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2 years ago

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The city subway is being extended to include a new stop in a large open plaza. The entrance to the new stop needs to be both arc

hitecturally distinctive and energy efficient, and it must fit within the budget. As part of your summer job with the Department of Transportation (DOT), you've been asked to review three design ideas and recommend which firm should develop its idea into a full design proposal

1 answer:

rodikova [14]2 years ago

4 0

What is the question you are asking?

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Wouldn't it be -1 I keep getting me wrong answer

AVprozaik [17]

Have u tryed 1 insted of -1 or it might be o solution

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2 years ago

What is the formula for finding the height of a cylinder?

andrezito [222]

H= volume/ pi(r)^2 is the height of a cylinder

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2 years ago

Read 2 more answers

Para embalar una caja se emplea 4,2 m de cinta adhesiva. ?Cuántas cajas se podrán embalar con tres rollos que tienen 3 hm, 7 dam

kompoz [17]

For this case, we perform the conversions:

First roll:

$1 \ Hectometer --------> 100 \ meters\\3 \ Hectometer --------> x$

$x = \frac {3 * 100} {1}\\x = 300 \ meters.$

We make a rule of three to determine the number of "c" boxes that can be packed with 300 meters of adhesive tape.

1 -----------> 4.2

c -----------> 300

$c = \frac {300 * 1} {4.2}\\c = 71.42857143\\c = 71$

You can pack 71 boxes.

Second roll:

$1 \ Decametro --------> 10 \ meters\\7 \ Decameter --------> x\\x = \frac {7 * 10} {1}\\x = 70 \ meters.$

We make a rule of three to determine the number of "c" boxes that can be packed with 70 meters of adhesive tape.

1 -----------> 4.2

c -----------> 70

$c = \frac {70 * 1} {4.2}\\c = 16.66666667\\c = 16$

You can pack 16 boxes.

Third roll:

1 -----------> 4.2

c -----------> 50

$c = \frac {50 * 1} {4.2}\\c = 11.9047619\\c = 11$

You can pack 11 boxes.

Thus, in total you can pack $11 + 16 + 71 = 98 \ boxes$

Answer:

98 boxes

4 0

2 years ago

An athlete ran 10 times around the circular track shown below. Approximately how many meters did the athlete run?

Anon25 [30]

Answer:

Step-by-step explanation:

Since the picture isn't given. I have no idea what the radius or diameter is and as such, have no definite answer for you.

However, from the question, if he's to run round the circular track once, then he must have run a total of

2πr meters, with r being the radius.

Now, the question says that he ran round the track 10 times, this means that whatever value we get there, for 1 trial, we multiply it by 10, to get the value for 10 trials. Essentially,

10 * 2πr

The value gotten is the needed answer. All you have to do is substitute r for the value you have in your diagram. We already know that π is 3.142

5 0

2 years ago

A real estate agent has 14 properties that she shows. She feels that there is a 50% chance of selling any one property during a

Ratling [72]

Answer:

91.02% probability of selling more than 4 properties in one week.

Step-by-step explanation:

For each property, there are only two possible outcomes. Either it is sold, or it is not. The chance of selling any one property is independent of selling another property. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

$n = 14, p = 0.5$

Compute the probability of selling more than 4 properties in one week.

Either you sell 4 or less properties in one week, or you sell more. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 4) + P(X > 4) = 1$

We want to find $P(X > 4)$ . So

$P(X > 4) = 1 - P(X \leq 4)$

In which

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{14,0}.(0.5)^{0}.(0.5)^{14} = 0.000061$

$P(X = 1) = C_{14,1}.(0.5)^{1}.(0.5)^{13} = 0.000854$

$P(X = 2) = C_{14,2}.(0.5)^{2}.(0.5)^{12} = 0.0056$

$P(X = 3) = C_{14,3}.(0.5)^{3}.(0.5)^{11} = 0.0222$

$P(X = 4) = C_{14,4}.(0.5)^{4}.(0.5)^{10} = 0.0611$

So

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.000061 + 0.000854 + 0.0056 + 0.0222 + 0.0611 = 0.0898$

Finally

$P(X > 4) = 1 - P(X \leq 4) = 1 - 0.0898 = 0.9102$

91.02% probability of selling more than 4 properties in one week.

8 0

2 years ago