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2 years ago

Cheese sticks that were previously priced at "10 for $1" are now "2 for $1". Find each percent change.

Mathematics

1 answer:

AysviL [449]2 years ago

7 0

Answer:

The percent decrease in the number of cheese sticks you can buy for $1 is 80%.

Step-by-step explanation:

Given : Cheese sticks that were previously priced at "10 for $1" are now "2 for $1".

To find : The percent decrease in the number of cheese sticks you can buy for $1 ?

Solution :

The formula used to find percent decrease is given by,

$\%\text{ change}=\frac{\text{Amount of change}}{\text{Original amount }}\times 100$

The price change from 10 to 2,

$\%\text{ change}=\frac{10-2}{10}\times 100$

$\%\text{ change}=\frac{8}{10}\times 100$

$\%\text{ change}=80\%$

The percent decrease in the number of cheese sticks you can buy for $1 is 80%.

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Ibrahim has two lists of numbers. The mean of the numbers in the first list is p. The mean of the numbers in the second list is

Nata [24]

Step-by-step explanation:

The two conditions that must be satisfied for Ibrahim to be correct are:

1. The range of numbers in each list must also be the same.

2. The number of numbers in both list must also be same.

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2 years ago

A worker drives a 0.500 kg spike into a rail tie with a 2.50 kg sledgeham-

Marianna [84]

Total internal energy increases by 1760 J

Step-by-step explanation:

The kinetic energy of an object is the energy possessed by the object due to its motion.

It is calculated as

$KE=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

For the hammer in this problem:

m = 2.50 kg

v = 65.0 m/s

So its kinetic energy is

$KE=\frac{1}{2}(2.50)(65)^2=5281 J$

Then the problem says that 1/3 of the hammer's kinetic energy is converted into internal energy: therefore, the total internal energy increases by

$\frac{1}{3}KE=\frac{1}{3}(5281)=1760 J$

Learn more about kinetic energy:

brainly.com/question/6536722

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8 0

2 years ago

What is the simplified form of the following expression? Assume x not-equals 0. RootIndex 5 StartRoot StartFraction 10 x Over 3

SVEN [57.7K]

Answer:

$\dfrac{\sqrt[5]{810x^3}}{3x}$

Step-by-step explanation:

As a rule, the "simplified form" means there are no fractions under a radical.

$\sqrt[5]{\dfrac{10x}{3x^3}}=\sqrt[5]{\dfrac{10x(3^4x^2)}{3x^3(3^4x^2)}}=\sqrt[5]{\dfrac{810x^3}{(3x)^5}}=\boxed{\dfrac{\sqrt[5]{810x^3}}{3x}}$

5 0

2 years ago

According to the WHO MONICA Project the mean blood pressure for people in China is 128 mmHg with a standard deviation of 23 mmHg

Sati [7]

Answer:

a) Mean blood pressure for people in China.

b) 38.21% probability that a person in China has blood pressure of 135 mmHg or more.

c) 71.30% probability that a person in China has blood pressure of 141 mmHg or less.

d) 8.51% probability that a person in China has blood pressure between 120 and 125 mmHg.

e) Since Z when X = 135 is less than two standard deviations from the mean, it is not unusual for a person in China to have a blood pressure of 135 mmHg

f) 157.44mmHg

Step-by-step explanation:

When the distribution is normal, we use 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.

If X is two standard deviations from the mean or more, it is considered unusual.

In this question:

$\mu = 128, \sigma = 23$

a.) State the random variable.

Mean blood pressure for people in China.

b.) Find the probability that a person in China has blood pressure of 135 mmHg or more.

This is 1 subtracted by the pvalue of Z when X = 135.

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

$Z = \frac{135 - 128}{23}$

$Z = 0.3$

$Z = 0.3$ has a pvalue of 0.6179

1 - 0.6179 = 0.3821

38.21% probability that a person in China has blood pressure of 135 mmHg or more.

c.) Find the probability that a person in China has blood pressure of 141 mmHg or less.

This is the pvalue of Z when X = 141.

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

$Z = \frac{141 - 128}{23}$

$Z = 0.565$

$Z = 0.565$ has a pvalue of 0.7140

71.30% probability that a person in China has blood pressure of 141 mmHg or less.

d.)Find the probability that a person in China has blood pressure between 120 and 125 mmHg.

This is the pvalue of Z when X = 125 subtracted by the pvalue of Z when X = 120. So

X = 125

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

$Z = \frac{125 - 128}{23}$

$Z = -0.13$

$Z = -0.13$ has a pvalue of 0.4483

X = 120

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

$Z = \frac{120 - 128}{23}$

$Z = -0.35$

$Z = -0.35$ has a pvalue of 0.3632

0.4483 - 0.3632 = 0.0851

8.51% probability that a person in China has blood pressure between 120 and 125 mmHg.

e.) Is it unusual for a person in China to have a blood pressure of 135 mmHg? Why or why not?

From b), when X = 135, Z = 0.3

Since Z when X = 135 is less than two standard deviations from the mean, it is not unusual for a person in China to have a blood pressure of 135 mmHg.

f.) What blood pressure do 90% of all people in China have less than?

This is the 90th percentile, which is X when Z has a pvalue of 0.28. So X when Z = 1.28. Then

X = 120

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

$1.28 = \frac{X - 128}{23}$

$X - 128 = 1.28*23$

$X = 157.44$

157.44mmHg

6 0

2 years ago

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:

Component Reliabilities:

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:

Reliability of the Robot with backup of component 1 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

Similarly, to find out the reliability of component 2:

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

Reliability of the Robot with backup of component 3 can be computed as:

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

Reliability of the Robot with backup of component 4 can be computed as:

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 highest reliability can be achieved by adding a backup of component 4 with a value of 0.8664. So, Component 4 should get the backup in order to achieve the highest reliability.

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 chances of failure 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 reliability for each of the component & its backup 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

The reliability of the robot with backups 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

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

4 0

2 years ago