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Arisa [49]
3 years ago
8

Insert grouping symbols in 5 x 2 + 8 - 4 - 2 so that its value is 15

Mathematics
1 answer:
Arlecino [84]3 years ago
5 0
5×2+8-4+1=15
cause
5×2=10
10+8=18
18-4=14
14+1=15
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There were 20 gallons of gasoline in your car's gas tank. You used 4 gallons and 1 quart. How much gasoline remains in the tank?
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25 gallons and 3 quarts
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A discounted amusement park ticket costs $12.95 less than the original price (p). Write and solve an equation to find the origin
Elena-2011 [213]
P = the original price of the ticket.

Let x =  the discounted price.
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3 0
3 years ago
Are the equations 3x = -9 and 4x = -12 equivalent? Explain
denis23 [38]

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3 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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