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3 years ago

WILL GIVE A BRAINLIEST!!

2 answers:

4 0

If the vertex is at the maximum point they dy/dx=0 at that point...

dA/dx=-2x+4

dB/dx=-2x+8

dC/dx=2x-4

dD/dx=2x-8

So both B and D have a zero velocity at x=4, but only B is equal to 3 when x=4

B. is the correct quadratic with the vertex at (4,3)

scoray [572]3 years ago

3 0

Answer:

Step-by-step explanation:

We know that the vertex of the parabola is

$V(4,3)$

That means this points is being intercept by the parabola. In other words, the function that represents this parabola has to have this relation, when $x=4$ , $y=3$ .

So, you can observe that the second choice has a function that follows this relation. Let's see

$y=-x^{2} +8x-13$

So, for $x=4$

$y=-(4)^{2} +8(4)-13\\y=-16+32-13\\y=3$

Therefore, function B is the answer because it includes the vertex at its relation.

You might be interested in

(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

3 years ago

How do you do this? I am really confused

german

Answer:

Step-by-step explanation:

ok, so this is an infinitly repeating function, so you can write it as:

sqrt(12-x)

Although, x is also equal to sqrt(12-x), so

x = sqrt(12-x), and

x^2 = 12-x

x^2-12+x = 0

now just apply the quadratic formula and you're good

hope i helped :D

8 0

2 years ago

Let x and y be positive integers and x be greater than y. 11x/11y =

navik [9.2K]

$\frac{11x}{11y} =\frac{x}{y}$

7 0

3 years ago

PLEASE HELP FAST!!!!

kakasveta [241]

Answer:

i think c. sorry if its wrong

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

10.

Delicious77 [7]

Answer:

A. 4.5

Step-by-step explanation:

According to my graphing calculator, e^1.5=4.5 which is rounded to the nearest tenth.

Answer: e^1.5 = 4.482

Step-by-step explanation:

Given:

We need to find the value of function at x=1.5

Put x=1.5 into f(x)

Now we will verify the result using graph.

First we draw the graph of f(x) and a vertical line x=1.5

Please see the attachment for graph.

Hence, The value of

6 0

3 years ago