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Stolb23 [73]
3 years ago
5

Find the value of x?

Mathematics
2 answers:
jarptica [38.1K]3 years ago
7 0

using the angle sum property of triangle.

According to this, sum of all angles of a triangle is equal to 180°

x° + 116° + 34° = 180°

x° + 150° = 180°

x° = 180° - 150°

x° = 30°

Bess [88]3 years ago
6 0

Answer:

30

Step-by-step explanation:

The sum of the angles of a triangle is 180

x+34 +116 = 180

Combine like terms

x+150 = 180

Subtract 150 from each side

x+150-150 = 180-150

x = 30

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The probability of getting heads on the first toss is  1/2 .

Then, the probability of getting heads on the second toss is  1/2 .

Then, the probability of getting heads on the third toss is  1/2 .

The probability of all three things happening is ...

                 (1/2) x (1/2) x (1/2)  =  1/8  =  0.125  =  12.5% .



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2 years ago
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Q6) Renee buys 5kg of sweets to sell. She pays £10 for the sweets. Renee puts all the sweets into bags. She puts 250g of sweets
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Answer:

30% percent profit

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100 pence in a pound.

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She sells the bags for 65 pence each.

She sells them all for 20×65=1300 pence.

She gets a profit of 300 pence or 3 pounds.

She gets a percent profit of 30%.

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2 years ago
What is the unit rate for meters per second if a car travels 450 meters in 25 ​seconds?
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M/s =meters per second
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On a number line, a number, b, is located the same distance from 0 as another number, a, but in the opposite direction. The
oksian1 [2.3K]

Answer:

b=-a

Step-by-step explanation:

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3 0
3 years ago
An article contained the following observations on degree of polymerization for paper specimens for which viscosity times concen
devlian [24]

Answer:

(a) A 95% confidence interval for the population mean is [433.36 , 448.64].

(b) A 95% upper confidence bound for the population mean is 448.64.

Step-by-step explanation:

We are given that article contained the following observations on degrees of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range:

420, 425, 427, 427, 432, 433, 434, 437, 439, 446, 447, 448, 453, 454, 465, 469.

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 = \frac{\sum X}{n} = 441

            s = sample standard deviation = \sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }  = 14.34

            n = sample size = 16

            \mu = population mean

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

<em />

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

P(-2.131 < t_1_5 < 2.131) = 0.95  {As the critical value of t at 15 degrees of

                                             freedom are -2.131 & 2.131 with P = 2.5%}  

P(-2.131 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 2.131) = 0.95

P( -2.131 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 2.131 \times {\frac{s}{\sqrt{n} } } ) = 0.95

P( \bar X-2.131 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+2.131 \times {\frac{s}{\sqrt{n} } } ) = 0.95

<u>95% confidence interval for</u> \mu = [ \bar X -2.131 \times {\frac{s}{\sqrt{n} } } , \bar X +2.131 \times {\frac{s}{\sqrt{n} } } ]

                                      = [ 441-2.131 \times {\frac{14.34}{\sqrt{16} } } , 441+2.131 \times {\frac{14.34}{\sqrt{16} } } ]

                                      = [433.36 , 448.64]

(a) Therefore, a 95% confidence interval for the population mean is [433.36 , 448.64].

The interpretation of the above interval is that we are 95% confident that the population mean will lie between 433.36 and 448.64.

(b) A 95% upper confidence bound for the population mean is 448.64 which means that we are 95% confident that the population mean will not be more than 448.64.

6 0
3 years ago
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