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LenKa [72]
2 years ago
11

A circle with center Q located at (1, 2) is drawn on the coordinate plane below.

Mathematics
1 answer:
Vlada [557]2 years ago
5 0
<h2>Use this link to find your answer</h2>

https://lalinks.org/linksweb

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The original selling price for a dress was D dollars. If the selling price is increased 30%. After that the price dropped 15%. W
hodyreva [135]
15% increase
15/100xD
D=100/15
D=6.67
D+6.67
6 0
3 years ago
g A lawyer commutes daily from his suburban home to his midtown office. The average time for a one-way trip is 24 minutes, with
rodikova [14]

Answer:

a) 85.31%

b) 56 minutes

c) 17 days

d) 0.1721

Step-by-step explanation:

In order to make the calculations easier, let's <em>standardize the curve</em> by doing the change

Z=\frac{X-\mu}{\sigma}

where \mu is the average trip-time and \sigma the standard deviation and

P(X≤ t) is the probability that the trip takes less than t  minutes.

a)

Here we are looking for the value of P(X>20).

For X=20 we have Z = (20-24)/3.8 = -1.05

So, we want the area under the standard normal curve for Z > -1.05

that we can compute either using a table or a computer and we find this area equals 0.8531

So, he arrives late to work 85.31% of the times.

(See picture 1 attached)

b)

In this case we are looking for a value t of time such that

P(X≥ t) = 20% = 0.2

So, we are seeking a value Z such that the area under the normal curve to the left of Z equals 0.2

By using a table or a computer we find Z = 0.842

(See picture 2 attached)

By inserting this value in the equation on standardization  

0.842=\frac{X-24}{38}\Rightarrow X=38*0.842+24=55.996

So 20% of the longest trips takes 55.996 ≅ 56 minutes

c)

Since the average of days he arrives late is 85.31%, it is expected that in 20 work days he arrives late 85.31% of 20, which equals 17 days.

d)

Since the probability that he does not arrive late is 1-0.8531 = 0.1469 and the probability of arriving early is independent of the previous trip,

the probability of arriving early n days in a row is

0.1496*0.1496*...*0.1496 n times and

the probability of being early most than 10 trips in 20 days is

(0.1469)+(0.1469)^2+(0.1469)^3+...+(0.1469)^10=0.1721

3 0
3 years ago
What is the process 9.04 turning into 9.10?
Yuki888 [10]
You would add 0.06.

Hope This Helps You!
Good Luck Studying :)
3 0
2 years ago
Honon spent 3 1/4 hours on homework while Sequoia spent 2 5/6 hours on homework. How much more time did Honon spend on homework
Rzqust [24]

Answer:

5/12 hours

Step-by-step explanation:

Honon = 3 1/4 hours

Sequoia = 2 5/6 hours

How much more time did Honon spend on homework than Sequoia

= number of hours Honon spent - number of hours Sequoia spent

= 3 1/4 hours - 2 5/6 hours

= 13/4 - 17/6

= (39-34) /12

= 5/12 hours

6 0
3 years ago
829 to the nearest hundred
Ivenika [448]
Hi There!

<span>829 to the nearest hundred?


</span><span>Nearest 100th = 800</span><span>

</span>
6 0
3 years ago
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