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Gnoma [55]
3 years ago
15

Dora drove 80 km in 40 minutes then 120 km in 1 hour and then the final 200km took him 1 hour 20 minutes. what is his average sp

eed for the whole journey​
Mathematics
1 answer:
bonufazy [111]3 years ago
8 0

9514 1404 393

Answer:

  133 1/3 km/hour

Step-by-step explanation:

Average speed is computed as ...

  average speed = (total distance)/(total time)

  = (80 km +120 km +200 km)/(2/3 + 1 + 1 1/3 hour) = 400 km/(3 hour)

  = 133 1/3 km/hour

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Trigonometry Story Problem? A gondola on an amusement park ride spins at speed of 13 revolutions per minute. If the gondola is 2
Neko [114]
<span>23.2 mph What you do is figure out how far the gondola is moving. Given that it's moving in a circle with a radius of 25 feet, the circumference is: 2 * 3.14159 * 25 = 157.08 feet. Now it travels that distance 13 times per minute giving: 157.08 * 13 = 2042.04 feet per minute. Finally, assuming it keeps up that speed for 1 hour, you get 2042.04 * 60 = 122522.4 feet per hour. Now convert that to miles per hour by dividing by 5280 122522.4 / 5280 = 23.2</span>
7 0
4 years ago
Solve for a:8-4(a-1)=2+3(4-a)
OLga [1]
8-4(a-1)=2+3(4-a) \\&#10;8-4a+4=2+12-3a\\&#10;-3a+4a=12-2-12\\&#10;a=-2
5 0
3 years ago
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 259.2-cm and a standard dev
Varvara68 [4.7K]

Answer:

The probability that the average length of rods in a randomly selected bundle of steel rods is greater than 259 cm is 0.65173.

Step-by-step explanation:

We are given that a company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 259.2 cm and a standard deviation of 2.1 cm. For shipment, 17 steel rods are bundled together.

Let \bar X = <u><em>the average length of rods in a randomly selected bundle of steel rods</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean length of rods = 259.2 cm

           \sigma = standard deviaton = 2.1 cm

           n = sample of steel rods = 17

Now, the probability that the average length of rods in a randomly selected bundle of steel rods is greater than 259 cm is given by = P(\bar X > 259 cm)

 

     P(\bar X > 259 cm) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{259-259.2}{\frac{2.1}{\sqrt{17} } } ) = P(Z > -0.39) = P(Z < 0.39)

                                                                = <u>0.65173</u>

The above probability is calculated by looking at the value of x = 0.39 in the z table which has an area of 0.65173.

8 0
3 years ago
An amusement park sells child and adult tickets at a ratio of 3:1. On Saturday, they sold 63 adult tickets. How many child ticke
nalin [4]

Answer:

The number of child tickets sold by the amusement park is 189.

Step-by-step explanation:

Let A represent the number of adult tickets and C represents the number of child tickets, therefore we have:

3:1 = C:A .................... (1)

Where;

C = ?

A = 63

Substituting for the value into equation (1), we have:

3:1 = C:63

This can be converted to solve for C as follows:

3 / (3 + 1) = C / (C + 63)

3 / 4 = C / (C + 63)

0.75 = C / (C + 63)

0.75(C + 63) = C

0.75C + (0.75 * 63) = C

0.75C + 47.25 = C

47.25 = C - 0.75C

47.25 = 0.25C

C = 47.25 / 0.25

C = 189

Therefore, the number of child tickets sold by the amusement park is 189.

4 0
3 years ago
15 POINTS!
ollegr [7]
How often do they pay? If they only pay once a year, 90000/100=900.  900 * .52 = 468
6 0
3 years ago
Read 2 more answers
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