The perimeter of a rectangle, P, is given by P = 2L + 2W, where L is its length and W is its width. What is the perimeter of a rectangle of length 9 ft and width 13 ft?

Answers

Answer 1

Answer: Answer: 44 ft

Step-by-Step Explanation:

Length (L) = 9 ft
Width (W) = 13 ft
Perimeter (P) = 2L + 2W

Substitute values of ‘L’ and ‘W’ :-

= 2L + 2W
= 2(L + W)
= 2(9 + 13)
= 2(22)
= 2 * 22
=> 44

Hence, Perimeter (P) = 44 ft

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A digital camcorder repair service has set a goal not to exceed an average of 5 working days from the time the unit is brought in to the time repairs are completed. A random sample of 12 repair records showed the following repair times (in days): 5, 7, 4, 6, 7, 5, 5, 6, 4, 4, 7, 5.(a) H0: μ ≤ 5 days versus H1: μ > 5 days. At α = .05, choose the right option. Reject H0 if tcalc > 1.7960
Reject H0 if tcalc < 1.7960

b. Calculate the Test statistic.

c-1. The null hypothesis should be rejected.
i. TRUE
ii. FALSE

c-2. The average repair time is longer than 5 days.
i. TRUE
ii. FALSE

c-3 At α = .05 is the goal being met?
i. TRUE
ii. FALSE

Answers

Answer:

a) Reject H0 if tcalc > 1.7960

b) $t=(5.42-5)/((1.16)/(√(12)))=1.239$

c-1) ii. FALSE

c-2) ii.FALSE

c-3)i. TRUE

Step-by-step explanation:

1) Data given and notation

$\bar X=5.42$ represent the mean time for the sample

$s=1.16$ represent the sample standard deviation for the sample

$n=12$ sample size

$\mu_o =5$ represent the value that we want to test

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

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

a) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less than 5 days, the system of hypothesis would be:

Null hypothesis: $\mu \leq 5$

Alternative hypothesis: $\mu > 5$

We don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=(\bar X-\mu_o)/((s)/(√(n)))$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Rejection zone

On this case we need a critical value that accumulates 0.05 of the area on the right tail. The degrees of freedom are given by 11. And we can use the following excel code to find the critical value : "T.INV(1-0.95,11)" and the critical value would be given by $t_(\alpha/2)=1.795$ .

And the rejection zone is given by:

Reject H0 if tcalc > 1.7960

b) Calculate the statistic

We can replace in formula (1) the info given like this:

$t=(5.42-5)/((1.16)/(√(12)))=1.239$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=12-1=11$

Since is a one side test the p value would be:

$p_v =P(t_((11))>1.239)=0.121$

c-1. The null hypothesis should be rejected.

ii. FALSE

c-2. The average repair time is longer than 5 days.

ii. FALSE

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, and the true mean is not significantly higher than 5.

c-3 At α = .05 is the goal being met?

i. TRUE

We fail to reject the null hypothesis so then the goal is met.

Use a table to multiply (–5a)(2a – 1). A) –15a B) 5a2 + 10a C) –10a2 – 5a D) –10a2 + 5a

Answers

Answer:

-10a² + 5a

Step-by-step explanation:

Given the expression (–5a)(2a – 1)

Open the bracket

(–5a)(2a – 1)

= -5a(2a) -5a(-1)

= -10a² + 5a

hence the equivalent expression is -10a² + 5a

Simplify the expression below.(43+1)^3
A. 1223 + 1
B. 6432 + 1
C. 6413 + 48y2 + 16y + 3
D. 6413 + 4812 + 12y + 1

Answers

I believe the answer is C

Find parametric equations for the line. (Enter your answers as a comma-separated list of equations. Let x, y, and z be functions of t.) The line in the direction of the vector 5 i + 5 j − 6k and through the point (−4, 4, −2).

Answers

The parametric equations are:

x = -4 + 5t

y = 4 + 5t

z = -2 - 6t

The given direction vector is:

$\bar{V} = 5i + 5j - 6k$

The direction vector can also be written as:

$\bar{V} = <a, b, c> = <5, 5, -6>$

The point X₀ = (x₀, y₀, z₀) = (-4, 4, -2)

The parametric equation is of the form:

$X = X_(0) + \bar{V}t$

This is:

$\left[\begin{array}{ccc}x\ny\nz\end{array}\right] = \left[\begin{array}{ccc}x_0\ny_0\nz_0\end{array}\right] + \left[\begin{array}{ccc}a\nb\nc\end{array}\right]t$

$\left[\begin{array}{ccc}x\ny\nz\end{array}\right] = \left[\begin{array}{ccc}-4\n4\n-2\end{array}\right] + \left[\begin{array}{ccc}5\n5\n-6\end{array}\right]t$

The parametric equations are therefore:

x = -4 + 5t

y = 4 + 5t

z = -2 - 6t

Learn more here: brainly.com/question/13072659

Answer:

x=5t-4 , y=5t+4 , z=-6t-2

Step-by-step explanation:

So we are going to use (-4,4,-2) as an initial point, p.

The direction vector is v=5i+5j-6k or <5,5,-6>.

The vector equation is r=vt+p.

