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givi [52]
2 years ago
9

An airline charges an extra fee if a carry-on bag weighs more than 30 pounds. After packing, Lorrie’s carry-on weighs 21.2 pound

s. The inequality represents the amount of weight Lorrie can add to the carry-on without going over the 30-pound limit.
21.2 + w<(With a line underneath it) 30

How much weight can Lorrie add to the carry-on without going over the 30-pound limit?
less than 8.8 pounds
more than 8.8 pounds
no less than 8.8 pounds
no more than 8.8 pounds
Mathematics
2 answers:
Goryan [66]2 years ago
6 0
To solve this you subtract 21.2 from 30.
30 - 21.2 = 8.8.

If she goes over 8.8, her bag will weigh more than 30 pounds. So the answer is: no more than 8.8 pounds. 
Andrews [41]2 years ago
6 0

Answer:

Option D:No more than 8.8 pounds.

Step-by-step explanation:

We are given that an airline charges an extra fee if a carry bag weighs more than 30 pounds.

Lorrie's carry weighs= 21.2 pounds

We are  given the inequality that represents amount of weight Lorrie can add without going over the 30- pound limit.

21.2+w\leq30

We have to find how much weight Lorrie can add to the carry on-without going over the 30

By solving inequality we find the value of w

Subtract 21.2 from both sides of inequality

21.2+w-21.2\leq 30-21.2

w\leq8.8

Therefore, Lorrie can add  weight less than or equal to 8.8 pounds.

Hence, option D  is true.

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Mateus’s bank issued an advertisement saying that 90\%90%90, percent of its customers are satisfied with the bank’s services. Si
densk [106]

Answer:

The probability of getting a sample with 80% satisfied customers or less is 0.0125.

Step-by-step explanation:

We are given that the results of 1000 simulations, each simulating a sample of 80 customers, assuming there are 90 percent satisfied customers.

Let \hat p = <u><em>sample proportion of satisfied customers</em></u>

The z-score probability distribution for the sample proportion is given by;

                                Z  =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, p = population proportion of satisfied customers = 90%

            n = sample of customers = 80

Now, the probability of getting a sample with 80% satisfied customers or less is given by = P( \hat p \leq 80%)

  P( \hat p \leq 80%) = P( \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } \leq \frac{0.80-0.90}{\sqrt{\frac{0.80(1-0.80)}{80} } } ) = P(Z \leq -2.24) = 1 - P(Z < 2.24)

                                                                  = 1 - 0.9875 = <u>0.0125</u>

The above probability is calculated by looking at the value of x = 2.24 in the z table which has an area of 0.9875.

8 0
2 years ago
Question 1 a
Savatey [412]

Answer:

B) A one-sample t-test for population mean would be used.

Step-by-step explanation:

The complete question is shown in the image below.

The marketing executive is interested in comparing the mean number of sales of this year to that of previous year.

The marketing executive already has the value of mean from previous year and uses a sample to calculate the mean and standard deviation of sales for the current year.

Since, data is being collected for one sample only this limits us to chose between one sample test for mean. So now the possible options are one sample t-test for population mean and one sample t-test for population mean.

If we read the statement we can see that we have the value of sample mean and sample standard deviation. Value of population standard deviation is unknown. In cases where value of population standard deviation is not known and sample standard deviation is given, t-test is used.

Therefore, we can conclude that A one-sample t-test for population mean would be used.

5 0
2 years ago
The perimeter of a triangle equals 8 inches. Two of its sides are equal and each of them is 1.3 inches bigger than the third sid
mafiozo [28]
Let's call the lengths of our two types of sides <em />x and y.

The two sides will that our 1.3 inches bigger than the third side will be have length x, and the length of the other side will be known as y. Thus, x = y + 1.3.

Considering this, we can add our sides together and set this value equal to 8, given the information in the problem:
(y + 1.3) + (y + 1.3) + y = 3y + 2.6 = 8

Now, let's solve for y.
3y + 2.6 = 8
3y = 5.4
y = 1.8

Now, we are not done yet. We must determine the true lengths of all of our sides. Using the equation we found earlier, the length of the two bigger sides is y + 1.3 = 1.8 + 1.3 = \boxed{3.1} inches and the length of our smaller side is simply \boxed{1.8} inches.

To verify, we can add these sides together and check that they equal 8:
3.1 + 3.1 + 1.8 = 8 ✔
5 0
2 years ago
At UF, there are always a few days between the end of classes and the beginning of final exams. These days are meant as a study
Natalija [7]

Answer:

This sample is not representative of all UF students, since those who are not local are not considered.

Step-by-step explanation:

This is a common statistics practice, when we want to study something from a population, we find a sample of this population.

However, the sample has to be representative

For example:

I want to estimate the proportion of New York state residents who are Buffalo Bills fans. So i ask, lets say, 1000 randomly selected Buffalo residents wheter they are Buffalo Bills fans, and expand this to the entire population of New York State residents. This is not representative of all New York State residents, just Buffalo residents.

In this problem, we have that:

They conduct a phone survey (with local numbers selected at random from the student directory) calling people during "dead week". Will this sample be representative of all UF students?

They only call those students with local numbers.

However, in an university, it is expected that there will be a good percentage of non local students.

So this sample is not representative of all UF students, since those who are not local are not considered.

8 0
2 years ago
Which number produces a rational number when multiplied by 0.5
ivanzaharov [21]
10 is your answer because it is a terminating or repeating decimals
5 0
2 years ago
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