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Montano1993 [528]
3 years ago
13

Al, Pablo, and Marsha shared the driving on a 1,500-mile trip. Which of the three drove the greatest distance on the trip?(1) Al

drove 1 hour longer than Pablo but at an average rate of 5 miles per hour slower than Pablo.(2) Marsha drove 9 hours and averaged 50 miles per hour.
Mathematics
1 answer:
Svetach [21]3 years ago
6 0

Answer:

Pablo drove the greatest distance of 527 miles.

Missing Content:

Since information on Pablo is not given, it is impossible to solve this problem. Let us assume some scenario for Pablo. Say:

(3)Pablo averaged 52.7 miles per hour and drove for 10 hours.

Step-by-step explanation:

Using this information, we can find out the data on Al,

Al drove 1 hour longer than Pablo so,

Time Al drove= 10+1 =11 hours

Average rate of Al=Pablo's average rate-5= 52.7-5 = 47.7 miles per hour.

We know that,

Average Speed = \frac{Distance}{Time}

Distance =Average Speed x Time

Distance travelled by Al = 47.7 x 11 =524.7 miles

Distance Travelled by Marsha = 50 x 9=450 miles

Distance Travelled by Pablo= 52.7 x 10= 527 miles

<u>As we can see that Pablo drove the greatest distance which is 527 miles.</u>

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Answer:

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Step-by-step explanation:

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r/ 2 + − 2r ⋅ 2/ 2 + 6 − 3

r − 2 r ⋅ 2 /2 + 6 ⋅ 2/ 2 + − 3 ⋅ 2/ 2

r − 2 r ⋅2 + 6⋅ 2 − 3 ⋅ 2 /2

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7 0
3 years ago
Divide. (6.496×10−6)÷(2.8×105) Express your answer in scientific notation. Enter your answer in the boxes.
ss7ja [257]

we have been asked to perform (6.496*10^{-6})\div (2.8*10^5)

The given expression can be written as

\frac{(6.496*10^{-6})}{2.8*10^5}

Now it can be re-written as below

=\frac {(6.496*10^{-6})}{2.8*10^5}=\frac{6.496}{2.8}*\frac{10^{-6}}{10^5}\\ \\ \text{Now using the exponent rule }  \\ \\ \frac{a^m}{a^n}=a^{m-n}\\ \\ \text{we can re-write as below}\\ \\  \frac{6.496}{2.8}*\frac{10^{-6}}{10^5}= \frac{6.496}{2.8}*10^{-6-5}= \frac{6.496}{2.8}*10^{-11}\\ \\ \text{Now divide we get}\\ \\ =2.32*10^{-11}

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3 years ago
I need help with number 7 and 8. This problem is from chapter 14.1 advanced 1 big ideas math. Someone please help ASAP.
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A flashlight battery manufacturer makes a model of battery whose mean shelf life is three years and four months, with a standard
Vladimir [108]

Answer:

17,065 of those batteries can be expected to last between three years and one month and three years and seven months

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem:

I will calculate the time in months. Each year has twelve months.

Mean shelf life is three years and four months, with a standard deviation of three months. So

\mu = 3*12 + 4 = 40

\sigma = 3

Proportion lasting between three years and one month and three years and seven months:

This is the pvalue of Z when X = 3*12 + 7 = 43 subtracted by the pvalue of Z when X = 3*12 + 1 = 37

X = 43

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

Z = \frac{43 - 40}{3}

Z = 1

Z = 1 has a pvalue of 0.8413.

X = 37

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

Z = \frac{37 - 40}{3}

Z = -1

Z = -1 has a pvalue of 0.1587.

0.8413 - 0.1587 = 0.6826

Out of 25,000 batteries:

68.26% of the batteries are expected to last between three years and one month and three years and seven months.

0.6826*25000 = 17,065

17,065 of those batteries can be expected to last between three years and one month and three years and seven months

6 0
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