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igomit [66]
3 years ago
12

Choose the abbreviation of the postulate or theorem that supports the conclusion that the triangles are congruent.

Mathematics
2 answers:
Naddik [55]3 years ago
7 0

Answer: HA

Step-by-step explanation:

Given:  ∠C, ∠F are right angles ∠'s

Then the given triangles in picture Δ ABC and ΔDEF are right triangles, such that the sides opposite to the right angles AB and and DE are hypotenuse of triangles Δ ABC and ΔDEF respectively.

Also it is given that AB = DE; ∠A = ∠D

Then by HA theorem ,

Δ ABC ≅ ΔDEF

  • HA Theorem says that if the hypotenuse and an angle of a right triangle are congruent to the hypotenuse and an angle of another triangle, then the two triangles are said to be congruent.
miss Akunina [59]3 years ago
4 0

Answer:

C. HA (Hypotenuse angle congruence).

Step-by-step explanation:

We have been given a diagram of two right triangles. We are asked to choose the correct congruence theorem for our given triangles.

In our triangles ABC and DEF we have,

m\angle C=m\angle F=90^o

AB=DE

\angle A=\angle D

We can see that side length AB is hypotenuse of triangle ABC and side length DE is hypotenuse of triangle DEF.

Angle A and angle D are acute angles (less than 90 degrees) as we have one right angle in our both triangles and measure of other two angles will be 90 degrees in both triangles by angle sum property of triangles.

Hypotenuse angle theorem states that two right triangles are congruent, if the hypotenuse and an acute angle of one right triangle are congruent to the hypotenuse and an acute angle of another right triangle.

Therefore, by HA congruence triangle ABC is congruent to triangle DEF and option C is the correct choice.

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Let, price of fudge and gum is x and y respectively.

Converting given statements into mathematical equation.

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Price of buying 20 pieces of fudge and 20 pieces of bubble gum is :

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Allegiant Airlines charges a mean base fare of $89. In addition, the airline charges for making a reservation on its website, ch
lara31 [8.8K]

Answer:

a) (<em>μ</em>) = $128

b) 0.9472

c)  0.6671

Step-by-step explanation:

Given that:

Allegiant Airlines charges a mean base fare of $89.

this implies that: mean base fare = $89.

The question proceeds by stating the additional charge on  its website, checking bags, and inflight beverages.

so , additional charges turns out to be = $39 per passenger

Now, Suppose a random sample of 60 passengers is taken

random sample (n) = 60

The population standard deviation of total flight cost is known to be $40

standard deviation (σ) = 40

Question (a) says; we should find the population mean cost per flight

To determine that; we have to consider the total sum(<em>μ</em>) of the mean base fare with the mean additional charges.

Population mean cost per flight (<em>μ</em>)  =  mean base fare + mean additional charges

(<em>μ</em>) = $89 + $39

(<em>μ</em>) = $128

b)

What is the probability the sample mean will be within $10 of the population mean cost per flight (to 4 decimals)?

To determine that; we have:

P(128 - 10 ≤ Х ≤ 128 +10)

= P(118 ≤ Х ≤  138)

= P[\frac{118-128}{40/\sqrt{60} }\leq  \frac{X- \delta }{\alpha /\sqrt{n} }\leq  \frac{138-128}{40\sqrt{60} }]               (where \delta  =  <em>μ  </em>and ∝ = σ )

= P[-1.9365\leq z +1.9365]

= P[z\leq 1.9365]-P[z\leq -1.9365]

Using Excel Command to approach this process, we have;

= 0.9736 - 0.0264

= 0.9472     (to four decimal places)

∴ the probability that the sample mean will be within $10 of the population mean cost per flight = 0.9472

c)

What is the probability the sample mean will be within $5 of the population mean cost per flight (to 4 decimals)?

We wll have to go through the process like the one attempted above in question (b);

So;

P(128 - 5 ≤ Х ≤ 128 + 5)

= P(123 ≤ Х ≤  133)

=  P[\frac{123-128}{40/\sqrt{60} }\leq  \frac{X- \delta }{\alpha /\sqrt{n} }\leq  \frac{133-128}{40\sqrt{60} }]                         (where \delta  =  <em>μ  </em>and ∝ = σ )

= P[-0.9682\leq z +0.9682]

= P[z\leq 0.9682]-P[z\leq-0.9682]

Computing these data in Excel; we have

= 0.8335 -0.1665

= 0.6671         (to 4 decimal places)

∴  the probability that the sample mean will be within $5 of the population mean cost per flight.

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