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exis [7]
3 years ago
10

The number of fans who attended a softball game increased from 1200 for the first game to 1300 for the second game. How many fan

s will attennd if the same percent increase occurs?
Mathematics
1 answer:
kondaur [170]3 years ago
4 0
1400 fans because it is increasing by 100 for each game
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Solve step by step! Please answer correctly !!!!!!!!!!! Will mark brainliest !!!!!!!!!!!!!!!!!!
Colt1911 [192]

Answer:

X is an acute angle, which means it is less than 90*, so it is probably 45*

Step-by-step explanation:

I was confused at first but then, I’m just like, I can solve this!

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3 years ago
Read 2 more answers
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lesantik [10]

Answer:

56

Step-by-step explanation:

7 0
3 years ago
Twenty-eight athletes participate in a 100-meter race. The time of each athlete is measured during the qualification round. The
andriy [413]

The average time of the 20 athletes who did not qualify for the final is 11.2 seconds .

<u>Step-by-step explanation:</u>

Here we have , Twenty-eight athletes participate in a 100-meter race. The time of each athlete is measured during the qualification round. The average time is 11 seconds. The 8 athletes with the fastest time qualified for the final. The average time of these 8 athletes is 10.5 seconds. Let's have equations for this:

\frac{x_1+x_2+.....+x_2_8}{28} = 11 where x_1, x_2, ....., x_2_8 are athletes

⇒ x_1 + x_2 + .. + x_2_8 = 11(28)= 308

⇒ (x_1 + x_2 + .. + x_8 )+ (x_9+x_1_0+...+x_2_8) = 308

⇒ (\frac{x_1 + x_2 + .. + x_8}{8}  )(8)+ (\frac{x_9+x_1_0+...+x_2_8}{20} )(20) = 308

⇒ 10.5(8)+m(20) = 308 , where m is average time of the 20 athletes who did not qualify for the final.

⇒ 84+m(20) = 308

⇒ 20m= 224

⇒ m=11.2

∴ The average time of the 20 athletes who did not qualify for the final is 11.2 seconds .

8 0
3 years ago
A statistician selected a random sample of 125 observations from a population with a known standard deviation equal to 16 and co
serious [3.7K]

Answer:

Step-by-step explanation:

Given that a statistician selected a random sample of 125 observations from a population with a known standard deviation equal to 16 and computed a sample mean equal to 77

We can use Z critical values since population std deviation is known. Also sample size >30

We find std error of mean = \frac{16}{\sqrt{125} } \\=1.4311

Margin of error = Z critical value * 1.4311

Z critical values for 93% = 1.81

for 89% = 1.60

n Std error Z critical  Conf interval  

            89%    

     

125 1.4311     1.6  (74.71024 79.28976 )

     

     

          93%    

            1.81  (74.409709 79.590291 )

We find that when confidence level increases interval width increses.

c) When sigma changes to 441, std error changes to \frac{16}{\sqrt{441} } \\=0.7619

So we get

n Std error Z critical  Conf interval  

           89%    

     

441  0.7619     1.6  (75.781 78.219 )

     

     

         93%    

           1.81  (75.621 78.37904)

d) When n = 625, std error changes to 16/25 = 0.64

n Std error Z critical  Conf interval  

        89%    

     

441 0.64 1.6        (75.976 78.024 )

     

     

       93%    

        1.81      (75.842 78.1584)

When sample size increases, confidence interval width decreases.

8 0
3 years ago
You are selling concessions at a local swim meet. Hot dogs are being sold for $1.00 and soda is being sold for $0.50. At the end
melomori [17]

Answer:

use this equation: 150=1x+0.5y

Step-by-step explanation:

a) standard form equation above

b) 150=1x+0.5y

subtract 1x from both sides

then divide all by 0.5 to get slope intercept form

c) plug it in: 150=1(40)+0.5y and solve to get 220

d) take the y intercept from answer b. and the y intercept shows how many you have to sell of each to reach you $150(I think, this part might be wrong...idk)

5 0
3 years ago
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