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2 years ago

Mr. Mendoza invests in five stocks. The table shows the growth in stock prices in dollars at the end of a trading day.

Mathematics

2 answers:

Morgarella [4.7K]2 years ago

8 0

Answer:

The correct answer is C, as growth on stock E is bigger from the growth of stock B

Step-by-step explanation:

In order to resolve this problem, we must have in mind that the negative numbers are smaller when they are more distant from 0, and that positive numbers are bigger when they are more distant from 0. So, the biggest number of growth is the one that is more distant from 0.

The correct answer is C, as growth on stock E is bigger from the growth of stock B.

NikAS [45]2 years ago

7 0

Answer:

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Step-by-step explanation:

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Which is the equation of the line with slope 0 passing through the point (-3,-1)?

madreJ [45]

When a line has a slope of zero it means that it is horizontal. This means that all the x values have the same y values. In this case it means that the y value is always -1.

The equation for this line is...

y = -1

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2 years ago

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Find the slope of the line passing through the points (5, 7) and (5, -9)

Fantom [35]

Answer:

undefined

Step-by-step explanation:

To find the slope, we can use the slope formula

m = (y2-y1)/(x2-x1)

= (-9-7)/(5-5)

= (-16)/0

When we divide by zero, the slope is undefined

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2 years ago

A manufacturer makes ball bearings out of molten steel. It takes 5.24 cubic centimeters of molten steel to make ? ball bearings

sammy [17]

<span>So we want to know how much ball bearings can be made with 5.24 cm^3 if one ball bearing has a diameter of 1 cm. We know that radius r=d/2=0.5cm So the volume V of one ball bearing is: V=(4/3)*pi*r^3 so V=(4/3)*3.14*(0.5)^3cm^3=0.524cm^3. Now we simply divide the volume of steel by the volume of the ball bearing: 5.24/0.524=10. So we can make 10 ball bearings from 5.24 cm^3 of steel</span>

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2 years ago

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I nee help. Can anyone help please.

Radda [10]

You have to take 60 degrees and add it to 90 degrees because that’s how much the box in the corner is worth

Then add those 2 together which is 150 after that’s subtract it by 270 which is 120

Answer: 120 degrees or 120

7 0

2 years ago

(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

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2 years ago