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JulijaS [17]
4 years ago
10

Two jets leave at the same time and fly in opposite directions. One is heading west 50 mph faster than the other flying east. Af

ter 2 hours they are 2500 miles apart. Find each of their speeds.
Mathematics
1 answer:
alexira [117]4 years ago
7 0

Let's assume

speed of second jet is x mph

we are given

One is heading west 50 mph faster than the other flying east

so, speed of second jet is x+50

they are moving in opposite direction

now, we will find distance travelled by them after 2 hours

Distance travelled by second jet after 2 hours:

we know that

distance =speed*time

so, d_1=x*2

so, d_1=2x

Distance travelled by first jet after 2 hours:

we know that

distance =speed*time

so, d_2=(x+50)*2

so, d_1=2x+100

now, we can find total distance

so, d=d_1+d_2

d=2x+2x+100

we are given that distance as 2500

so, we can set it to 2500

2500=2x+2x+100

now, we can solve for x

4x=2400

x=600

so,

speed of second jet is 600 mph

speed of first jet is 650 mph.................Answer

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Step-by-step explanation:

Hello!

The distance can be positive or negative, as we are just counting the number of units they are apart.

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<h3>Find G</h3>
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An Article a Florida newspaper reported on the topics that teenagers most want to discuss with their parents. The findings, the
Sergio039 [100]

Answer:

The estimate is  P__{hat}} \pm E  = 0.37 \pm 0.0348

Step-by-step explanation:

From the question we are told that  

    The sample size is  n =  522

    The sample proportion of students  would like to talk about school is  \r p__{hat}} =  0.37

  Given that the confidence level is  90 % then the level of significance can be mathematically evaluated as

                  \alpha  =  100 - 90

                  \alpha  =  10\%

                  \alpha  =  0.10

Next we obtain the critical value of  \frac{\alpha }{2} from the normal distribution table, the value is  

               Z_{\frac{\alpha }{2} } =Z_{\frac{0.10}{2} } =  1.645

Generally the margin of error can be mathematically represented as

               E =  Z_{\frac{\alpha }{2} } *  \sqrt{\frac{\r P_{hat}(1- \r P_{hat} )}{n } }

=>            E = 1.645 *  \sqrt{\frac{0.37 (1- 0.37  )}{522 } }

=>             E = 0.0348

Generally the estimate the proportion of all teenagers who want more family discussions about school at 90% confidence level is  

                       P__{hat}} \pm  E

substituting values

                     0.37 \pm 0.0348

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3 years ago
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