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Phoenix [80]
1 year ago
11

Molly was on a long 136 mile road trip. The first part of the trip there was lots of traffic, she only averaged 16 mph. The seco

nd part of the trip there was no traffic so she could drive 44 mph. If the trip took her 5 hours, how long did she travel at each speed? In traffic she drove for _____ hours After the traffic cleared she drove for ____ hours.
Mathematics
1 answer:
Mazyrski [523]1 year ago
7 0

Answer:

In traffic, she drove for 3 hours

and After the traffic cleared she drove for 2 hours.

Explanation:

Given that the road trip was 136 miles;

d=136

The first part of the trip there was lots of traffic, she only averaged 16 mph;

v_1=16

The second part of the trip there was no traffic so she could drive 44 mph;

v_2=44

She traveled for a total of 5 hours;

t=5

let x represent the time in traffic when she traveled at 16 mph

t_1=x

the time the traffic is clear would be;

t_2=t-t_1=5-x

Recall that distance equals speed multiply by time;

d=v_1t_1_{}_{}^{}+v_2t_2

substituting the values;

136=16x+44(5-x)

solving for x;

\begin{gathered} 136=16x+220-44x \\ 44x-16x=220-136 \\ 28x=84 \\ x=\frac{84}{28} \\ x=3 \end{gathered}

So;

\begin{gathered} t_1=3\text{ hours} \\ t_2=5-x=5-3=2 \\ t_2=2\text{ hours} \end{gathered}

Therefore, In traffic, she drove for 3 hours

and After the traffic cleared she drove for 2 hours.

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2 years ago
Y+3=2(x-1) im not so sure that i got the answer correct.
Vikentia [17]
So just try it for x=1 how many will get for y 

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x=1
y=-3 so choice A.(1,-3) is right sure 

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8 0
3 years ago
Find the volume of a cube 8 1/8 cm.
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I hope this helps you



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6 0
3 years ago
Read 2 more answers
The difference of n and 7 is no more than 12. Find all possible values of n.
WINSTONCH [101]

Answer:

-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29

7-12= -5

7+12= 29

-5 ≤ n ≤ 29 or -5 ≤ n < 30 or -6< n ≤ 29 or -6 < n < 30

either one is acceptable. question was not clear but i did my best! :)

4 0
2 years ago
Read 2 more answers
The lengths of bolts in a batch are distributed normally with a mean of 3 cm and a standard
ValentinkaMS [17]

Answer:

0.8413 = 84.13% probability that a bolt has a length greater than 2.96 cm.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 3 cm and a standard deviation of 0.04 cm.

This means that \mu = 3, \sigma = 0.04

What is the probability that a bolt has a length greater than 2.96 cm?

This is 1 subtracted by the p-value of Z when X = 2.96. So

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

Z = \frac{2.96 - 3}{0.04}

Z = -1

Z = -1 has a p-value of 0.1587.

1 - 0.1587 = 0.8413

0.8413 = 84.13% probability that a bolt has a length greater than 2.96 cm.

3 0
2 years ago
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