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slava [35]
2 years ago
11

Geometry help please!! 20 points

Mathematics
1 answer:
vekshin12 years ago
6 0

Answer:

Step-by-step explanation:

Opposite angles of a cyclic quadrilateral are supplementary, so <u>(6x+72)+3x=180</u>

<u />(6x+72)+3x=180\\9x+72=180\\9x=108\\x=12

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Klio2033 [76]
3x - 2y = 6
-2y = -3x + 6
y = (-3/-2)x + 6/-2
y = 3/2x - 3
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3 years ago
Help me with this<br>Please Help​
Arte-miy333 [17]
Answer:
80mm

Step by step explanation:

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3 0
2 years ago
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John borrows $6.50 from his friends to pay for snacks, and $11.75 from his sister to go to the movies how much money does he nee
Nina [5.8K]

John has borrowed $6.50 from his friends and $11.75 from his sister. To find how much money he needs to repay you can just add up the amount of money he has borrowed.

Make an equation to solve for this, where x = amount of money John needs to repay. Make the amount of money John borrowed add up to x. Your equation should look like:

$6.50 + $11.75 = x

Combine like terms (add 6.50 and 11.75).

$18.25 = x

After solving the equation, you have figured out that John need to repay $18.25 in total.

6 0
3 years ago
Read 2 more answers
M A. 59<br> B. 31<br> C. 149<br> D. 62
lbvjy [14]
I think The answer is a.59
4 0
3 years ago
An automobile company wants to determine the average amount of time it takes a machine to assemble a car. A sample of 40 times y
aksik [14]

Answer:

A 98% confidence interval for the mean assembly time is [21.34, 26.49] .

Step-by-step explanation:

We are given that a sample of 40 times yielded an average time of 23.92 minutes, with a sample standard deviation of 6.72 minutes.

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 average time = 23.92 minutes

             s = sample standard deviation = 6.72 minutes

             n = sample of times = 40

             \mu = population mean assembly time

<em> Here for constructing a 98% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation. </em>

<u>So, a 98% confidence interval for the population mean, </u>\mu<u> is; </u>

P(-2.426 < t_3_9 < 2.426) = 0.98  {As the critical value of z at 1%  level

                                               of significance are -2.426 & 2.426}  

P(-2.426 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 2.426) = 0.98

P( -2.426 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 2.426 \times {\frac{s}{\sqrt{n} } } ) = 0.98

P( \bar X-2.426 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+2.426 \times {\frac{s}{\sqrt{n} } } ) = 0.98

<u>98% confidence interval for</u> \mu = [ \bar X-2.426 \times {\frac{s}{\sqrt{n} } } , \bar X+2.426 \times {\frac{s}{\sqrt{n} } } ]

                                     = [ 23.92-2.426 \times {\frac{6.72}{\sqrt{40} } } , 23.92+2.426 \times {\frac{6.72}{\sqrt{40} } } ]  

                                    = [21.34, 26.49]

Therefore, a 98% confidence interval for the mean assembly time is [21.34, 26.49] .

7 0
3 years ago
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