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Alex777 [14]
2 years ago
6

Samira ran 3 more miles than twice the number of miles Ada ran. Which statement makes this comparison using the correct variable

expression? If Ada ran m miles, then Samira ran m2+3 miles. If Ada ran m miles, then Samira ran 2m + 3 ⁢ miles. If Ada ran m miles, then Samira ran 2(3)+m ⁢ miles. If Ada ran m miles, then Samira ran 3m + 2 ⁢ miles.
Mathematics
2 answers:
Brilliant_brown [7]2 years ago
6 0
"Samira ran 3 miles more that twice the number of miles Ada ran." This means that to get the number of miles that Samira ran, we will multiply the number of miles that Ada ran and then add 3 to the product.

Assume that Ada ran m miles, the following the steps mentioned above, we can calculate the number of miles that Samira ran as follows:
Samira ran: 2m + 3 miles

Comparing this to the choices, we will find that the correct choice is the second one:
<span> If Ada ran m miles, then Samira ran 2m + 3 ⁢ miles.</span>
Irina18 [472]2 years ago
6 0

Answer:

2m+3 miles

Step-by-step explanation:

You might be interested in
One of the industrial robots designed by a leading producer of servomechanisms has four major components. Components’ reliabilit
Ivahew [28]

Answer:

a) Reliability of the Robot = 0.7876

b1) Component 1: 0.8034

    Component 2: 0.8270

    Component 3: 0.8349

    Component 4: 0.8664

b2) Component 4 should get the backup in order to achieve the highest reliability.

c) Component 4 should get the backup with a reliability of 0.92, to obtain the highest overall reliability i.e. 0.8681.

Step-by-step explanation:

<u>Component Reliabilities:</u>

Component 1 (R1) : 0.98

Component 2 (R2) : 0.95

Component 3 (R3) : 0.94

Component 4 (R4) : 0.90

a) Reliability of the robot can be calculated by considering the reliabilities of all the components which are used to design the robot.

Reliability of the Robot = R1 x R2 x R3 x R4

                                      = 0.98 x 0.95 x 0.94 x 0.90

Reliability of the Robot = 0.787626 ≅ 0.7876

b1) Since only one backup can be added at a time and the reliability of that backup component is the same as the original one, we will consider the backups of each of the components one by one:

<u>Reliability of the Robot with backup of component 1</u> can be computed by first finding out the chance of failure of the component along with its backup:

Chance of failure = 1 - reliability of component 1

                             = 1 - 0.98

                             = 0.02

Chance of failure of component 1 along with its backup = 0.02 x 0.02 = 0.0004

So, the reliability of component 1 and its backup (R1B) = 1 - 0.0004 = 0.9996

Reliability of the Robot = R1B x R2 x R3 x R4

                                         = 0.9996 x 0.95 x 0.94 x 0.90

Reliability of the Robot = 0.8034

<u>Similarly, to find out the reliability of component 2:</u>

Chance of failure of component 2 = 1 - 0.95 = 0.05

Chance of failure of component 2 and its backup = 0.05 x 0.05 = 0.0025

Reliability of component 2 and its backup (R2B) = 1 - 0.0025 = 0.9975

Reliability of the Robot = R1 x R2B x R3 x R4

                = 0.98 x 0.9975 x 0.94 x 0.90

Reliability of the Robot = 0.8270

<u>Reliability of the Robot with backup of component 3 can be computed as:</u>

Chance of failure of component 3 = 1 - 0.94 = 0.06

Chance of failure of component 3 and its backup = 0.06 x 0.06 = 0.0036

Reliability of component 3 and its backup (R3B) = 1 - 0.0036 = 0.9964

Reliability of the Robot = R1 x R2 x R3B x R4  

                = 0.98 x 0.95 x 0.9964 x 0.90

Reliability of the Robot = 0.8349

<u>Reliability of the Robot with backup of component 4 can be computed as:</u>

Chance of failure of component 4 = 1 - 0.90 = 0.10

Chance of failure of component 4 and its backup = 0.10 x 0.10 = 0.01

Reliability of component 4 and its backup (R4B) = 1 - 0.01 = 0.99

Reliability of the Robot = R1 x R2 x R3 x R4B

                                      = 0.98 x 0.95 x 0.94 x 0.99

Reliability of the Robot = 0.8664

b2) According to the calculated values, the <u>highest reliability can be achieved by adding a backup of component 4 with a value of 0.8664</u>. So, <u>Component 4 should get the backup in order to achieve the highest reliability.</u>

<u></u>

c) 0.92 reliability means the chance of failure = 1 - 0.92 = 0.08

We know the chances of failure of each of the individual components. The <u>chances of failure</u> of the components along with the backup can be computed as:

