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3 years ago

Two similar triangles are shown below: Which two sets of angles are corresponding angles?

Mathematics

2 answers:

fenix001 [56]3 years ago

6 0

W and V are corresponding as well as X and Y

your answer is A

you can tell by the amount of lines in each angle :)

vekshin13 years ago

5 0

The answer is aaaaaaaaaa

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CAN SOMEONE PLEASE HELP I WILL GIVE BRAINLIEST TO HOWEVER ANSWERS FIRST!!

igomit [66]

Answer: y+5= -1/7(x-7)

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4 years ago

OverLord2011 [107]

A research paper recommends using Poisson process to model the number of failures in commercial water pipes The paper also gives estimates the failure rate in units of failures per 100 miles of pipe per day; for four different types of pipe and for many different years_ For example, for cast iron pipe in 2005, the authors' estimate is the failure rate is 0.0864 failures per 100 miles per day: Suppose a town had 2500 miles of cast iron pipe underground in 2005 What is the probability of at least two failures

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3 years ago

Find the distance between ( -2, 5) (-2, 14).

WITCHER [35]

Answer:

Step-by-step explanation:

x didn't change in value, y changed by 9

the distance between the two points is 9

6 0

3 years ago

If 10 + a = 10, then a = 0

Lunna [17]

Answer:

B) identity property of addition.

Step-by-step explanation:

Given:

The statement given is:

$10+a=10,\ then\ a=0$

Here, we observe that 10 is added to a giving the number 10 itself as the answer. So, we know of a property which says that when 0 is added to a number, the result is the number itself. This property is called identity property of addition.

So, the value of $a$ must be 0 because on adding $a$ to 10, we are getting the number 10 only. So, the above statement is an example of identity property of addition.

5 0

4 years ago

A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x overbar, is found

joja [24]

Answer:

(a) 80% confidence interval for the population mean is [109.24 , 116.76].

(b) 80% confidence interval for the population mean is [109.86 , 116.14].

(c) 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed.

Step-by-step explanation:

We are given that a simple random sample of size n is drawn from a population that is normally distributed.

The sample mean is found to be 113 and the sample standard deviation is found to be 10.

(a) The sample size given is n = 13.

Firstly, the pivotal quantity for 80% 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 = 113

s = sample standard deviation = 10

n = sample size = 13

$\mu$ = population mean

Here for constructing 80% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 80% confidence interval for the population mean, $\mu$ is ;

P(-1.356 < $t_1_2$ < 1.356) = 0.80 {As the critical value of t at 12 degree

of freedom are -1.356 & 1.356 with P = 10%}

P(-1.356 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.356) = 0.80

P( $-1.356 \times }{\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

P( $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

80% confidence interval for $\mu$ = [ $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.356 \times }{\frac{10}{\sqrt{13} } }$ , $113+1.356 \times }{\frac{10}{\sqrt{13} } }$ ]

= [109.24 , 116.76]

Therefore, 80% confidence interval for the population mean is [109.24 , 116.76].

(b) Now, the sample size has been changed to 18, i.e; n = 18.

So, the critical values of t at 17 degree of freedom would now be -1.333 & 1.333 with P = 10%.

80% confidence interval for $\mu$ = [ $\bar X-1.333 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.333 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.333 \times }{\frac{10}{\sqrt{18} } }$ , $113+1.333 \times }{\frac{10}{\sqrt{18} } }$ ]

= [109.86 , 116.14]

Therefore, 80% confidence interval for the population mean is [109.86 , 116.14].

(c) Now, we have to construct 98% confidence interval with sample size, n = 13.

So, the critical values of t at 12 degree of freedom would now be -2.681 & 2.681 with P = 1%.

98% confidence interval for $\mu$ = [ $\bar X-2.681 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+2.681 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-2.681 \times }{\frac{10}{\sqrt{13} } }$ , $113+2.681 \times }{\frac{10}{\sqrt{13} } }$ ]

= [105.56 , 120.44]

Therefore, 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed because t test statistics is used only when the data follows normal distribution.

6 0

3 years ago

Two similar triangles are shown below: Which two sets of angles are corresponding angles?​

Two similar triangles are shown below: Which two sets of angles are corresponding angles?