All categories

3 years ago

Complete the inequality statement. 4 1/4___ 17/4

Mathematics

1 answer:

Mumz [18]3 years ago

8 0

Answer:

they are =

im sorry it was a little late

You might be interested in

Mark earns $300 per month plus an additional 2% on all of his sales. Last month he sold $59,000. What was Mark’s income for the

Lina20 [59]

Answer:

59,000 * .02= $1,180 commission plus $300 = $1480

3 0

2 years ago

25% of what number is 21

chubhunter [2.5K]

If we call the number "x" then:
x*25%=21
x*25/100=21
x=21*4
x=84

8 0

3 years ago

What is 5/9*2 4/ equal

Kobotan [32]

The answer is 1.11111

6 0

3 years ago

A store diplays will have 6 rows,12 contaners in each row.if 39 contaners has been set UPS so farsa explain how to find the numb

torisob [31]

Multiply:
6•12=72

Then, you subtract 39 from 72:
72-39=33

33 containers would still need to be set up.

4 0

3 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:

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

And if we replace we got:

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

Part 2

For this case the degreed of freedom are given by:

$df = c-1 = 2-1$

Where c represent the number of categories c=2

And we can calculate the p value given by:

$p_v = P(\chi^2_{1} >2)=0.157$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(2,1,TRUE)"

7 0

3 years ago