All categories

3 years ago

PLEASE HELP ME IM GIVING FORTY POINTS

1 answer:

4 0

Answer to part A: If the sun produces 3.9 * (10^33) ergs or radiant energy for every second, then how much energy does the sun produce in 3.25 * (10^3) seconds? Just multiply (3.25 * [10^3]) * (3.9[10^33]) = 1.2675 * 10^37 of ergs of radiation.

Answer to part B: 1.435 * 10^3 mm is a reasonable answer because it is much smaller than 1.435 * 10^3.

Answer to BONUS:
(2^8 * 5^-5 * 19^0) ^-2 (5^-2 / 2^3) * 2^28 <-- Start
(256 * 0.00032 * 1) ^-2 (0.04 / 8) * 268435456 <-- Simplify
0.08192^-2 * 0.005 * 268435456 <-- Simplify the terms in parentheses.


Final Answer: 200000000

Whew! That was hard!

You might be interested in

Blue Sky Jeans Company manufactures jeans.

tamaranim1 [39]

Answer: 1/4

Step-by-step explanation:

8 0

2 years ago

A recipe requires 1/3 cup of milk for each 1/4 cup of water. How many cups of water are needed for each cup of milk?

goldfiish [28.3K]

D) 1 1/3

(1/3) x reciprocal of 1/4 so (4/1) = 4/3 which simplifies to 1 1/3

8 0

3 years ago

Read 2 more answers

The author drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 200 times. Here are the observed

algol [13]

Answer:

$\chi^2 = \frac{(27-33.33)^2}{33.33}+\frac{(31-33.33)^2}{33.33}+\frac{(42-33.33)^2}{33.33}+\frac{(40-33.33)^2}{33.33}+\frac{(28-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33} =5.860$

$p_v = P(\chi^2_{5} >5.860)=0.32$

Since the p value is higher than the significance level assumed 0.05 we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we can assume that we have equally like results.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

Number: 1, 2 , 3 , 4 , 5 ,6

Frequency: 27, 31, 42, 40, 28, 32

We need to conduct a chi square test in order to check the following hypothesis:

H0: The outcomes are equally likely.

H1: The outcomes are not equally likely.

The level of significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The observed values are given:

$O_{1}=27$ $O_{2}=31$

$O_{3}=42$ $O_{4}=40$

$O_{5}=28$ $O_{6}=32$

The expected values are given by:

$E_{1} =\frac{1}{6}*200=33.33$ $E_{2} =\frac{1}{6}*200=33.33$

$E_{3} =\frac{1}{6}*200=33.33$ $E_{4} =\frac{1}{6}*200=33.33$

$E_{5} =\frac{1}{6}*200=33.33$ $E_{6} =\frac{1}{6}*200=33.33$

And now we can calculate the statistic:

$\chi^2 = \frac{(27-33.33)^2}{33.33}+\frac{(31-33.33)^2}{33.33}+\frac{(42-33.33)^2}{33.33}+\frac{(40-33.33)^2}{33.33}+\frac{(28-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33} =5.860$

Now we can calculate the degrees of freedom for the statistic given by:

$df=Categories-1=6-1=5$