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Artyom0805 [142]
3 years ago
5

Two side length of a triangle are given as 9 cm and 5 cm. What could be the length of its third side?

Mathematics
2 answers:
astraxan [27]3 years ago
5 0

Answer: Two sides are given 5 cm and 9 cm. .Therefore, the third side can not be greater than or equal to (9+5)=14 cm and it can not be less than or equal to (9–5)=4 cm

Step-by-step explanation:

Please Give me brainliest

Lana71 [14]3 years ago
5 0

Answer:

Step-by-step explanation:

the length to the third side can be

a) a^2+b^2=c^2

a=9, b=5

c^2=81+25=106

c=sqrt106=10.29

or

a=5, c=9

b^2=c^2-a^2

b^2=81-25=56

b=sqrt56=7.48

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A random sample of 10 parking meters in a resort community showed the following incomes for a day. Assume the incomes are normal
GenaCL600 [577]

Answer:

A 95% confidence interval for the true mean is [$3.39, $6.01].

Step-by-step explanation:

We are given that a random sample of 10 parking meters in a resort community showed the following incomes for a day;

Incomes (X): $3.60, $4.50, $2.80, $6.30, $2.60, $5.20, $6.75, $4.25, $8.00, $3.00.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                         P.Q.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean income = \frac{\sum X}{n} = $4.70

            s = sample standard deviation = \sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }  = $1.83

            n = sample of parking meters = 10

            \mu = population mean

<em>Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.</em>

<u>So, 95% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-2.262 < t_9 < 2.262) = 0.95  {As the critical value of t at 9 degrees of

                                            freedom are -2.262 & 2.262 with P = 2.5%}  

P(-2.262 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 2.262) = 0.95

P( -2.262 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu < 2.262 \times {\frac{s}{\sqrt{n} } } ) = 0.95

P( \bar X-2.262 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+2.262 \times {\frac{s}{\sqrt{n} } } ) = 0.95

<u>95% confidence interval for</u> \mu = [ \bar X-2.262 \times {\frac{s}{\sqrt{n} } } , \bar X+2.262 \times {\frac{s}{\sqrt{n} } } ]

                                         = [ 4.70-2.262 \times {\frac{1.83}{\sqrt{10} } } , 4.70+ 2.262 \times {\frac{1.83}{\sqrt{10} } } ]

                                         = [$3.39, $6.01]

Therefore, a 95% confidence interval for the true mean is [$3.39, $6.01].

The interpretation of the above result is that we are 95% confident that the true mean will lie between incomes of $3.39 and $6.01.

Also, the margin of error  =  2.262 \times {\frac{s}{\sqrt{n} } }

                                          =  2.262 \times {\frac{1.83}{\sqrt{10} } }  = <u>1.31</u>

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