Component 1 = 0.02 x 0.08 = 0.0016

Component 2 = 0.05 x 0.08 = 0.0040

Component 3 = 0.06 x 0.08 = 0.0048

Component 4 =  0.10 x 0.08 = 0.0080

So, the <u>reliability for each of the component & its backup</u> is:

Component 1 (R1BB) = 1 - 0.0016 = 0.9984

Component 2 (R2BB) = 1 - 0.0040 = 0.9960

Component 3 (R3BB) = 1 - 0.0048 = 0.9952

Component 4 (R4BB) = 1 - 0.0080 = 0.9920

<u>The reliability of the robot with backups</u> for each of the components can be computed as:

Reliability with Component 1 Backup = R1BB x R2 x R3 x R4

                                                              = 0.9984 x 0.95 x 0.94 x 0.90

Reliability with Component 1 Backup = 0.8024

Reliability with Component 2 Backup = R1 x R2BB x R3 x R4

                                                              = 0.98 x 0.9960 x 0.94 x 0.90

Reliability with Component 2 Backup = 0.8258

Reliability with Component 3 Backup = R1 x R2 x R3BB x R4

                                                               = 0.98 x 0.95 x 0.9952 x 0.90

Reliability with Component 3 Backup = 0.8339

Reliability with Component 4 Backup = R1 x R2 x R3 x R4BB

                                                              = 0.98 x 0.95 x 0.94 x 0.9920

Reliability with Component 4 Backup = 0.8681

<u>Component 4 should get the backup with a reliability of 0.92, to obtain the highest overall reliability i.e. 0.8681. </u>

4 0
2 years ago
If you know that a &lt; b, and both a and b are positive numbers, then what must be true about the relationship between the oppo
Ymorist [56]

Answer:

The number further left on a number line is the smaller number. For positive numbers, the number closest to zero is smaller. For negative numbers, the number closest to zero is larger. If a is less than b, and they are both positive, then a is closer to 0 than b. The opposite of a is also closer to zero than the opposite of b, so the opposite of a must be larger than the opposite of b.

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
If Julie invests $9,250 at a rate of 7%, compounded weekly, find the value of the investment after 5 years.
Reil [10]

Answer:

Option D is correct.

Step-by-step explanation

Principal =  $9250

rate of interest = 7%  or 0.07

time = 5 years or 260 weeks

      [ Since there are 52 weeks in a year . for 5 years it will be 5x52=260 weeks]

Applying the formula  

Amount after t years =       P(1+\frac{r}{n} )^{nt}

                                         where P = principal

                                                     r = rate % in decimals

                                                    n= number of times in a year

                                                     t = times ( in years)

                          plugging the values  in the formula

              Amount =  9250(1+\frac{0.07}{52} )^{(52X(5)}

                             =  9250(1+0.001346 )^{(260)}

                              = 9250(1.001346 )^{(260)}

                               = 9250(1.418733588)

                                =$13123.29

5 0
2 years ago
Suppose the average price for new cars in 2012 has a mean of $30,100 and a standard deviation of $5,600. Based on this informati
KatRina [158]

Answer:

($13,300,$46,900)

Step-by-step explanation:

We are given the following in he question:

Mean, μ = $30,100

Standard Deviation, σ = $5,600

Chebyshev's Theorem:

  • According to  theorem atleast 1 - \dfrac{1}{k^2}  percent of data lies within 2 standard deviations of mean.
  • For k = 3,

1 -\dfrac{1}{(3)^2} = 0.889 \approx 89\%

Thus, 89% of data lies within three standard deviation of mean.

\mu - 3(\sigma) = 30100 - 3(5600) = 13300\\\mu + 3(\sigma) = 30100 + 3(5600) = 46900

Thus, we expect at least 89% of new car prices to fall within ($13,300,$46,900)

3 0
2 years ago
A clothing store is analyzing merchandise prices to help make decisions on what to charge for clothing on clearance. The store o
Aleks [24]

The variable is Quantitative, has Interval level of measurement.

Variables which can be quantified & expressed numerically are Quantitative variables. Eg : as given , price

Variables which cant be qualified & expressed numerically are Qualitative variables. Eg : level of honesty, loyalty etc

Nominal & Ordinal are qualitative variables : signifying yes or no to a category (like men or women) , or ranks (x better than y) respectively. So price level is not such categorical & ordinal ratio.

Quantitative ratio variables are with reference to time , or are in forms of rate (like speed , growth per year). So, price level is not such ratio variable also.

Price is a quantitative variable, in which the ranking, its difference can be calculated. This is characteristic of a <u>Quantitative Interval Variable</u>.

4 0
2 years ago
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