Gabriel saves 40% of his monthly paycheck for college.He earned $270 last month.How much money did Gabriel save for college?

Select the exclusion that fits best with this problem.<br><br><br><br> 13x^6/51x^4

Damm [24]

ANSWER

$x \ne0$

EXPLANATION

The given problem is

$\frac{13 {x}^{6} }{ 51 {x}^{4} }$

This is defined if and only if

$x \ne0$

Even the given expression can be simplified to obtain:

$\frac{13 {x}^{2} }{ 51}$

The exclusion is

$x \ne0$

5 0

3 years ago

the length of a rectangle is five feet less than its width. If the area of the rectangle is 84 square feet, find its dimensions

KIM [24]

Let
x---------> the length of the rectangle
y--------> the width of the rectangle

we know that
A=84 ft²
[area of rectangle]=x*y-----> 84=x*y-----> equation 1
x=y-5------> equation 2
substitute 2 in 1
84=[y-5]*y-----> 84=y²-5y---------> y²-5y-84=0

using a graph tool-----> to resolve the second order equation
see the attached figure

the solution is
y=12
x=y-5-----> x=12-5-----> x=7

the answer is
the length of the rectangle is 7 ft
the width of the rectangle is 12 ft

6 0

3 years ago

There are 10000 people living in a certain city. Suppose that the rate of population growth in the city is proportional to the n

hichkok12 [17]

Answer:

.............30000.............

7 0

2 years ago

On Texas Avenue between University Drive and George Bush Drive, accidents occur according to a Poisson process at a rate of thre

Zarrin [17]

Answer:

(a) The probability is 0.6514

(b) The probability is 0.7769

Step-by-step explanation:

If the number of accidents occur according to a poisson process, the probability that x accidents occurs on a given day is:

$P(x)=\frac{e^{-at}*(at)^{x} }{x!}$

Where a is the mean number of accidents per day and t is the number of days.

So, for part (a), a is equal to 3/7 and t is equal to 1 day, because there is a rate of 3 accidents every 7 days.

Then, the probability that a given day has no accidents is calculated as:

$P(x)=\frac{e^{-3/7}*(3/7)^{x}}{x!}$

$P(0)=\frac{e^{-3/7}*(3/7)^{0}}{0!}=0.6514$

On the other hand the probability that February has at least one accident with a personal injury is calculated as:

P(x≥1)=1 - P(0)

Where P(0) is calculated as:

$P(x)=\frac{e^{-at}*(at)^{x} }{x!}$

Where a is equivalent to (3/7)(1/8) because that is the mean number of accidents with personal injury per day, and t is equal to 28 because 4 weeks has 28 days, so:

$P(x)=\frac{e^{-(3/7)(1/8)(28)}*((3/7)(1/8)(28))^{x}}{x!}$

$P(0)=\frac{e^{-(3/7)(1/8)(28)}*((3/7)(1/8)(28))^{0}}{0!}=0.2231$

Finally, P(x≥1) is:

P(x≥1) = 1 - 0.2231 = 0.7769

3 0

3 years ago

To estimate the mean height μ of male students on your campus,you will measure an SRS of students. You know from government data

nexus9112 [7]

Answer:

a) $\sigma = 0.167$

b) We need a sample of at least 282 young men.

Step-by-step explanation:

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

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

This Zscore is how many standard deviations the value of the measure X 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.

(a) What standard deviation must x have so that 99.7% of allsamples give an x within one-half inch of μ?

To solve this problem, we use the 68-95-99.7 rule. This rule states that:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviations of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we want 99.7% of all samples give X within one-half inch of $\mu$ . So $X - \mu = 0.5$ must have $Z = 3$ and $X - \mu = -0.5$ must have $Z = -3$ .

So

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

$3 = \frac{0.5}{\sigma}$

$3\sigma = 0.5$

$\sigma = \frac{0.5}{3}$

$\sigma = 0.167$

(b) How large an SRS do you need to reduce the standard deviationof x to the value you found in part (a)?

You know from government data that heights of young men are approximately Normal with standard deviation about 2.8 inches. This means that $\sigma = 2.8$