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Complete Question

The quality control manager at a computer manufacturing company believes that the mean life of a computer is 91 months with a standard deviation of 10 months if he is correct. what is the probability that the mean of a sample of 68 computers would differ from the population mean by less than 2.08 months? Round your answer to four decimal places. Answer How to enter your answer Tables Keypad

Answer:

[tex]P(-1.72<Z<1.72)=0.9146[/tex]

Step-by-step explanation:

From the question we are told that:

Population mean \mu=91

Sample Mean \=x =2.08

Standard Deviation \sigma=10

Sample size n=68

Generally the Probability that The  sample mean  would differ from the population mean

P(|\=x-\mu|<2.08)

From Table

[tex]P(|\=x-\mu|<2.08)=P(|z|<1.72)[/tex]

T Test

[tex]Z=\frac{\=x-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex]

[tex]Z=\frac{2.08}{\frac{10}{\sqrt{68} } }[/tex]

[tex]Z=1.72[/tex]

[tex]P(|\=x-\mu|<2.08)=P(|z|<1.72)[/tex]

[tex]P(-1.72<Z<1.72)[/tex]

Therefore From Table

[tex]P(-1.72<Z<1.72)=0.9146[/tex]

Answer:

Step-by-step explanation:

yes

Answer:

A) x = 0.

B) f is concave up for (-∞, 0).

C) f is concave down for (0, ∞).

Step-by-step explanation:

We are given the function:

[tex]f(x)=5+12x-x^3[/tex]

A)

We want to find the x-coordinates of all inflection points.

Recall that inflections points (may) occur when the second derivative equals zero. Hence, find the second derivative. The first derivative is given by:

[tex]f'(x) = 12-3x^2[/tex]

And the second:

[tex]f''(x) = -6x[/tex]

Set the second derivative equal to zero:

[tex]0=-6x[/tex]

And solve for x. Hence:

[tex]x=0[/tex]

We must test the solution. In order for it to be an inflection point, the second derivative must change signs before and after. Testing x = -1:

[tex]f''(-1) = 6>0[/tex]

And testing x = 1:

[tex]f''(1) = -6<0[/tex]

Since the signs change for x = 0, x = 0 is indeed an inflection point.

B)

Recall that f is concave up when f''(x) is positive, and f is concave down when f''(x) is negative.

From the testing in Part A, we know that f''(x) is positive for all values less than zero. Hence, f is concave up for all values less than zero. Our interval is:

[tex](-\infty, 0)[/tex]

C)

From Part A, we know that f''(x) is negative for all values greater than zero. So, f is concave down for that interval:

[tex](0, \infty)[/tex]

Answer:

a) 0 is stable when n = odd

b) 0 is semi-stable when n = even

c) 0 is unstable when n is odd

Step-by-step explanation:

Th differential equation for this question

dx/dt = x^n

n = positive integer

a) value of n where 0 is stable

0 is stable when x^n is replaced with -x^n

because considering n to be an odd number

-x^n > 0 when x < 0    while -x^n < 0 when x > 0

∴ In this scenerio we can conclude that 0 is stable when  n = odd number

b) Value of n where 0 is Semi-stable

assuming n is an even number

x^n > 0  for all the values of x

c) Value of n where 0 is unstable

lets assume n is odd

when n < 0,  xⁿ < 0

when n > 0,  xⁿ > 0

i.e. 0 is asymptotically unstable when n is an odd number

Answer:

The minimum sample size required to create the specified confidence interval is 295.

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1 - 0.95}{2} = 0.025[/tex]

Now, we have to find z in the Z-table as such z has a p-value of [tex]1 - \alpha[/tex].

That is z with a pvalue of [tex]1 - 0.025 = 0.975[/tex], so Z = 1.96.

Now, find the margin of error M as such

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

Variance of 0.49:

This means that [tex]\sigma = \sqrt{0.49} = 0.7[/tex]

They want to construct a 95% confidence interval with an error of no more than 0.08. What is the minimum sample size required to create the specified confidence interval?

The minimum sample size is n for which M = 0.08. So

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

[tex]0.08 = 1.96\frac{0.7}{\sqrt{n}}[/tex]

[tex]0.08\sqrt{n} = 1.96*0.7[/tex]

[tex]\sqrt{n} = \frac{1.96*0.7}{0.08}[/tex]

[tex](\sqrt{n})^2 = (\frac{1.96*0.7}{0.08})^2[/tex]

[tex]n = 294.1[/tex]

Rounding up:

The minimum sample size required to create the specified confidence interval is 295.

