BD is the bisector of angle ABC. Find the length of the segment AE.

anyanavicka [17]

Measurement of one angle and length of one side I think

8 0

3 years ago

A family drove a total of 5,300 miles.They drove 10 more miles in tj second half of the trip than the first half. How many did t

AVprozaik [17]

If they drove a total of 5,300 miles in 2 days, and on the second day they drove 10 more miles then the 1st day.

5,300 ÷ 2 = 2,650

they drove 10 miles less on the first day

2,650 - 10 = 2,640

They drove 2,640 miles on the first day.

Hope this helps. :)

7 0

3 years ago

A parking lot charges $3 to park a car for the first hour and $2 per hour after that. If you use more than one parking space, th

fenix001 [56]

Consider the charge for parking one car for t hours.

If t is more than 1, then the function is y=3+2(t-1), because 3 $ are payed for the first hour, then for t-1 of the left hours, we pay 2 $.

If t is one, then the rule y=3+2(t-1) still calculates the charge of 3 $, because substituting t with one in the formula yields 3.

75% is 75/100 or 0.75.

For whatever number of hours t, the charge for the first car is 3+2(t-1) $, and whatever that expression is, the price for the second car and third car will be

0.75 times 3+2(t-1). Thus, the charge for the 3 cars is given by:

3+2(t-1)+0.75[3+2(t-1)]+0.75[3+2(t-1)]=3+2(t-1)+0.75 × 2[3 + 2(t − 1)].

Thus, the function which total parking charge of parking 3 cars for t hours is:

f(t) = (3 + 2(t − 1)) + 0.75 × 2(3 + 2(t − 1))

Answer: C

4 0

3 years ago

How do you solve 7x-5=30?

Nadya [2.5K]

7x - 5 = 30

* add 5 to both sides

7x = 30 + 5

7x = 35

* divided both sides by 7

(7/7)x = 35/7

x = 5

8 0

3 years ago

Read 2 more answers

WILL MARK AS BRAINLIEST!!! 5. A 2011 study by The National Safety Council estimated that there are nearly 5.7 million traffic ac

Artyom0805 [142]

Answer:

Step-by-step explanation:

Hello!

Regarding the reasons that traffic accidents occur:

28% are caused by distracted drivers using cell phones or texting

11% of the drivers' user their phones at any time

The probability of a driver having an accident is 5.26%

a)

DC = event that a randomly selected driver is using a cell phone.

P(DC)= 0.11

b)

TA = event that a randomly selected driver has a traffic accident.

P(TA)= 0.0526

c) and f)

If both events are related, i.e. dependent, then you would expect that the occurrence of one of these events will affect the probability of the other one. If they are not related, i.e. independent events, then their probabilities will not be affected by the occurrence of one or another:

If both events are independent P(TA|DC)= P(TA)

If they are dependent, then:

P(TA|DC)≠ P(TA)

P(TA|DC)= 0.28

P(TA)= 0.0526

As you can see the probability of the driver having an accident given that he was using the cell phone is different from the probability of the driver having an accident. This means that both events are related.

d) and e)

You have to calculate the probability that "the driver was distracted with the phone given that he had an accident", symbolically P(DC|TA)

P(DC|TA) = $\frac{P(DCnTA)}{P(TA)}$

$P(TA|DC)= \frac{P(TAnDC}{P(DC)}$ ⇒ P(DC∩TA)= P(TA|DC)*P(DC)= 0.28 * 0.11= 0.0308

P(DC|TA) = $\frac{0.0308}{0.0526}= 0.585= 0.59$

I hope this helps!

3 0

3 years ago