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Ksenya-84 [330]
3 years ago
5

SOMEONE HELP ME WITH THIS PROOF

Mathematics
1 answer:
Pavlova-9 [17]3 years ago
6 0
Okay, for this proof, I'll write the steps out.

Statements:
1. ABCD is a parallelogram
2. FG bisects DB
3. <GEB ≡ (pretend congruent symbol) <FED
4. DE ≡ BE
5. <CDA ≡ <ABC
6. < CDB ≡ <DBA
7. triangle DFE ≡ triangle BGE
8. FE ≡ GE
9. DB bisects FG

Reasons:
1. Given
2. Given
3. vertical angles are congruent
4. If bisected, then split into congruent parts
5. opposite angles of a parallelogram are congruent
6. Subtraction Property
7. ASA
8. CPCTC
9. segment split into congruent parts by other segment is bisected.

Hope this helped!  :)
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A lumber company is making doors that are 2058.0 millimeters tall. If the doors are too long they must be trimmed, and if the do
USPshnik [31]

Answer:

No, we don't have sufficient evidence to support the claim that the doors are either too long or too short.

Step-by-step explanation:

We are given that a lumber company is making doors that are 2058.0 millimeters tall. If the doors are too long they must be trimmed, and if the doors are too short they cannot be used.

A sample of 11 is made, and it is found that they have a mean of 2043.0 millimeters with a standard deviation of 25.0.

Let \mu = <u><em>population mean length of doors</em></u>.

So, Null Hypothesis, H_0 : \mu = 2058.0 millimeters      {means that the mean length of doors is 2058.0 millimeters}

Alternate Hypothesis, H_A : \mu\neq 2058.0 millimeters     {means that the doors are either too long or too short}

The test statistics that will be used here is <u>One-sample t-test statistics</u> because we know about the population standard deviation;

                               T.S.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean = 2043.0 millimeters

             s = sample standard deviation = 25.0 millimeters

             n = sample of doors = 11

So, <u>the test statistics</u> =  \frac{2043.0-2058.0}{\frac{25.0}{\sqrt{11} } }

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The value of t-test statistics is -1.989.

Now, at a 5% level of significance, the t table gives a critical value between -2.228 and 2.228 at 10 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so <u><em>we have insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.

Therefore, we conclude that the mean length of doors is 2058.0 millimeters.

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Nataly_w [17]

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