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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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Is P 90 degrees from the origin, Q 180 degrees from the origin, and R 270 degrees from the origin?
krek1111 [17]

Answer:

Step-by-step explanation:

(x,y)→(-x,-y) (180° about the origin)

1.p(-3,2) ,in the second quadrant

P(-2,-3) is in 4th quadrant.

in the clockwise it is rotation of 270° about the origin.

2.

Q(-4,-5) is in 4th quadrant.

Q(4,5) is in 1st quadrant.

so it is 180° rotation in the clockwise direction.

3.

R(1,7) is in 1st quadrant.

R(7,-1) is in 4th quadrant.

Hence it is 90° rotation about the origin in clockwise direction.

4 0
3 years ago
What are the possible values of x in -11x^2 + 2x = 10?
vlabodo [156]
        -11x² + 2x = 10
-11x² + 2x - 10 = 10 - 10
-11x² + 2x - 10 = 0
x = <u>-(2) +/- √((2)² - 4(-11)(-10))</u>
                     2(-11)
x = <u>-2 +/- √(4 - 440)</u>
               -22
x = <u>-2 +/- √(-436)
</u>             -22<u>
</u>x = <u>-2 +/- 2i√(109)
</u>              -22
x = <u>-2 + 2i√(109</u>)        x = <u>-2 - 2i√(109)</u>
             -22                            -22
x = ¹/₁₁ - ¹/₁₁i√(109)    x = ¹/₁₁ + ¹/₁₁i√(109)
<u />
8 0
3 years ago
How far will a skier travel in 2 1/2 minutes? Explain how you figured it out
Kay [80]

3600 meters 60 seconds would be 1440 traveled

and 1440 + 1440 + 720= <u>3600</u>

Hope this helps :)

8 0
3 years ago
A data set includes 103 body temperatures of healthy adult humans having a mean of 98.1 F and a standard deviation of 0.56 F. Co
Ira Lisetskai [31]

Answer:

The 99​% confidence interval estimate of the mean body temperature of all healthy humans is (97.955F, 98.245F).

98.6F is above the upper end of the interval, which means that the sample suggests that the mean body temperature could be lower than 98.6F.

Step-by-step explanation:

The first step is finding the confidence interval

The sample size is 103.

The first step to solve this problem is finding how many degrees of freedom there are, that is, the sample size subtracted by 1. So

df = 103-1 = 102

Then, we need to subtract one by the confidence level \alpha and divide by 2. So:

\frac{1-0.99}{2} = \frac{0.01}{2} = 0.005

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 102 and 0.005 in the t-distribution table, we have T = 2.63.

Now, we need to find the standard deviation of the sample. That is:

s = \frac{0.56}{\sqrt{103}} = 0.055

Now, we multiply T and s

M = T*s = 2.63*0.055 = 0.145

For the lower end of the interval, we subtract the mean by M. So 98.1 - 0.145 = 97.955F.

For the upper end of the interval, we add the mean to M. So 98.1 + 0.145 = 98.245F.

The 99​% confidence interval estimate of the mean body temperature of all healthy humans is (97.955F, 98.245F).

98.6F is above the upper end of the interval, which means that the sample suggests that the mean body temperature could be lower than 98.6F.

5 0
3 years ago
Please help!! will give brainliest ASAP
soldier1979 [14.2K]

Answer:

c = 36.06

Step-by-step explanation:

a² + b² = c²

20² + 30² = c²

400 + 900 = c²

c² = 1300

\sqrt{c^2} =\sqrt{1300}

c = 36.06

8 0
3 years ago
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