That means we have r=<5,5,-6>t + <-4,4,-2>.

So the parametric equations are

x=5t-4

y=5t+4

z=-6t-2

Consider the following functions. f(x) = x − 3, g(x) = x2 Find (f + g)(x). Find the domain of (f + g)(x). (Enter your answer using interval notation.) Find (f − g)(x). Find the domain of (f − g)(x). (Enter your answer using interval notation.) Find (fg)(x). Find the domain of (fg)(x). (Enter your answer using interval notation.) Find f g (x). Find the domain of f g (x). (Enter your answer using interval notation.)

Answers

Answer:

$(f+g)(x)=x-3+x^2$ ; Domain = (-∞, ∞)

$(f-g)(x)=x-3-x^2$ ; Domain = (-∞, ∞)

$(fg)(x)=x^3-3x^2$ ; Domain = (-∞, ∞)

$((f)/(g))(x)=(x-3)/(x^2)$ ; Domain = (-∞,0)∪(0, ∞)

Step-by-step explanation:

The given functions are

$f(x)=x-3$

$g(x)=x^2$

$(f+g)(x)=f(x)+g(x)$

Substitute the values of the given functions.

$(f+g)(x)=(x-3)+x^2$

$(f+g)(x)=x-3+x^2$

The function $(f+g)(x)=x-3+x^2$ is a polynomial which is defined for all real values x.

Domain of (f+g)(x) = (-∞, ∞)

$(f-g)(x)=f(x)-g(x)$

Substitute the values of the given functions.

$(f-g)(x)=(x-3)-x^2$

$(f-g)(x)=x-3-x^2$

The function $(f-g)(x)=x-3-x^2$ is a polynomial which is defined for all real values x.

Domain of (f-g)(x) = (-∞, ∞)

$(fg)(x)=f(x)g(x)$

Substitute the values of the given functions.

$(fg)(x)=(x-3)x^2$

$(fg)(x)=x^3-3x^2$

The function $(fg)(x)=x^3-3x^2$ is a polynomial which is defined for all real values x.

Domain of (fg)(x) = (-∞, ∞)

$((f)/(g))(x)=(f(x))/(g(x))$

Substitute the values of the given functions.

$((f)/(g))(x)=(x-3)/(x^2)$

The function $((f)/(g))(x)=(x-3)/(x^2)$ is a rational function which is defined for all real values x except 0.

Domain of (f/g)(x) = (-∞,0)∪(0, ∞)

$(f + g)(x) = x^2 + x - 3$ , domain: all real numbers.

$(f - g)(x) = -x^2 + x - 3$ , domain: all real numbers.

$(fg)(x) = x^3 - 3x^2$ , domain: all real numbers.

$f(g(x)) = x^2 - 3$ , domain: all real numbers.

To find (f + g)(x), we need to add the functions f(x) and g(x).

The function f(x) = x - 3 and the function $g(x) = x^2.$

So, $(f + g)(x) = f(x) + g(x) = (x - 3) + (x^2).$

Expanding this equation, we get $(f + g)(x) = x^2 + x - 3.$

To find the domain of (f + g)(x), we need to consider the domain of the individual functions f(x) and g(x).

Since both f(x) = x - 3 and $g(x) = x^2$ are defined for all real numbers, the domain of (f + g)(x) is also all real numbers.

To find (f - g)(x), we need to subtract the function g(x) from f(x).

So, $(f - g)(x) = f(x) - g(x) = (x - 3) - (x^2).$

Expanding this equation, we get $(f - g)(x) = -x^2 + x - 3.$

The domain of (f - g)(x) is also all real numbers, since both f(x) and g(x) are defined for all real numbers.

To find (fg)(x), we need to multiply the functions f(x) and g(x).

So, $(fg)(x) = f(x) * g(x) = (x - 3) * (x^2).$

Expanding this equation, we get $(fg)(x) = x^3 - 3x^2.$

The domain of (fg)(x) is all real numbers, since both f(x) and g(x) are defined for all real numbers.

To find f(g(x)), we need to substitute g(x) into the function f(x).

So, $f(g(x)) = f(x^2) = x^2 - 3.$

The domain of f(g(x)) is also all real numbers, as $g(x) = x^2$ is defined for all real numbers, and f(x) = x - 3 is defined for all real numbers.

In summary:

- $(f + g)(x) = x^2 + x - 3$ , domain: all real numbers.

- $(f - g)(x) = -x^2 + x - 3$ , domain: all real numbers.

- $(fg)(x) = x^3 - 3x^2$ , domain: all real numbers.

- $f(g(x)) = x^2 - 3$ , domain: all real numbers.

To Learn more about real numbers here:

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Solve for x. Write both solutions, separated
by a comma.
5x2 + 2x - 7 = 0

Answers

Answer:

it equals 1

Step-by-step explanation:

(5)(2)+2x−7=5

Step 1: Simplify both sides of the equation.

(5)(2)+2x−7=5

10+2x+−7=5

(2x)+(10+−7)=5(Combine Like Terms)

2x+3=5

Step 2: Subtract 3 from both sides.

2x+3−3=5−3

2x=2

Step 3: Divide both sides by 2.

x=1

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