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3 years ago

Given: AD = BC and AD || BC

Mathematics

1 answer:

o-na [289]3 years ago

5 0

Answer:

See below

Step-by-step explanation:

Proof:

Statements | Reasons

AD ≅ BC | Given

AD ║ BC | Given

AC ≅ AC | Reflexive Property

∠DAC ≅ ∠ACB | If 2 || lines are cut by a trans., the | alternate interior ∠s are congruent.

ΔADC ≅ ΔBCA | S.A.S Postulate

BA ≅ DC | Corresponding sides of congruent Δs

So, quad. ABCD is a ║gm | If a quad. has its opposite sides

| congruent, the quad. is a parallelogram.

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Answer:

Step-by-step explanation:

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3 years ago

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What of b is a solution to this equation? 6b=6 A) b=1 B)b=8

weqwewe [10]

Answer:

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Step-by-step explanation:

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3 years ago

What is x² + 7x - 30

Aleonysh [2.5K]

Are you looking for the value of x?

We can complete the square/factor:

x^2+7x-30=0

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By the zero product property rule, we have two equations, x+10=0 and x-3=0.

So the zeroes of x are -10 and 3

Hope this helped. Comment below if this didn't make sense.

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4 years ago

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mars1129 [50]

Answer:

The 80% confidence interval for difference between two means is (0.85, 1.55).

Step-by-step explanation:

The (1 - α) % confidence interval for difference between two means is:

$CI=(\bar x_{1}-\bar x_{2})\pm t_{\alpha/2,(n_{1}+n_{2}-2)}\times SE_{\bar x_{1}-\bar x_{2}}$

Given:

$\bar x_{1}=M_{1}=6.1\\\bar x_{2}=M_{2}=4.9\\SE_{\bar x_{1}-\bar x_{2}}=0.25$

Confidence level = 80%

$t_{\alpha/2, (n_{1}+n_{2}-2)}=t_{0.20/2, (5+5-2)}=t_{0.10,8}=1.397$

*Use a t-table for the critical value.

Compute the 80% confidence interval for difference between two means as follows:

$CI=(6.1-4.9)\pm 1.397\times 0.25\\=1.2\pm 0.34925\\=(0.85075, 1.54925)\\\approx(0.85, 1.55)$

Thus, the 80% confidence interval for difference between two means is (0.85, 1.55).

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3 years ago