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kolezko [41]
2 years ago
9

The traffic engineer for a large city is conducting a study on traffic flow at a certain intersection near the city administrati

on building. The engineer will collect data on different variables related to the intersection each day for ten days. Of the following variables, which will be measured using continuous data?
(A)-The number of cars passing through the intersection in one hour
(B)-The number of pedestrians crossing the intersection in one hour
(C)-The number of bicyclists crossing the intersection in one hour
(D)-The number of food trucks that park within four blocks of the intersection
(E)-The number of minutes for a car to get from the intersection to the administration building
Mathematics
1 answer:
barxatty [35]2 years ago
8 0

Answer:

e) The number of minutes for a car to get from the intersection to the administration building

Step-by-step explanation:

Continuous Data:

Data that can take any value (an infinite number of values) within a certain range.

For example, the statistics of a group of people form continuous data, but the number of people in that group form discrete data.

Inglés

In this case, counting items such as cars, bicycles, people, are considered discrete data. They exclusively take integer values.

But time data can take continuous values.

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For the level 3 course, exam hours cost twice as much as workshop hours, workshop hours cost twice as much as lecture hours. How
natulia [17]
<h2>Answer</h2>

Cost of lectures = $7.33 per hour

<h2>Explanation</h2>

Let e the cost of the exam hours

Let w be the cost of the workshop hours

Let l be the cost of the lecture hours.

We know from our problem that exam hours cost twice as much as workshop, so:

e=2w equation (1)

We also know that workshop hours cost twice as much as lecture hours, so:

w=2l equation (2)

Finally, we also know that 3hr exams 24hr workshops  and 12hr lectures cost $528, so:

3e+24w+12l=528 equation (1)

Now, lets find the value of l:

Step 1.  Solve for l in equation (3)

3e+24w+12l=528

12l=528-3e-24w equation (4)

Step 2. Replace equation (1) in equation (4) and simplify

12l=528-3e-24w

12l=528-3(2w)-24w

12l=528-6w-24w

12l=528-30w equation (5)

Step 3. Replace equation (2) in equation (5) and solve for l

12l=528-30w

12l=528-30(2l)

12l=528-60l

72l=528

l=\frac{528}{72}

l=\frac{22}{3}

l=7.33

Cost of lectures  = $7.33 per hour



3 0
2 years ago
Read 2 more answers
An English teacher has equal numbers of science fiction, biography, and
swat32

Answer

1/9

Step-by-step explanation:

the chance that she gets it one day is 1/3 so you multiply that by 1/3 because there are three days.

4 0
2 years ago
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The area of a trapezoid is 140 cm2 and the height is 20 cm. What is the sum of the bases?
Reil [10]
D 7
A million apologies if I’m wrong half of my brain is still on vacation!
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2 years ago
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To the nearest degree, what is the measure of the central angle for faucets? 37° 24° 48° 43°
Masteriza [31]

Answer:

\large\boxed{43^o}

Step-by-step explanation:

\text{Faucets}\to12\%\\\\p\%=\dfrac{p}{100}\to12\%=\dfrac{12}{100}=0.12\\\\12\%\ \text{of}\ 360^o\to0.12\cdot360^o=43.2^o\approx43^o

4 0
2 years ago
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A machine fills boxes weighing Y lb with X lb of salt, where X and Y are normal with mean 100 lb and 5 lb and standard deviation
Bas_tet [7]

Answer:

Option b. None is the correct option.

The Answer is 63%

Step-by-step explanation:

To solve for this question, we would be using the z score formula

The formula for calculating a z-score is given as:

z = (x-μ)/σ,

where

x is the raw score

μ is the population mean

σ is the population standard deviation.

We have boxes X and Y. So we will be combining both boxes

Mean of X = 100 lb

Mean of Y = 5 lb

Total mean = 100 + 5 = 105lb

Standard deviation for X = 1 lb

Standard deviation for Y = 0.5 lb

Remember Variance = Standard deviation ²

Variance for X = 1lb² = 1

Variance for Y = 0.5² = 0.25

Total variance = 1 + 0.25 = 1.25

Total standard deviation = √Total variance

= √1.25

Solving our question, we were asked to find the percent of filled boxes weighing between 104 lb and 106 lb are to be expected. Hence,

For 104lb

z = (x-μ)/σ,

z = 104 - 105 / √25

z = -0.89443

Using z score table ,

P( x = z)

P ( x = 104) = P( z = -0.89443) = 0.18555

For 1061b

z = (x-μ)/σ,

z = 106 - 105 / √25

z = 0.89443

Using z score table ,

P( x = z)

P ( x = 106) = P( z = 0.89443) = 0.81445

P(104 ≤ Z ≤ 106) = 0.81445 - 0.18555

= 0.6289

Converting to percentage, we have :

0.6289 × 100 = 62.89%

Approximately = 63 %

Therefore, the percent of filled boxes weighing between 104 lb and 106 lb that are to be expected is 63%

Since there is no 63% in the option, the correct answer is Option b. None.

3 0
2 years ago
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