor the analysis of iron levels in human blood. The lines on the control chart were obtained

sergejj [24]

Answer:

Step-by-step explanation:

Hello!

The variable is X: iron level on human blood.

It has a mean of μ= 51.50 mg/dl and a standard deviation of σ= 3.50 mg/dl.

According to the control chart, the process should be shut down for troubleshooting when the analysis shows values X[bar]≥ 53.42 mg/dl and X[bar]≤ 49.58 mg/dl

Warnings are received at levels X[bar]≥52.78 mg/dl and X[bar]≤ 50.22 mg/dl

The system works between levels 49.58<X[bar]<53.42 and works without warnings between 50.22<X[bar]<52.78.

Using these parameters you have to analyze if the lists of sample means to see which ones are within the working values are wich ones are outside this interval.

Sample 1

Min= 50.15

Max= 51.99

Mean= 51.61

Without warning 50.22<X[bar]<52.78

Without Action interval 49.58<X[bar]<53.42

This sample's min value is below the lower limit of the warning interval but not low enough to reach action levels, the max value is within the working range.

Sample 2

Min= 50.32

Max= 52.56

Mean= 51.16

Without warning 50.22<X[bar]<52.78

Without Action interval 49.58<X[bar]<53.42

Both the max and min values of the sample are within the working range without warning.

Sample 3

Min= 50.25

Max= 53.12

Mean= 51.83

Without warning 50.22<X[bar]<52.78

Without Action interval 49.58<X[bar]<53.42

The max value of this sample is above the upper limit of the warning interval but does not surpass the upper bond of the troubleshoot interval. Min value is within working values.

Sample 4

Min= 50.05

Max= 53.01

Mean= 51.70

Without warning 50.22<X[bar]<52.78

Without Action interval 49.58<X[bar]<53.42

Both min and max values surpass the warning interval but do not reach action levels.

Sample 5

Min= 50.35

Max= 52.71

Mean= 51.37

Without warning 50.22<X[bar]<52.78

Without Action interval 49.58<X[bar]<53.42

Both min and max values are within working levels without warnings.

Considering that samples reaching warning levels should be shut down as a precaution, they are classified as:

Shutdown: Sample 1, 3 and 4

Do not shutdown: Sample 2 and 5.

I hope it helps!

8 0

2 years ago

On your first day as an apprentice at the oil refinery, you are asked to create usable graphs and tables for the other workers w

Naya [18.7K]

Answer:

89,250 gallons per hour

Step-by-step explanation:

4 0

2 years ago

What is the value of k such that x-5 is a factor of x3 – x2 + kx - 30?

Paraphin [41]

Answer:

k = - 14

Step-by-step explanation:

given that (x - 5) is a factor of the polynomial then x = 5 is a root and

x³ - x² + kx - 30 = 0 for x = 5, that is

5³ - 5² + 5k - 30 = 0

125 - 25 + 5k - 30 = 0

70 + 5k = 0 ( subtract 70 from both sides )

5k = - 70 ( divide both sides by 5 )

k = - 14

5 0

2 years ago

Which graph represents the function of f(x) = the quantity of 9 x squared plus 9 x minus 18, all over 3 x plus 6

lora16 [44]

Answer:

The answer is the option C

graph of $3x$ minus $3$ , with discontinuity at negative $2$ , negative $9$

Step-by-step explanation:

we have

$f(x)=\frac{9x^{2}+9x-18}{3x+6}$

Simplify

$f(x)=9\frac{(x^{2}+x-2)}{3(x+2)}$

$f(x)=3\frac{(x^{2}+x-2)}{(x+2)}$

Step 1

Convert to a factored form the numerator

$x^{2}+x-2=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+x=2$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+x+0.25=2+0.25$

$x^{2}+x+0.25=2.25$

Rewrite as perfect squares

$(x+0.5)^{2}=2.25$

Square root both sides

$x+0.5=(+/-)1.5$

$x=-0.5(+/-)1.5$

$x=-0.5+1.5=1$

$x=-0.5-1.5=-2$

$x^{2}+x-2=(x-1)(x+2)$

Step 2

Simplify the function f(x)

$f(x)=3\frac{(x^{2}+x-2)}{(x+2)}=3\frac{(x-1)(x+2)}{(x+2)}$

The domain of the function f(x) is all real numbers except the number $x=-2$

Because the denominator can not be zero

$f(x)=3\frac{(x-1)(x+2)}{(x+2)}=3(x-1)=3x-3$

$f(x)=3x-3$ ------> with a discontinuity at $x=-2$

$f(-2)=3(-2)-3=-9$

The discontinuity is at point $(-2,-9)$

the answer in the attached figure

8 0

2 years ago

Read 2 more answers

Evaluate 10m +\dfrac {n^2}410m+ 4 n 2 10, m, plus, start fraction, n, squared, divided by, 4, end fraction when m=5m=5m, equal

Readme [11.4K]

Answer: $54$

Step-by-step explanation:

Given the following expresion provided in the exercise:

$10m+\frac{n^2}{4}$

You can follow these steps in order to evaluate it when $m=5$ and $n=4$ :

1. You need to substitute $m=5$ and $n=4$ into the given expression:

$10(5)+\frac{(4)^2}{4}$

2. Now you can solve the mutiplication:

$=50+\frac{(4)^2}{4}$

3. Since $4^2=4*4$ , you get:

$=50+\frac{16}{4}$

4. You must solve the division. Divide the numerator 16 by the denominator 4. Then:

$=50+4$

5. And finally, you must solve the addition. So, you get this result:

$=54$

3 0

2 years ago