Refer to the article "Fire Up the Sun."

1 answer:

8 0

Based on the information given, it should be noted the central idea is C. Making and using a solar oven is a learning experience.

<h3>What is a central idea?</h3>

A central idea simply means the main idea that's in a literary work. It's the unifying element that's in a story.

The central idea is that making and using a solar oven is a learning experience and the detail that supports the central idea is that "by completing this project, you will make an inexpensive, ecologically friendly piece of equipment and learn about the science behind what you’re doing".

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In a recent survey, three out of every five teenagers said they listen to music while studying.

Virty [35]

Answer:

225

Step-by-step explanation:

3 out of 5 is .6 is u multiply .6 by 375 you get 225

7 0

3 years ago

What is the slope of the following graph? (write your answer as a fraction unless divided by 1. Do not use spaces in your answer

alexira [117]

The slope of the graph is 4/5

<h3>How to determine the slope</h3>

Using the formula

Slope = Δ y-axis / Δ x- axis

y2 = -4

y1 = 0

x2 = -5

x1 = 0

Slope = $\frac{y2 - y1}{x2 - x1}$

Substitute the values into the formula

<em>Slope = </em> $\frac{-4 - 0}{- 5 - 0}$

Slope = $\frac{-4}{-5}$

Slope = $\frac{4}{5}$

Therefore, the slope of the graph is 4/5

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5 0

2 years ago

A local bank needs information concerning the savings account balances of its customers. A random sample of 15 accounts was chec

PSYCHO15rus [73]

Answer:

(513.14, 860.36) is a 98% confidence interval for the true mean

Step-by-step explanation:

We have a small sample size of n = 15, the mean balance was $\bar{x} = 686.75$ with a standard deviation of s = 256.20. The confidence interval is given by $\bar{x}\pm t_{\alpha/2}(\frac{s}{\sqrt{n}})$ where $t_{\alpha/2}$ is the 100 $(\alpha/2)$ th quantile of the t distribution with n-1=15-1=14 degrees of freedom. As we want a $100(1-\alpha)$ % = 98% confidence interval, we have that $\alpha = 0.02$ and the confidence interval is $686.75\pm t_{0.01}(\frac{256.20}{\sqrt{15}})$ where $t_{0.01}$ is the 1st quantile of the t distribution with 14 df, i.e., $t_{0.01} = -2.6245$ . Then, we have $686.75\pm (-2.6245)(\frac{256.20}{\sqrt{15}})$ and the 98% confidence interval is given by (513.1379, 860.3621)