1. A manufacturer claims that the mean lifetime of its fluorescent bulbs is 1500 hours. A homeowner selects 40 bulbs and finds t

Answer:

Step-by-step explanation:

Given that:

Sarah, Jose, and you all reside on the same street:

The distance between you and Jose is 5 blocks

The distance between Jose and Sarah is 2 blocks

Calculating the distance (b) between you and Sarah:

Since it is not specified if Sarah is closer to you or to Jose;

Thus;

Distance b from you to Sarah will be:

(The distance to Jose ± the distance to Sarah)

b = (5 ± 2)

b = (5 + 2) or (5 - 2)

b = 7 blocks or 3 blocks

Important details about isosceles triangle ABC:

• The median CD, which is drawn to the base AB, also acts as an altitude to that base in the isosceles triangle (CD⊥AB). This indicates that triangles ACD and BCD are congruent right triangles, each with hypotenuses AC and BC.
• In isosceles triangle ABC, the sides AB and BC are equal, meaning AC=BC.
• The base angles at AB are equal, m∠A=m∠B=30°.

1. Consider the right triangle ACD. The angle adjacent to side AD is 30°, which dictates that the hypotenuse AC is double the length of the opposite side CD relating to angle A.

AC=2CD.

2. Now, for right triangle BCD, the angle next to side BD is also 30°, so hypotenuse BC is twice the opposite leg CD linked to angle B.

BC=2CD.

3. To calculate the perimeters of triangles ACD, BCD, and ABC:

4. If the total of the perimeters of triangles ACD and BCD is 20 cm greater than the perimeter of triangle ABC, then

5. Given that AC=BC=2CD, the lengths of legs AC and BC of the isosceles triangles are 20 cm.

Answer: 20 cm.

Answer:

Step-by-step explanation:

There are 15 antennas in total.

Out of these, 3 are defective.

This means that 12 antennas are functioning: 15-3=12.

To ensure that no two defective antennas are adjacent, we need to have only one defective at a time placed between the functional ones.

So,

We align the 13 functional antennas, then look for the spaces where the defective antennas can fit

__G __ G __ G __ G __ G __G __ G __ G __ G __ G __ G __ G __G __

Each gap represented by an underscore indicates a possible location for a defective antenna, allowing for just one per space.

Consequently, there are 14 potential spots for the defective antennas. With 3 defectives, we are dealing with a combinatorial arrangement.

ⁿCr= n!/(n-r)!r!

The total number of arrangements possible is

14C3=14!/(14-3)!3!

14C3=14×13×12×11!/11!×3×2

14C3=14×13×12/6

That gives us 364 distinct ways to arrange them.