A square field with an area of 2000 square ft is to be enclosed by a fence. The slabs are 5ft long.

Each line represents a proportional relationship. Write an equation for each line. Show your work.

Semenov [28]

Answer:

The answer is in the picture. I hope it is helpful.

6 0

3 years ago

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Which equation represents y = −x2 + 6x + 7 in vertex form? y = −(x + 3)2 + 4 y = −(x − 3)2 + 10 y = −(x + 3)2 + 2 y = −(x − 3)2

Colt1911 [192]

$\text{The vertex form:}\ y=a(x-h)^2+k\\\\f(x)=ax^2+bx+c\to h=\dfrac{-b}{2a},\ k=f(h)\\\\\text{We have}\ y=-x^2+6x+7\to f(x)=-x^2+6x+7\\\\a=-1,\ b=6,\ c=7\\\\h=\dfrac{-6}{2(-1)}=\dfrac{-6}{-2}=3\\\\k=f(3)=-3^2+6(3)+7=-9+18+7=16\\\\Answer:\ \boxed{y=-(x-3)^2+16}$

8 0

3 years ago

If f(x) = x-1/3 and g(x)= 3x+1, what is (f o g)(x)?

N76 [4]

Answer:

$(f o g)(x) = 3x + \frac{2}{3}$

Step-by-step explanation:

We have the function $f(x) = x-\frac{1}{3}$ and we have the function $g(x) = 3x + 1$ . We want to find g(x) composed with f(x)

Then, the function (f o g)(x) is the same since f(g(x))

That is, you must do x = g(x) and then enter g(x) into the function f(x).

$f(g(x)) = (g(x)) -\frac{1}{3}$

$f(g(x)) = (3x + 1) -\frac{1}{3}$

Simplifying, we obtain:

$(f o g)(x) = 3x + 1 -\frac{1}{3}\\\\(f o g)(x) = 3x + \frac{2}{3}$

Finally. The composite function is:

$(f o g)(x) = 3x + \frac{2}{3}$

6 0

3 years ago

A bicyclist travels at a constant speed of 18 mi/h. How long does it take the bicyclist to travel 90 mi

Stells [14]

If the bicyclists is traveling 18 miles every hour and needs to travel a sum of 90 miles, you need to divide 90/18 equaling the total time (in hours) needed to travel 90 miles. Hope this helps

6 0

3 years ago

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(A) A small business ships homemade candies to anywhere in the world. Suppose a random sample of 16 orders is selected and each

Leviafan [203]

Following are the solution parts for the given question:

For question A:

$\to (n) = 16$

$\to (\bar{X}) = 410$

$\to (\sigma) = 40$

In the given question, we calculate $90\%$ of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:

$\to \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{S}{\sqrt{n}}$

$\to C.I= 0.90\\\\\to (\alpha) = 1 - 0.90 = 0.10\\\\ \to \frac{\alpha}{2} = \frac{0.10}{2} = 0.05\\\\ \to (df) = n-1 = 16-1 = 15\\\\$

Using the t table we calculate $t_{ \frac{\alpha}{2}} = 1.753$ When $90\%$ of the confidence interval:

$\to 410 \pm 1.753 \times \frac{40}{\sqrt{16}}\\\\ \to 410 \pm 17.53\\\\ \to392.47 < \mu < 427.53$

So $90\%$ confidence interval for the mean weight of shipped homemade candies is between $392.47\ \ and\ \ 427.53$ .

For question B:

$\to (n) = 500$

$\to (X) = 155$

$\to (p') = \frac{X}{n} = \frac{155}{500} = 0.31$

Here we need to calculate $90\%$ confidence interval for the true proportion of all college students who own a car which can be calculated as

$\to p' \pm Z_{\frac{\alpha}{2}} \times \sqrt{\frac{p'(1-p')}{n}}$

$\to C.I= 0.90$

$\to (\alpha) = 0.10$

$\to \frac{\alpha}{2} = 0.05$

Using the Z-table we found $Z_{\frac{\alpha}{2}} = 1.645$

therefore $90\%$ the confidence interval for the genuine proportion of college students who possess a car is

$\to 0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}\\\\ \to 0.31 \pm 0.034\\\\ \to 0.276 < p < 0.344$

So $90\%$ the confidence interval for the genuine proportion of college students who possess a car is between $0.28 \ and\ 0.34.$

For question C:

In question A, We are $90\%$ certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
In question B, We are $90\%$ positive that the true percentage of college students who possess a car is between 0.28 and 0.34.

Learn more about confidence intervals:

brainly.in/question/16329412

7 0

2 years ago