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3 years ago

How to do proof problems in geometry

1 answer:

Crank3 years ago

6 0

First, start with your given. If your given has the words: bisect or midpoint, then the next step will be the definition of that.

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What is the solution to the system of equations? One-fourth x minus one-half y = 8 One-half x + three-fourths y = negative 5 (–8

kiruha [24]

Answer:

Step-by-step explanation:

x/4-y/2=8, x-2y=32, x=32+2y

x/2+3y/4=-5, 2x+3y=-20, x=(-20-3y)/2

32+2y=(-20-3y)/2

64+4y=-20-3y

7y=-84

y=-12, since x=32+2y

x=32+2(-12), x=32-24=8

So the solution is the point (8, -12)

3 0

3 years ago

Read 2 more answers

A machine is used to fill containers with a liquid product. Fill volume can be assumed to be normally distributed. A random samp

motikmotik

Answer:

There is no statistical evidence at 1% level to accept that the mean net contents exceeds 12 oz.

Step-by-step explanation:

Given that a random sample of ten containers is selected, and the net contents (oz) are as follows: 12.03, 12.01, 12.04, 12.02, 12.05, 11.98, 11.96, 12.02, 12.05, and 11.99.

We find mean = 11.015

Sample std deviation = 3.157

a) $H_0: \bar x= 12 oz\\H_a: \bar x >12$

(Right tailed test)

Mean difference /std error = test statistic

$\frac{11.015-12}{\frac{3.157}{\sqrt{10} } } \\=-0.99$

p value =0.174

Since p >0.01, our alpha, fail to reject H0

Conclusion:

There is no statistical evidence at 1% level to accept that the mean net contents exceeds 12 oz.

3 0

3 years ago

In a sample of 41 temperature readings taken from the freezer of a restaurant, the mean is 29.7 degrees and the population stand

yKpoI14uk [10]

Answer: $=(29.1596\text{ degrees},\ 30.2404\text{ degrees})$

Step-by-step explanation:

Given: Sample size: n = 41

Sample mean $\overline{x}= 29.7$ degrees

Population standard deviation $\sigma=2.7$ degrees

Confidence level (c) = 80% =0.80

Significance level (a) = 1- c = 1-0.80 = 0.20

z-score for 80% confidence level : z = 1.2816 [from z-table]

Confidence level for population mean :-

$\overline{x}\pm z\dfrac{\sigma}{\sqrt{n}}$

$=29.7\pm ( 1.2816)\dfrac{2.7}{\sqrt{41}}$

$=29.7\pm ( 1.2816)\dfrac{2.7}{6.403124}$

$=29.7\pm ( 1.2816)(0.42167)$

$=29.7\pm 0.5404$

$=(29.7-0.5404,\ 29.7+0.5404)$

$=(29.1596,\ 30.2404)$

Hence, 80% confidence interval for the temperatures in the freezer $=(29.1596\text{ degrees},\ 30.2404\text{ degrees})$

7 0

2 years ago

tangare [24]

Answer:

Step-by-step explanation:

If you multiply by 5 to get it over 100, or the percent, you get 55/100 which can just be written as 0.55 or 55%.

7 0

3 years ago

Mr. McDowell invested $14,000 in equipment to print yearbooks for Ardrey Kell High School. Each yearbook costs $7 to print and s

Feliz [49]

14000+7x=35x (x being the number of books) solve for x and you have your answer. 14000=35x-7x=28x therefore x=14000/28=500 so he needs to sell 500 books before he breaks even.

8 0

3 years ago