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yKpoI14uk [10]
3 years ago
13

Three people are running for class president. From a poll that was taken, the probability that Candidate A will win is estimated

to be 0.3. The probability that Candidate B will win is estimated to be 0.55. Given a reasonable estimate of the probability that Candidate C will win.
Mathematics
1 answer:
Firdavs [7]3 years ago
3 0
It will be 80 candidate C will get 80 because every time that you add 25 it turns to a greater number
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Good at area pls helppp <br><br> Find the area. The figure is not drawn to scale.
Sauron [17]

Answer:

A=13.8

Step-by-step explanation:

The area equals 13.8

8 0
3 years ago
Della bought a tree seedling that was 2 1/4 feet tall. During the first year, it grew 1 1/6 Feet. After two years, it was 5 feet
Margarita [4]

Answer:

Step-by-step explanation:

Alright, lets get started.

Della bought a tree seeding that was 2\frac{1}{4} feet tall means \frac{9}{4} feet.

First year it grew 1\frac{1}{6} feet means it grew \frac{7}{6} feet.

It means after 1 year, the height of tree will be = \frac{9}{4}+\frac{7}{6}=\frac{41}{12}

After 2 years, the height of tree is = 5 feet

So, it grew in second year = 5-\frac{41}{12}

So, it grew in second year = \frac{60-41}{12}=\frac{19}{12}

So, it grew in second year=1\frac{7}{12} feet.   :  Answer

Hope it will help :)

3 0
3 years ago
9*10^7 is how many times as large as 3*10^-2
Elanso [62]

Answer:

A

Step-by-step explanation:

9 is 3 times larger than 3

10^7 is 10^5 times larger than 10^-2.

(just subtract the exponent)

I hope this helps!

pls ❤ and mark brainliest pls!

7 0
2 years ago
Read 2 more answers
What decimal is represented by this expanded form? 3X1000 + 6X100 + 1X10 + 2X1/10 + 1X 1/1000
Nata [24]
3000 + 600 + 10 + 0.2 + 0.001

= 3,610.201  answer
5 0
3 years ago
Read 2 more answers
The caffeine content (in mg) was examined for a random sample of 50 cups of black coffee dispensed by a new machine. The mean an
Genrish500 [490]

Answer:

 We are 98% confident interval for the mean caffeine content for cups dispensed by the machine between 107.66 and 112.34 mg .

Step-by-step explanation:

Given -

The sample size is large then we can use central limit theorem

n = 50 ,  

Standard deviation(\sigma) = 7.1

Mean \overline{(y)} = 110

\alpha = 1 - confidence interval = 1 - .98 = .02

z_{\frac{\alpha}{2}} = 2.33

98% confidence interval for the mean caffeine content for cups dispensed by the machine = \overline{(y)}\pm z_{\frac{\alpha}{2}}\frac{\sigma}\sqrt{n}

                     = 110\pm z_{.01}\frac{7.1}\sqrt{50}

                      = 110\pm 2.33\frac{7.1}\sqrt{50}

       First we take  + sign

   110 +  2.33\frac{7.1}\sqrt{50} = 112.34

now  we take  - sign

110 -  2.33\frac{7.1}\sqrt{50} = 107.66

 We are 98% confident interval for the mean caffeine content for cups dispensed by the machine between 107.66 and 112.34 .

               

5 0
3 years ago
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