Answer:

A sample of 18 is required.

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1 - 0.92}{2} = 0.04[/tex]

Now, we have to find z in the Z-table as such z has a p-value of [tex]1 - \alpha[/tex].

That is z with a pvalue of [tex]1 - 0.04 = 0.96[/tex], so Z = 1.88.

Now, find the margin of error M as such

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

A previous study indicated that the standard deviation was 2.2 days.

This means that [tex]\sigma = 2.2[/tex]

How large a sample must be selected if the company wants to be 92% confident that the true mean differs from the sample mean by no more than 1 day?

This is n for which M = 1. So

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

[tex]1 = 1.88\frac{2.2}{\sqrt{n}}[/tex]

[tex]\sqrt{n} = 1.88*2.2[/tex]

[tex](\sqrt{n})^2 = (1.88*2.2)^2[/tex]

[tex]n = 17.1[/tex]

Rounding up:

A sample of 18 is required.

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Answer:

  (a)  y = −50x + 250

Step-by-step explanation:

In case you don't realize that the graph starts at 250 and decreases by 50 for each increase of 1 in x, you can see if any of the equations match the given points. The only one that does is the first one:

  y = -50x +250

Answer:

(a)  y = −50x + 250

Step-by-step explanation:

Answer:

Following are the response to this questions:

Step-by-step explanation:

Please find the graph file in the attachment.

Given:

P=2

B=-5

H=?

[tex]H=\sqrt{P^2+B^2}[/tex]

    [tex]=\sqrt{2^2+(-5)^2}\\\\=\sqrt{4+25}\\\\=\sqrt{29}\\\\[/tex]

Using formula:

[tex]\to \ cosec \theta \ or\ \ csco \theta =\frac{H}{P}\\\\\to \cos \theta=\frac{B}{H}\\\\\to \tan \theta=\frac{p}{B}\\\\[/tex]

So,

[tex]\to \ cosec \theta \ or\ \ csco \theta =\frac{\sqrt{29}}{2}\\\\\to \cos \theta=\frac{-5}{\sqrt{29}} =\frac{-5}{\sqrt{29}}\times \frac{\sqrt{29}}{\sqrt{29}}=-\frac{5\sqrt{29}}{29}\\\\\to \tan \theta=\frac{2}{-5}= -\frac{2}{5}\\\\[/tex]

Answer:

X^2 + 3x - 10=0

Answer:

A significant negative relationship exists between the variables

Step-by-step explanation:

Base on the information given in the question which goes thus : For every unit increase in x, y decreases by 12.8094. The value 12.8094 is the slope which is the rate of change in y variable per unit change in the independent variable. The sign or nature of the slope Coefficient gives an hint about the relationship between the x and y variables. The slope Coefficient in this case is negative and thus we'll have a negative relationship between the x and y variables (an increase in x leads to a corresponding decrease in y). This is a negative association.

Answer:

5x+20y=425

Step-by-step explanation:

Its 5 bucks for x pairs of skates

Its 20 dollars for y bikes

x+y rentals have to equal 25

all of this is equal to 425. All that is left to do is test with number until the statement is true.

try :

5(5)+(20)(20)=425

x + y do equal 25, and the total is equal to 425.

Answer:

-11

Step-by-step explanation:

I am going to assume that it is -x^2-5y^3.

-(4^2)-5(-1^3)

-16-5(-1)

-16+5

-11

Answer:

- 11

Step-by-step explanation:

If x = 4,  y = -1

then,

        - x^2 - 5y^3 = - (4)^2 - 5(-1)^3

                            = - 16 + 5

                            = - 11

Answer:

The sample size necessary is of 168.

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1 - 0.95}{2} = 0.025[/tex]

Now, we have to find z in the Z-table as such z has a p-value of [tex]1 - \alpha[/tex].

That is z with a pvalue of [tex]1 - 0.025 = 0.975[/tex], so Z = 1.96.

Now, find the margin of error M as such

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

Based on a previous study, arrival delay times have a standard deviation of 39.6 minutes.

