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Otrada [13]
2 years ago
5

Help pls :’( ASAPP!!! “Complete the proof”

Mathematics
1 answer:
nata0808 [166]2 years ago
6 0

1) \overline{AB} \cong \overline{CD}, \overline{AD} \cong \overline{CB}, \overline{AX} \perp \overline{BD}, \overline{CY} \perp\overline{BD} (given)

2) \overline{BD} \cong \overline{BD} (reflexive property)

3) \triangle ABD \cong \triangle ACDB (SSS)

4) \angle ADB \cong \angle CBY (CPCTC)

5) \angle CYB and \angle AXD are right angles (perpendicular lines form right angles)

6) \triangle CYB and \triangle AXD are right triangles (a triangle with a right angle is a right triangle)

7) \triangle AXD \cong \triangle CYB (HA)

8) \overline{AX} \cong \overline{CY} (CPCTC)

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Step-by-step explanation:

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A number x divided by 36
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Alisa had 1 & 1/2 liter of juice in a bottle. She drank three eighths liters of juice. What fraction of the juice in the bot
emmasim [6.3K]

Answer:

1/4

Step-by-step explanation:

The fraction of the juice Alisa drank was

   3/8 liter

--------------------

  3/2 liters

and this fraction can be simplified by inverting the divisor (3/2 liters) and multiplying instead:

 3       2

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  8       3

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7 0
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Some transportation experts claim that it is the variability of speeds, rather than the level of speeds, that is a critical fact
scZoUnD [109]

Answer:

Explained below.

Step-by-step explanation:

The claim made by an expert is that driving conditions are dangerous if the variance of speeds exceeds 75 (mph)².

(1)

The hypothesis for both the test can be defined as:

<em>H</em>₀: The variance of speeds does not exceeds 75 (mph)², i.e. <em>σ</em>² ≤ 75.

<em>Hₐ</em>: The variance of speeds exceeds 75 (mph)², i.e. <em>σ</em>² > 75.

(2)

A Chi-square test will be used to perform the test.

The significance level of the test is, <em>α</em> = 0.05.

The degrees of freedom of the test is,

df = n - 1 = 55 - 1 = 54

Compute the critical value as follows:

\chi^{2}_{\alpha, (n-1)}=\chi^{2}_{0.05, 54}=72.153

Decision rule:

If the test statistic value is more than the critical value then the null hypothesis will be rejected and vice-versa.

(3)

Compute the test statistic as follows:

\chi^{2}=\frac{(n-1)\times s^{2}}{\sigma^{2}}

    =\frac{(55-1)\times 94.7}{75}\\\\=68.184

The test statistic value is, 68.184.

Decision:

cal.\chi^{2}=68.184

The null hypothesis will not be rejected at 5% level of significance.

Conclusion:

The variance of speeds does not exceeds 75 (mph)². Thus, concluding that driving conditions are not dangerous on this highway.

7 0
3 years ago
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