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VikaD [51]
4 years ago
6

Jason is traveling by car from his home to his office. He has gone 3.5 miles so far. From this point on, he can cover 100 miles

every 2 hours. What is the equation of a line that models the total miles traveled, y, in x hours after this point?
A.
y = 100x + 3.5
B.
y = 100x – 3.5
C.
y = 50x + 3.5
D.
y = -50x + 3.5
Mathematics
2 answers:
Novay_Z [31]4 years ago
8 0
Your answer is 50x + 3.5
igor_vitrenko [27]4 years ago
5 0

answer: 50x +3.5

Step-by-step explanation:

100 miles divided by 2 hours is 50 miles per hour, and he has already gone 3.5 miles, so the answer would be y= 50x +3.5

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Answer:

x = 2 1/ 2 , − 1 1/2

Step-by-step explanation:

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What is the expression of 1/4/1/7​
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4 years ago
Suppose that the speed at which cars go on the freeway is normally distributed with mean 68 mph and standard deviation 5 miles p
denis23 [38]

Answer:

A)X \sim N(68 , 25)

B) the probability that is traveling more than 70 mph is 0.3446

C) the probability that it is traveling between 65 and 75 mph is 0.6449

D) 90% of all cars travel at least 56.85 mph fast on the freeway

Step-by-step explanation:

The speed at which cars go on the freeway is normally distributed with mean 68 mph and standard deviation 5 miles per hour.

Mean = \mu = 68 mph

Standard deviation = \sigma = 5 mph

A) X ~ N( _____, _______ )

In general X \sim N( \mu , \sigma^2)

\mu = 68 mph

\sigma = 5 mph

\sigma^2 = 5^2 = 25

So, X \sim N(68 , 25)

B) If one car is randomly chosen, find the probability that is traveling more than 70 mph.i.e.P(X>70)

So,Z = \frac{x-\mu}{\sigma}\\Z=\frac{70-68}{5}

Z=0.4

Using Z table

P(Z>70)=1-P(Z<70)=1-0.6554=0.3446

Hence the probability that is traveling more than 70 mph is 0.3446

C) If one of the cars is randomly chosen, find the probability that it is traveling between 65 and 75 mph.

P(65<X<75)

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

AT x = 65

Z=\frac{65-68}{5}

Z=-0.6

AT x = 75

Z=\frac{75-68}{5}

Z=1.4

Using Z table

P(65<X<75)=P(-0.6<Z<1.4)=P(Z<1.4)-P(Z<-0.6)=0.9192-0.2743=0.6449

Hence the probability that it is traveling between 65 and 75 mph is 0.6449

D)90% of all cars travel at least how fast on the freeway?

Since we are supposed to find at least how fast on the freeway

So,P(X>x)=0.9

1-P(X<x)=0.9

1-0.9=P(X<x)

0.1=P(X<x)

Z value at 10% =-2.23

So, Z=\frac{x-\mu}{\sigma}\\-2.23=\frac{x-68}{5}\\-2.23 \times 5 =x-68\\(-2.23 \times 5)+68=x

56.85 = x

Hence 90% of all cars travel at least 56.85 mph fast on the freeway

8 0
3 years ago
Student enrollment at a local school is concerning the community because the number of students has dropped to 504, which is a 2
Trava [24]

Answer: The student enrollment the previous year was 630.

Step-by-step explanation: What we have currently is 504 students which represents 80 percent of the previous total number of students. We can conclude this because the question states clearly that the student enrollment dropped by 20 percent. In other words, we need to add back 20 percent to 80 percent to get the total 100 percent student enrollment from the previous year. The total number of students can be derived as follows;

If 504 equals 80 percent or (0.8), then 504 divided by 4 gives us 20 percent or (0.2)

20 percent times 5 equals 100 percent (or 0.2 times 5 equals 1)

Hence the calculation becomes;

(504/4) x 5 = Total

126 x 5 = 630

Therefore the total number of students enrollment the previous year was 630.

3 0
3 years ago
3. Paul starts with $1200 in his savings account. Each month Paul spends $40. What is the equation
REY [17]

Answer:

y = -40x + 1200

Step-by-step explanation:

let x be the # of months

4 0
3 years ago
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