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3 years ago

Help me with this problem please

Mathematics

1 answer:

Brums [2.3K]3 years ago

6 0

It makes more sense to start with parts (c) and (d), then answer parts (a) and (b).

(c) The plot is below.

(d) Mixtures that are the same color have the same ratio of blue to yellow. They fall on the same line.

(a) There are two (2) different shades of paint mixtures.

(b) Shades A and C are the bluest. Their ratio of blue to yellow is the highest. (The slope of the line connecting their dots to the origin is the steepest.)

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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

Find the general solution of the given differential equation. x2y' + xy = 8 y(x) = give the largest interval over which the gene

topjm [15]

$x^2y'+xy=8$
$xy'+y=\dfrac8x$
$(xy)'=\dfrac8x\implies xy=8\ln|x|+C\implies y=\dfrac{8\ln|x|}x+\dfrac Cx$

which wouldn't be a valid solution over any interval containing $x=0$ .

5 0

2 years ago

use the substitution method to solve the system of equations choose the correct order pair. 3x+y=10 y=x-2

shutvik [7]

3x+(x-2)=10
4x=12
x=12/4=3

y=x-2=3-2=1

so, x=3 and y=1

Answer: (3;1)

3 0

3 years ago

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Steven works at a large home appliance store. He earns a base salary of $20,000 plus commission which can be modeled by the equa

Lerok [7]

It is his base salary

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2 years ago

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Ashley recently opened a store that sells only natural ingredients. She wants to advertise her products by distributing bags of

melisa1 [442]

With 6 people preparing bags, they would be able to prepare 6 bags every 2 minutes.

900/6 * 2 = the total minutes

Divide that answer by 60 to get the numbe of hours.

The answer is 5 hours.

Hope that helps.- UF aka Nadia

8 0

2 years ago

Read 2 more answers