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2 years ago

Gabe can mow 2 1/2 acres of lawn in 1 day. How many acres of lawn can he mow in 2 1/3 days?

Mathematics

1 answer:

Basile [38]2 years ago

8 0

Answer:

5 5/6 acres

Step-by-step explanation:

(2 1/2 acres/day)·(2 1/3 days) = (5/2·7/3) acres = 35/6 acres = 5 5/6 acres

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A hospital finds that 22% of its accounts are at least 1 month in arrears. A random sample of 425 accounts was taken. What is th

GenaCL600 [577]

Answer:

8.85% probability that fewer than 82 accounts in the sample were at least 1 month in arrears

Step-by-step explanation:

For each account, there are only two possible outcomes. Either they are at least 1 month in arrears, or they are not. The probability of an account being at least 1 month in arrears is independent from other accounts. So the binomial probability distribution is used to solve this question.

However, we are working with a large sample. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using 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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$p = 0.22, n = 425$

$\mu = E(X) = np = 425*0.22 = 93.5$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{425*0.22*0.78} = 8.54$

What is the probability that fewer than 82 accounts in the sample were at least 1 month in arrears

This probability is the pvalue of Z when X = 82. So

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

$Z = \frac{82 - 93.5}{8.54}$

$Z = -1.35$

$Z = -1.35$ has a pvalue of 0.0885.

8.85% probability that fewer than 82 accounts in the sample were at least 1 month in arrears

8 0

2 years ago

In a completely randomized experimental design involving three assembly methods, 30 employees were randomly selected and 10 were

AnnZ [28]

Answer:

$F = \frac{MSR}{MSE} =\frac{45.89}{6.27}=7.32$

So then the best option is:

a. 7.32

Step-by-step explanation:

Previous concepts

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"

If we assume that we have $3$ groups and on each group from $j=1,\dots,10$ we have $10$ individuals on each group we can define the following formulas of variation:

$SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2$

$SS_{between=Treatment}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2$

$SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2$

And we have this property

$SST=SS_{between}+SS_{within}$

Where SST represent the total sum of squares.

The degrees of freedom for the numerator on this case is given by $df_{num}=k-1=3-1=2$ where k =3 represent the number of groups.

The degrees of freedom for the denominator on this case is given by $df_{den}=df_{between}=N-K=30-3=27$ .

And the total degrees of freedom would be $df=N-1=30 -1 =29$

From the info given we know that $MSR=\frac{SSR}{2}=45.89$

And $MSE=\frac{SSE}{27}=6.27$

From definition the F statisitc is defined as:

$F = \frac{MSR}{MSE} =\frac{45.89}{6.27}=7.32$

So then the best option is:

a. 7.32

3 0

2 years ago

Aaron wants to buy 2 concert tickets that cost $25.50 each. Aaron's grandfather pays him $4.25 an hour to rake leaves. How many

alexandr402 [8]

You are going to times 25.50 together and that's 51 so then you are going to divide 51.00 by 4.25 and that's 12

So the answer is 12

8 0

2 years ago

Read 2 more answers

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:

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

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

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

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