All categories

3 years ago

CREATE Write an equation that you would use the Addition Property of Equality to solve.

Mathematics

1 answer:

lidiya [134]3 years ago

4 0

1+1=2 that’s my guess

You might be interested in

You are dealt one card from a standard 52-card deck. Find the probability of being dealt an ace or a 6.

Ainat [17]

Answer:

2/13

Step-by-step explanation:

Since there's 4 of each card in the deck this means that there're 4 sixes and 4 aces, so 8 cards in a 52 card deck gives the problem 8/52 which equals 2/13

4 0

3 years ago

Solve for c.

Vanyuwa [196]

Answer:

Step-by-step explanation:

solve for c by simplifying both sides of the equation then isolate the variable

6 0

2 years ago

I NEED HELP FAST!

Degger [83]

Answer:

Step-by-step explanation:

4.5×5.5=24.75

so 24.75 is the area of the room but you have to buy whole tiles so you would need to but 25 tiles

8 0

3 years ago

5×143=use distributive property

Alex777 [14]

5*(143)
=5*(100+40+3)
=5(100)+5(40)+5(3)
=500+200+15
=715

Hope this helps!
Your fortune is "Be careful of who you choose to be with."

5 0

3 years ago

Read 2 more answers

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