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3 years ago

Help me with this thank you

Mathematics

1 answer:

nikklg [1K]3 years ago

6 0

It’s a frequency table, you find the middle number from the prices of mail received and you would multiply it. Not sure what with but if you search on Google Frequency tables there should be a good explanation.

Hope this helped you!

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In a particular triathlon the athletes cover 1/24 of the total distance by

pav-90 [236]

Answer:

15/8

Step-by-step explanation:

Let "d" be the total distance, that is, the distance for swimming (s), running (r) and going by bike (b).

d = s + r + b [1]

1/24 of the total distance is covered by swimming.

s = 1/24 d [2]

1/3 of the distance is covered by running.

r = 1/3 d [3]

If we replace [2] and [3] in [1], we get

d = 1/24 d + 1/3 d + b

b = d - 1/24 d - 1/3 d

b = 24/24 d - 1/24 d - 8/24 d

b = 15/24 d = 5/8 d

The ratio of the distance covered by bike to the distance covered by running is:

b/r = (5/8 d)/(1/3 d) = 15/8

5 0

2 years ago

I need help pleaseee stattt (11 points )

Harlamova29_29 [7]

Answer:

If you were to get a different piece of paper and trace the same shape and flip it over to the same position that one of them is in it will be the exact same measurements!

3 0

2 years ago

Assume that a cross is made between a heterozygous tall pea plant and a homozygous short pea plant. Fifty offspring are produced

Butoxors [25]

Answer:

1) $\chi^2 = \frac{(30-25)^2}{25} +\frac{(20-25)^2}{25} =1 +1 =2$

2) $df = c-1 = 2-1$

Where c represent the number of categories c=2

Step-by-step explanation:

Previous concepts

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:

Tall =30 , Short =20

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

H0: The deviations from a 1:1 ratio (25 tall and 25 short) are due to chance.

H1: The deviations from a 1:1 ratio (25 tall and 25 short) are NOT due to chance.

Part 1

So then we know that the expected values would be 25 for each case

The statistic to check the hypothesis is given by:

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