This means that [tex]\sigma = 39.6[/tex]

Find the sample size necessary to estimate the mean arrival delay time for all American Airlines flights from Dallas to Sacramento to within 6 minutes with 95% confidence.

This is n for which M = 6. So

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

[tex]6 = 1.96\frac{39.6}{\sqrt{n}}[/tex]

[tex]6\sqrt{n} = 1.96*39.6[/tex]

[tex]\sqrt{n} = \frac{1.96*39.6}{6}[/tex]

[tex](\sqrt{n})^2 = (\frac{1.96*39.6}{6})^2[/tex]

[tex]n = 167.34[/tex]

Rounding up:

The sample size necessary is of 168.

Answer:

A. y<-2x-1

Step-by-step explanation:

not C or D because it is a dashed line meaning the linear equation will either have the symbol ≥ or ≤.

when y is less than, you shade below

thus, the answer is A

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Answer:

  1. see below

  2. g(x) is f(x) translated left 2 and down 3

Step-by-step explanation:

1. The graphs are attached. F(x) is in red; g(x) is in blue.

__

2. The graph of g(x) = f(x -h) +k is the parent function translated by (h, k). Here we have (h, k) = (-2, -3), so g(x) is f(x) translated left 2 and down 3.

Answer:

The projected world population in 2015 was 8,705,121,030 people.

Step-by-step explanation:

Given that the population of the world in 1987 was 5 billion and the annual growth rate was estimated at 2 percent per year, assuming that the world population follows an exponential growth model, to find the projected world population in 2015 the following calculation must be performed :

5,000,000,000 x 1.02 ^ (2015-1987) = X

5,000,000,000 x 1.02 ^ 28 = X

5,000,000,000 x 1.741024 = X

8,705,121,030 = X

Therefore, the projected world population in 2015 was 8,705,121,030 people.

Answer:

The 1st,Thrid, Fifth Option

Step-by-step explanation:

The first option is true. We can move the orginal square root function to get g(x).

The second option is false. Function g(x) which equals

[tex] \sqrt{x - 3} - 1[/tex]

Domain is all real numbers greater than or equal to 3.

The third option is true. Since minimum point we can get is 0 in a square root function. We have a vertical shift so our new minimum point is

[tex]0 - 1 = - 1[/tex]

We can take the sqr root of 0 so

So all real numbers that are greater than or equal to -1 is true.

The fourth option is false, we need to add 3 instead of subtract 3.

The fifth option is true, we can do that to get back to our original function

Answer:

Step-by-step explanation:

If we are looking for the time(s) that the ball is at a height of 15, we simply sub in a 15 for the height in the position equation and solve for t:

[tex]15=-16t^2+23t+7[/tex] and

[tex]0=-16t^2+23t-8[/tex]

Factor this however you factor a quadratic in class to get

t = .59 seconds and t = .85 seconds.

This means that .59 seconds after the ball was thrown into the air it was 15 feet off the ground. Then the ball reached its max height, gravity took over, and began pulling it back down to earth. The ball passes the height of 15 feet again on its way down after .85 seconds.

Answer:

y = 7cos bx

Step-by-step explanation:

For a cosine function without pahse shift and vertical shift, but with amplitude given, it will also have period and thus , the formula for the cosine function is;

y = Acos bx

Where;

A is the amplitude

Period = 2π/b

Now, we are told that the amplitude is 7. Thus;

y = 7cos bx

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Answer:

  see attached

Step-by-step explanation:

The graph of g(x) is a vertically scaled version of the graph of f(x). The scale factor is 1/2, so vertical height at a given value of x is 1/2 what it is for f(x). This will make the graph appear shorter and fatter than for f(x).

The graph of g(x) is attached.

Answer:

i think 4 4 kilograms if im wrong sorry

Step-by-step explanation:

Answer:

The three answers I can think of are migration, immigration, and emigration.

Step-by-step explanation:

Hope this helps!

9514 1404 393

Answer:

  4/5

Step-by-step explanation:

The wording is ambiguous, as it often is when math expressions are described in English. We assume you intend ...

  [tex]\dfrac{3n+8}{n+7}=\dfrac{4}{3}\\\\3(3n+8)=4(n+7)\qquad\text{multiply by $3(n+7)$}\\\\9n+24=4n+28\qquad\text{eliminate parentheses}\\\\5n=4\qquad\text{subtract $4n+24$}\\\\\boxed{n=\dfrac{4}{5}}\qquad\text{divide by 5}[/tex]

The number is 4/5.

