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3 years ago

8

Ella spends 2/3 hour practicing the piano each day. She also spends 1/2 hour jogging. What is the common denominator of the frac

tions?

1 answer:

Marianna [84]3 years ago

4 0

The common denominator is actually 6.

1/2 is the same as 3/6,
2/3 is the same as 4/6,
and therefore the least common denominator (LCM) is 6.

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3 years ago

Read 2 more answers

Given the following observed and expected data (total of 1000), using chi-squared calculate a p-value that corresponds with this

finlep [7]

Answer:

The answer is " $0.90>p>0.75.$ "

Step-by-step explanation:

$\text{Cinnabar vestigial} \ \ \ \ \ \ \ \ \ \ \ 384 \ \ \ \ \ \ \ \ \ \ \ 390 \ \ \ \ \ \ \ \ \ \ \ -6 36 \ \ \ \ \ \ \ \ \ \ \ 0.092308\\\\$

$roof \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 408 \ \ \ \ \ \ \ \ \ \ \ 390 \ \ \ \ \ \ \ \ \ \ \ 18 \ \ \ \ \ \ \ \ \ \ \ 324 \ \ \ \ \ \ \ \ \ \ \ 0.830769\\\\\text{Cinnabar vestigial roof} \ \ \ \ \ \ \ \ \ \ \ \ \ 63 \ \ \ \ \ \ \ \ \ \ \ \ \ 70\ \ \ \ \ \ \ \ \ \ \ \ -7 \ \ \ \ \ \ \ \ \ \ \ 49 \ \ \ \ \ \ \ \ \ \ \ 0.7\\\\\text{wild type} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 72 \ \ \ \ \ \ \ \ \ \ \ 70 \ \ \ \ \ \ \ \ \ \ \ 2 \ \ \ \ \ \ \ \ \ \ \ 4 \ \ \ \ \ \ \ \ \ \ \ 0.057143\\\\$

$vestigial \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 32 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 35 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 9 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.257143\\\\$

$\text{Cinnabar roof} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 34\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 35 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.028571\\\\\text{roof vestigial} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5 \ \ \ \ \ \ \ \ \ \ \ \ \ \ -1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.2\\\\$

$\text{cinnabar} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.8 \\\\$

$Total \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1000 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2.965934$

Eight phenotypes were present.

Df is provided also by a number of phenotypes -1 The degree of freedom

$\to df = 8-1= 7$

For p-value 0,9, Chi-square is 2.83;

The p-value of 0.75 is 4.5. Chi-square

Chi-sqaure value is observed at 2.965.

That means 0.90>p-value>0.75.

7 0

3 years ago

What is 0.8637 rounded to the nearest hundredth?

Wittaler [7]

Answer:

0.87

Step-by-step explanation:

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3 years ago

Read 2 more answers