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2 years ago

3 plus 6 times the number of white marbles for a total of 15

1 answer:

5 0

The problem is 3+6x = 15 with x being the number of white numbers.
Isolate variable by subtracting 3 on both sides: 6x = 12.
Isolate variable by dividing 6 on both sides: x = 2
The number of white marbles is 2.

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If a body moves in a straight line according to the law s = 24t + 3t^2 - t^3, where s is the distance measured in meters from th

Anastasy [175]

As the comments state, the velocity is the derivative of the position.

Therefore, the velocity as function of time is:

ds / dt = 24 + 6t - 3t^2.

That is a parabola whose maximum is (1,27). With that you know that the velocity will never be either 63 m/s or 81 m/s.

Also, you know that the velocity at t = 1 s is 27 m/s.

And, you can also find that the velocity at t = 3 is 15 m/s.

I am confident on that this analysis solves your question. Else, insert a comment.

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3 years ago

A triangle is shown below what is the measure of the missing angle of the triangle note the triangle is not drawn to scale?

Nina [5.8K]

Answer: 65 degrees

Step-by-step explanation:180- 74+41 is 65 so

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3 years ago

PLEASE HELP AND EXPLAIN! A. y=60x B. y=80x C. y=10x D. y= 6x

Rudik [331]

Answer:

C).

Step-by-step explanation:

One of the points on the graph that the path of bike b crosses is (6,60).

When you divide this to find the unit rate you get 10/1 therefore the equation of bike b= y+10X

8 0

3 years ago

Help help help help

Crank

Answer:

C = 18

Step-by-step explanation:

Put the numbers in place of their symbols and do the arithmetic. The diagram is showing you r = 3.

C = 2πr

C = 2(3)(3) . . . . . using 3 for π, as directed

C = 18

3 0

2 years ago

Read 2 more answers

(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