Answer:

[tex]Cold = \frac{1}{6}\ mins[/tex]

Step-by-step explanation:

The correct given parameters are:

[tex]Both = \frac{1}{4}\ mins[/tex]

[tex]Hot = \frac{1}{12}\ mins[/tex]

Required

Time taken by the cold water faucet

We have:

[tex]Cold + Hot = Both[/tex]

Make Cold the subject

[tex]Cold = Both -Hot[/tex]

So, we have:

[tex]Cold = \frac{1}{4}-\frac{1}{12}[/tex]

Take LCM

[tex]Cold = \frac{3-1}{12}[/tex]

[tex]Cold = \frac{2}{12}[/tex]

Divide by 2

[tex]Cold = \frac{1}{6}[/tex]

Answer:

(a) [tex]H_o:\mu_1 = \mu_2[/tex]     [tex]H_a:\mu_1 > \mu_2[/tex]

(b) [tex]t = 0.74[/tex]

(c) [tex]p =0.2331[/tex]

(d) [tex]CI = (-2.095,4.095)[/tex]

Step-by-step explanation:

Given

[tex]\bar x_1=21,\ s_1=4,\ n_1=12,\\ \bar x_2=20,\ s_2=3,\ n_2=15[/tex]

Solving (a): The hypotheses

The test is right-tailed, means that the alternate hypothesis will contain greater than sign.

So, we have:

[tex]H_o:\mu_1 = \mu_2[/tex]

[tex]H_a:\mu_1 > \mu_2[/tex]

Solving (b); The test statistic (t)

This is calculated as:

[tex]t = \frac{\bar x_1 - \bar x_2}{\sqrt{\frac{s_1^2(n_1 - 1) + s_2^2(n_2 - 1)}{n_1 + n_2 - 2} * (\frac{1}{n_1} + \frac{1}{n_2})}}[/tex]

So, we have:

[tex]t = \frac{21 - 20}{\sqrt{\frac{4^2(12 - 1) + 3^2(15 - 1)}{12 + 15 - 2} * (\frac{1}{12} + \frac{1}{15})}}[/tex]

[tex]t = \frac{1}{\sqrt{\frac{302}{25} * (0.15)}}[/tex]

[tex]t = \frac{1}{\sqrt{12.08 * 0.15}}[/tex]

[tex]t = \frac{1}{\sqrt{1.812}}[/tex]

[tex]t = \frac{1}{1.346}[/tex]

[tex]t = 0.74[/tex]

Solving (c): The P-value

First, we calculate the degrees of freedom

[tex]df = n_1 + n_2 -2[/tex]

[tex]df = 12+15 -2[/tex]

[tex]df = 25[/tex]

Using the t distribution, the p-value is:

[tex]p =TDIST(0.74,25)[/tex]

[tex]p =0.2331[/tex]

Solving (d): The 90% confidence interval

Calculate significance level

[tex]\alpha = 1 - CI[/tex]

[tex]\alpha = 1 - 90\%[/tex]

[tex]\alpha = 0.10[/tex]

Calculate the t value (t*)

[tex]t^* = (\alpha/2,df)[/tex]

[tex]t^* = (0.10/2,25)[/tex]

[tex]t^* = (0.05,25)[/tex]

[tex]t^* = 1.708[/tex]

The confidence interval is calculated using:

[tex]CI = (\bar x - \bar x_2) \± t^* *\sqrt{\frac{s_1^2(n_1 - 1) + s_2^2(n_2 - 1)}{n_1 + n_2 - 2} * (\frac{1}{n_1} + \frac{1}{n_2})}[/tex]

[tex]CI = (21 - 20) \± 1.708 *\sqrt{\frac{4^2(12 - 1) + 3^2(15 - 1)}{12 + 15 - 2} * (\frac{1}{12} + \frac{1}{15})}[/tex]

[tex]CI = 1 \± 1.708 *1.812[/tex]

[tex]CI = 1 \± 3.095[/tex]

Split

[tex]CI = 1 - 3.095 \ or\ 1 + 3.095[/tex]

[tex]CI = -2.095 \ or\ 4.095[/tex]

[tex]CI = (-2.095,4.095)[/tex]