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VLD [36.1K]
2 years ago
15

Dave walked to his friend's house at a rate of 4 mph and returned back biking at a rate of 10 mph. If it took him 18 minutes lon

ger to walk than to bike, what was the total distance of the round trip?
I need help on it
Mathematics
1 answer:
katrin [286]2 years ago
6 0

Answer:

4 miles

Step-by-step explanation:

<u>Walking:</u>

Distance  = d miles

Rate = 4 mph

Time = t hours

d=4\cdot t

<u>Biking:</u>

Distance = d miles

Rate = 10 mph

Time =t-\dfrac{18}{60}=t-0.3 hours (convert minutes to hours)

d=10\cdot (t-0.3)

Hence,

4t=10(t-0.3)\\ \\4t=10t-3\\ \\4t-10t=-3\\ \\-6t =-3\\ \\6t=3\\ \\t=\dfrac{3}{6}=\dfrac{1}{2}=0.5\ hour

Therefore, the distance to friend's house is

d=4\cdot 0.5=2\ miles

and the total distance of the round trip is

2+2=4\ miles

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Which expression can be used to find the price of a $400 telescope after a 32% markup? Select all that apply.
blagie [28]
<h3>Option D</h3><h3>The expression can be used to find the price of a $400 telescope after a 32% markup is: 400 + 400 (0.32 )</h3>

<em><u>Solution:</u></em>

Given that,

Price of telescope = $ 400

Mark up = 32 %

To find: Cost after mark up

The formula used is:

<h3>Cost after mark up = Price of telescope + 32 % of Price of telescope</h3>

Therefore,

Cost after mark up = 400 + 32 % of 400

\text{Cost after mark up } = 400 + \frac{32}{100} \times 400\\\\\text{Cost after mark up } = 400 + 0.32 \times 400\\\\\text{Cost after mark up } = 400 + 128\\\\\text{Cost after mark up } = 528

Thus cost after mark up is $ 528 and expression is 400 + 400 (0.32 )

7 0
2 years ago
11) The Cost of maintaing a
dezoksy [38]

Answer:

a). Cost of 44 pupils = $14265

b). Least number of pupils = 31

Step-by-step explanation:

The given question is incomplete; here is the complete question.

The cost of maintaining a school is partly constant and partly varies as the number of pupils. With 50 pupils, the cost is $15,705.00 and with 40 pupils, it is $13,305.00.

(a) Find the cost when there are 44 pupils.

(b) If the fee per pupil is $360.00, what is the least number of pupils for which the school can run without a loss?

Let the equation representing the total cost of maintaining a school is,

C = ax + b

Where C = Total cost of maintaining a school

a = Fee per pupil

b = Fixed running cost

x = number of pupils

a). Cost of 50 pupils = $15705

    Equation will be,

    15705 = 50a + b -------(1)

    Cost of 40 pupils = $13305

    Equation will be,

    13305 = 40a + b --------(2)

    By subtracting equation (2) from equation (1),

    15705 - 13305 = (50a + b) - (40a + b)

    2400 = 10a

    a = 240

    From equation (1),

    b = 3705

    Equation representing the total cost will be,

    C = 240x + 3705

    If x = 44

    C = 240(44) + 3705

    C = $14265

b). If the fee per pupil 'a' = $360

    Let the number of pupils = p

    Total fee of 'p' pupils = $360p

    Total cost to run the school will be = 3705 + 240p

    For the school not to be in the loss,

    360p ≥ 3705 + 240p

    360p - 240p ≥ 3705

    120p ≥ 3705

    p ≥ \frac{3705}{120}

    p ≥ 30.875

    Therefore, to run the school without loss, number pupils should be at least 31.

3 0
2 years ago
Which triangle defined by three points on the coordinate plane is congruent with the triangle illustrated, and why?
HACTEHA [7]
The triangle defined by three points on the coordinate plane is congruent with the triangle illustrated:

C) (4,2); (8,2); (4,8) because the corresponding pairs of sides and corresponding pairs of angles are congruent. 

If we plot these points we can observe that they are congruent, we should also solve for the distance of each point between each other to conclude their congruency. 
4 0
2 years ago
Read 2 more answers
A major department store chain is interested in estimating the mean amount its credit card customers spent on their first visit
viva [34]

Answer: D) \$50\pm\$11.08

Step-by-step explanation:

As per given , we have

Sample size : n= 15

sample mean : \overline{x}=\$50.50

Sample standard deviation: s= $20

Since population standard deviation is unknown , so we use t-test.

Significance level for 95% confidence : \alpha=1-0.95=0.05

Critical t-value : t_{n-1, \alpha/2}=t_{014,0.025}=2.145  [Using students' t-value table]

Required 95% Confidence interval :-

\overline{x}\pm t_{n-1, \alpha/2}\dfrac{s}{\sqrt{n}}\\\\ =\$50.50\pm(2.145)\dfrac{\$20}{\sqrt{15}}\\\\\approx \$50\pm\$11.08

Hence, the required 95% confidence interval for the mean amount its credit card customers spent on their first visit to the chain's new store in the mall assuming that the amount spent follows a normal distribution.:

\$50\pm\$11.08

5 0
2 years ago
Nikita ran a 5-kilometer race in 39 minutes (0.65 of an hour) without training beforehand. In the first part of the race, her av
Sergeeva-Olga [200]
First part of the race

speed1, s1 = 8.75 km/h
distance1 = 5 km - x
time 1 = 0.65 h - t

Second part of the race

speed2, s2 = 6 km/h
distance2 = x
time2 = t

s2 = x/t = 6 ⇒ t = x / 6

s1 = [5 - x] / [0.65 - t] = 8.75

5 - x = 8.75 [0.65 - t]

5 -x = 8.75(0.65) - 8.75t

5 - x = 5.6875 - 8.75 (x/6)

5 - x = 5.6875 - 1.4583x

1.4583 x - x = 5.6875 - 5

0.4583 x = 0.6875

x = 0.6875 / 0.4583 = 1.5





5 0
2 years ago
Read 2 more answers
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