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fgiga [73]
3 years ago
15

A sport arena covers 710,430 square feet of ground a news paper reported that the arena covers about 700,000 square feet of grou

nd to what place value was the number rounded
Mathematics
1 answer:
n200080 [17]3 years ago
4 0
700,000 is your final answer so the number was rounded to the hundred thousands value/place.
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9 + 15m<br><br><br><br>factorisation of algebraic expression that contain common factor​
Svet_ta [14]
The common factor would be 3 because 3 is the highest multiple of 15 and 9
6 0
3 years ago
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Evalute 27⋅((3^3)^−1)
shtirl [24]

Answer: 0.81

Step-by-step explanation:

=27⋅((3^3)^−1)

Do the brackets first:

3^3 is the same as 3³ so it is <u>27</u>

then:

27^-1 equals <u> 0.03</u>

And lastly:

multiply 27 and 0.03

27(0.03)=<em><u>0.81</u></em>

8 0
3 years ago
What is the domain and range​
kogti [31]

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The domain is all the x-values, and the range is all the y-values

Step-by-step explanation:

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3 years ago
Angle p is an obtuse angle. which could be the measure of angle p
Lunna [17]
Between 90 and 180 degrees
4 0
2 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
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