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navik [9.2K]
3 years ago
12

Bertha has 6 marbles, and Sheniqwa has 9 marbles. What is the difference between the amount of marbles Bertha has and the amount

that Sheniqwa has?
Mathematics
2 answers:
Evgesh-ka [11]3 years ago
8 0
The answer is 3 marbles
atroni [7]3 years ago
3 0
3marbles is the difference
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A factory received an order for 1,758 laptops. For the first 6 days, it produced 125 laptops per day. If it was asked to finish
BartSMP [9]

Answer:

  • 252 laptops

Step-by-step explanation:

Target number is 1758

<u>Produced in 6  days:</u>

  • 6*125 = 750

<u>Remaining to produce:</u>

  • 1758 - 750= 1008

Days left- 4

<u>Should produce per day:</u>

  • 1008/4 = 252 laptops
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How do you tell a triangle is not a right triangle
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You can tell when you cant draw an even L shaped figure at the intersects of the bottom and the side. 90 degrees. But there are many different ways that may make more sense. Happy to help:)
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3 years ago
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What is the unit rate in miles per hour if Kelly walks 6 miles in 1.5 hours?
sammy [17]

Answer:

4 miles per hour.

Step-by-step explanation:

6/1.5 = 4

6 0
3 years ago
Can someone help me on this plz
Studentka2010 [4]

Answer:

y=4/3x+2.5

Step-by-step explanation:

im not 100% sure if i get it wrong reply

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
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