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gogolik [260]
3 years ago
10

Help me please thank you

Mathematics
1 answer:
dimaraw [331]3 years ago
8 0

Answer:

You have to upload a pic of you're work?

Step-by-step explanation:

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Cuánto es 12/25 mas 13/25
Arturiano [62]

Answer:

the answer is stop cheating

Step-by-step explanation:

Speak english

6 0
3 years ago
Read 2 more answers
The graph of f(x) and g(x) are shown below
Jlenok [28]

Answer:

c

Step-by-step explanation:

7 0
2 years ago
WILL CROWN BRAINLIEST, 5/5 RATING, AND LIKE!
Genrish500 [490]

Hey there!

If we have two white marbles and seven purple marbles, our first probability is out of nine.

First of all, we have the probability of selecting a purple marble, 7/9.

Now, if we do not replace it, there are only eight marbles left. Therefore, getting a white marble is 2/8, or 1/4.

We multiply these probabilities together to get our answer to a.

7/9(1/4)= 7/36

a. 7/36

Now, let's do b. We have 2/9, and then 1/8, giving us 1/36.

b. 1/36

Now, let's calculate selecting two purple to help us with C.

We have 7/9, and then 6/8, or 3/4.

7/9(3/4)= 7/12

Since there is a majority of purple marbles, there is a C) greater probability of selecting two purple marbles in a row.

I hope this helps!

4 0
3 years ago
Read 2 more answers
All cars can be classified into one of four​ groups: the​ subcompact, the​ compact, the​ midsize, and the​ full-size. There are
Olegator [25]

Answer:

p value = 0.302

Step-by-step explanation:

Given that all cars can be classified into one of four​ groups: the​ subcompact, the​ compact, the​ midsize, and the​ full-size. There are five cars in each group. Head injury data​ (in hic) for the dummies in the​ driver's seat are listed below.

H_0: All cars have same mean values

H_a: atleast two cars have different mean values

(Two tailed anova test)

Anova: Single Factor      

     

SUMMARY      

Groups Count Sum Average Variance  

Subcompact 5 3444 688.8 48502.7  

compact 5 2879 575.8 4582.7  

Midsize 5 2534 506.8 18720.2  

Full size 5 2689 537.8 23905.2  

     

     

ANOVA      

Source of Variation SS df MS            F             P-value F crit

Between Groups 94825 3 31608.33 1.321    0.302           3.24

Within Groups 382843.2 16 23927.7    

     

Total 477668.2 19    

Since p value of 0.302 is greater than 0.05 significance level we accept null hypothesis.

p value = 0.302

6 0
3 years ago
Evaluate M/N for M=10, N=-5, and P =-2
Leno4ka [110]

The answer is 2.

You would replace the letters with number which it would be 10/5

8 0
3 years ago
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