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2 years ago

6

Lalo has 1,500 minutes per month on his cell phone plan. How many more minutes can he use if he has already talked for 785 minut

es in an inequality? Explain.

1 answer:

lozanna [386]2 years ago

6 0

Answer: The answer is 715 minutes more.

Step-by-step explanation:

1500-785=715

Double check::::::::::::

715+785=1500

You might be interested in

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

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

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

$X - 128 = 1.28*23$

$X = 157.44$

So

157.44mmHg

6 0

2 years ago

A campus deli serves 300 customers over its busy lunch period from 11:30 a.m. to 1:30 p.m. A quick count of the number of custom

Irina-Kira [14]

Complete question is;

A campus deli serves 300 customers over its busy lunch period from 11:30 am to 1:30 pm. A quick count of the number of customers waiting in line and being served by the sandwich makers shows that an average of 10 customers are in process at any point in time. What is the average amount of time that a customer spends in process?

Answer:

4 minutes

Step-by-step explanation:

For this question, we will apply Little's law which is is a theorem that determines the average number of items in a stationary queuing system, based on the average waiting time of an item within a system and the average number of items arriving at the system per unit of time.

The formula for the law is:

Inventory = flow rate × flow time.

We are given;

Inventory = 300 customers

flow time is from 11: 30am to 1:30pm which is 2 hours = 120 minutes

Flow rate = 300/120 = 2.5 persons/minute

Now, Making flow rate the subject of the formula earlier given, we have;

flow rate = inventory/ flow time

Flow time is the time each person spends in the process

Thus, plugging in the relevant values, we get ;

We are told that an average of 10 customers are in process at any point in time.

Thus;

Average flow time = average inventory/flow rate = 10/2.5 = 4 minutes

3 0

2 years ago

From Statistics and Data Analysis from Elementary to Intermediate by Tamhane and Dunlop, pg 265. A thermostat used in an electri

Leviafan [203]

Answer:

$t=\frac{201.77-200}{\frac{2.41}{\sqrt{10}}}=2.32$

The degrees of freedom are given by:

$df=n-1=10-1=9$

The p value for this case is given by:

$p_v =2*P(t_{(9)}>2.32)=0.0455$

For this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly different from 200 F.

Step-by-step explanation:

Information given

data: 202.2 203.4 200.5 202.5 206.3 198.0 203.7 200.8 201.3 199.0

We can calculate the sample mean and deviation with the following formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

$\sigma=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=201.77$ represent the sample mean

$s=2.41$ represent the sample standard deviation

$n=10$ sample size

$\mu_o =200$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic

$p_v$ represent the p value for the test

Hypothesis to test

We want to determine if the true mean is equal to 200, the system of hypothesis are :

Null hypothesis: $\mu = 200$

Alternative hypothesis: $\mu = 200$

The statistic for this case is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

The statistic is given by:

$t=\frac{201.77-200}{\frac{2.41}{\sqrt{10}}}=2.32$

The degrees of freedom are given by:

$df=n-1=10-1=9$

The p value for this case is given by:

$p_v =2*P(t_{(9)}>2.32)=0.0455$

For this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly different from 200 F.

4 0

2 years ago

The spelling contest awards four prizes Tim,Sam,Jen, and Kate all made it to the final round. In how many ways can first, second

topjm [15]

It is the 3 one correct me if im wrong

7 0

2 years ago

Which of the following terms appear in the expression of (x+y)^10? The letter a in each term represents a real constant.

Ipatiy [6.2K]

A,B,D because the sum of the powers of variables in each term of the expansion is equal to the power the binomial is raised to.

5 0

2 years ago

Read 2 more answers