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malfutka [58]
3 years ago
12

Which angle has a measure equal to the sum

Mathematics
1 answer:
jarptica [38.1K]3 years ago
4 0

Answer:< RSC

Step-by-step explanation:

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A migrating bird flies 341 miles in 11 hours. How many miles can the bird fly in 6 hours?
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186 because you would divide 341 by 11 and you get 31 then multiply that by 6 and you get 186. Did this help?
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3 years ago
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I'm not sure ic I just leave like since I can't combine
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2\sqrt{12} - 7\sqrt{3} 
= 4\sqrt{3} - 7\sqrt{3}
= Since, we have same bases, we can subtract!
= To get,
= -3\sqrt{3} Answer
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3 years ago
Find the standard deviation of
Kruka [31]

Answer:

11.5

The sample standard deviation measures the spread of a data distribution of the sample. It is usually used to estimate the population standard deviation

7 0
3 years ago
A laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighings. Scale readings in repeated we
weqwewe [10]

Answer:

99% confidence interval for the given specimen is [3.4125 , 3.4155].

Step-by-step explanation:

We are given that a laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighing. Scale readings in repeated weighing are Normally distributed with mean equal to the true weight of the specimen.

Three weighing of a specimen on this scale give 3.412, 3.416, and 3.414 g.

Firstly, the pivotal quantity for 99% confidence interval for the true mean specimen is given by;

        P.Q. = \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \bar X = sample mean weighing of specimen = \frac{3.412+3.416+3.414}{3} = 3.414 g

            \sigma = population standard deviation = 0.001 g

            n = sample of specimen = 3

            \mu = population mean

<em>Here for constructing 99% confidence interval we have used z statistics because we know about population standard deviation (sigma).</em>

So, 99% confidence interval for the population​ mean, \mu is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99  {As the critical value of z at 0.5% level

                                                            of significance are -2.5758 & 2.5758}

P(-2.5758 < \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } < 2.5758) = 0.99

P( -2.5758 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X - \mu} < 2.5758 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.99

P( \bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.99

<u>99% confidence interval for</u> \mu = [ \bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } } ]

                                             = [ 3.414-2.5758 \times {\frac{0.001}{\sqrt{3} } } , 3.414+2.5758 \times {\frac{0.001}{\sqrt{3} } } ]

                                             = [3.4125 , 3.4155]

Therefore, 99% confidence interval for this specimen is [3.4125 , 3.4155].

6 0
3 years ago
The graph shows the relationship between the number of months different students practiced baseball and the number of games they
ratelena [41]

Answer:

A) from the line of best fit, the approximately y-intercept is (0,1.8). This means without any practice, 1h.8 games are won.

B) slope: (5.6-1.8)/(2-0) = 1.9

y = 1.9x + 1.8

(Line of best fit)

x = 13,

y = 1.9(13) + 1.8 = 26.5

Predicted no. of games won after 13 months of practice is 26.5

7 0
3 years ago
Read 2 more answers
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