Ask question

Ask question

All categories

3 years ago

14

What is the average speed of a biker if she biked 4 1/4 of a mile in 25 minutes?

2 answers:

mezya [45]3 years ago

6 0

Answer:

$\frac{10.2\text{ miles}}{\text{ hour}}$

Step-by-step explanation:

We are asked to find the speed of a biker, who biked 4 1/4 of a mile in 25 minutes.

To find average speed of the biker, we will divide total distance by total time taken.

$4\frac{1}{4}\text{ miles}\div 25\text{ min}$

Convert mixed fraction into improper fraction:

$4\frac{1}{4}\Rightarrow \frac{4*4+1}{4}=\frac{16+1}{4}=\frac{17}{4}$

$\frac{17}{4}\text{ miles}\div \frac{25}{1}\text{ min}$

Convert to multiplication problem by flipping the 2nd fraction:

$\frac{17}{4}\text{ miles}\times \frac{1}{25\text{ min}}$

$\frac{17\text{ miles}}{100\text{ min}}$

Convert minutes to hour:

$\frac{17\text{ miles}}{100\text{ min}}\times \frac{\text{60 min}}{\text{1 hour}}$

$\frac{17\text{ miles}}{10}\times \frac{6}{\text{1 hour}}$

$\frac{102\text{ miles}}{\text{10 hour}}$

$\frac{51\text{ miles}}{\text{5 hour}}$

$\frac{10.2\text{ miles}}{\text{ hour}}$

Therefore, the average speed of biker is 10.2 miles per hour.

nata0808 [166]3 years ago

5 0

Answer:

Step-by-step explanation:

What is the average speed of a biker if she biked 4 1/4 of a mile in 25 minutes?

-----------------

rate = distance/time

-----

rate = (17/4)/(25/60) = (17/4)(60/25) = (17*3/5) = 51/5 = 10.2 mph

----------------

You might be interested in

Simplify 9i(2+3i)

viktelen [127]

Simplify 9i(2+3i):
9i×2=18i
9i×3i=27i

18i+27i

3 0

3 years ago

Read 2 more answers

A hot air balloon is flying at and altitude of 2101 feet. If the angle of depression from the pilot in the balloon to a house on

alekssr [168]

The problem is modelled as right angle triangle, shown in the diagram below

The distance between the pilot and the house is labelled as $x$

Using the trigonometry ratio of sine
$sin(31)= \frac{opposite}{hypotenuse}$
$sin(31)= \frac{2010}{x}$
$x= \frac{2101}{sin(31)} = 4079.3 ft$

3 0

3 years ago

Please help! PLSSS! Its for a math test.

Maslowich

Answer:

330(D)

Step-by-step explanation:

So lets go over what we know:

#1: 44 out of 80 people think they have a average fitness level.

#2: 80 out of the 600 people in the school were surveyed.

Using what we know, lets try and find the answer.

So lets divide the total amount of people, by the amount of people that took the survey:

600/80

=

7.5

Ok, so now we know that we need to take 7.5 surveys to find out what all the students think of their fitness level.

Now lets take teh amount of people in a single survey, with average fitness(44), and multpliy it by how many surveys are needed for the whole school(7.5):

44*7.5

=

330

This is the amount of people in the whole school who think they have an average fitness level.

I hope this helps! :)

<u>Answer:</u>

<u>330</u>

8 0

3 years ago

A tank contains 60 kg of salt and 1000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain

MissTica

Answer:

(a) 60 kg; (b) 21.6 kg; (c) 0 kg/L

Step-by-step explanation:

(a) Initial amount of salt in tank

The tank initially contains 60 kg of salt.

(b) Amount of salt after 4.5 h

$\text{Let A = mass of salt after t min}\\\text{and }r_{i} = \text{rate of salt coming into tank}\\\text{and }r_{0} =\text{rate of salt going out of tank}$

(i) Set up an expression for the rate of change of salt concentration.

$\dfrac{\text{d}A}{\text{d}t} = r_{i} - r_{o}\\\\\text{The fresh water is entering with no salt, so}\\ r_{i} = 0\\r_{o} = \dfrac{\text{3 L}}{\text{1 min}} \times \dfrac {A\text{ kg}}{\text{1000 L}} =\dfrac{3A}{1000}\text{ kg/min}\\\\\dfrac{\text{d}A}{\text{d}t} = -0.003A \text{ kg/min}$

(ii) Integrate the expression

$\dfrac{\text{d}A}{\text{d}t} = -0.003A\\\\\dfrac{\text{d}A}{A} = -0.003\text{d}t\\\\\int \dfrac{\text{d}A}{A} = -\int 0.003\text{d}t\\\\\ln A = -0.003t + C$

(iii) Find the constant of integration

$\ln A = -0.003t + C\\\text{At t = 0, A = 60 kg/1000 L = 0.060 kg/L} \\\ln (0.060) = -0.003\times0 + C\\C = \ln(0.060)$

(iv) Solve for A as a function of time.

$\text{The integrated rate expression is}\\\ln A = -0.003t + \ln(0.060)\\\text{Solve for } A\\A = 0.060e^{-0.003t}$

(v) Calculate the amount of salt after 4.5 h

a. Convert hours to minutes

$\text{Time} = \text{4.5 h} \times \dfrac{\text{60 min}}{\text{1h}} = \text{270 min}$

b.Calculate the concentration

$A = 0.060e^{-0.003t} = 0.060e^{-0.003\times270} = 0.060e^{-0.81} = 0.060 \times 0.445 = \text{0.0267 kg/L}$

c. Calculate the volume

The tank has been filling at 6 L/min and draining at 3 L/min, so it is filling at a net rate of 3 L/min.

The volume added in 4.5 h is

$\text{Volume added} = \text{270 min} \times \dfrac{\text{3 L}}{\text{1 min}} = \text{810 L}$

Total volume in tank = 1000 L + 810 L = 1810 L

d. Calculate the mass of salt in the tank

$\text{Mass of salt in tank } = \text{1810 L} \times \dfrac{\text{0.0267 kg}}{\text{1 L}} = \textbf{21.6 kg}$

(c) Concentration at infinite time

$\text{As t $\longrightarrow \, -\infty,\, e^{-\infty} \longrightarrow \, 0$, so A $\longrightarrow \, 0$.}$

This makes sense, because the salt is continuously being flushed out by the fresh water coming in.

The graph below shows how the concentration of salt varies with time.

3 0

3 years ago

alisha [4.7K]

Subtract 10 and divide by 7

5 0

3 years ago

Read 2 more answers