An airplane travels at 950 km/h. how long does it take to travel 1.00km? in hours

Answer:

(a) 4i iterations

(b) "i × n" iterations

Step-by-step explanation:

(a) The provided algorithm segment shows:

for i:= 1 to 4, (Outer loop)

for j:= 1 to i (Inner loop)

next j,

next i

The inner loop executes i times while the outer loop completes 4 cycles.

The inner loop’s total execution when the full algorithm runs is:

= i × 4

= 4i iterations

(b) In the given algorithm segment;

for i:= 1 to n, (Outer loop)

for j:= 1 to i (Inner loop)

next j,

next i

where n denotes a set of positive integers.

<pthe inner="" loop="" also="" runs="" for="" times="" and="" the="" outer="" times.=""><pthus the="" total="" iterations="" of="" inner="" loop="" for="" entire="" algorithm="" is:="">

= i × n

= "i × n" iterations

</pthus></pthe>

Hindi ito palaging totoo.

An obtuse angle is one that measures more than 90 degrees but less than 180 degrees.

Response:

x = -y-12/5 + 4k

Detailed explanation:

Solve for x in the equation 12 - 5x - 4kx = y

12 - 5x - 4kx = y

Subtract 12 from both sides

-5x - 4kx = y - 12

Factor out -5x - 4kx to get -x(5 + 4k)

-x(5 + 4k) = y - 12

Now isolate x

-x = (y - 12)/(5 + 4k)

Thus, x = -y - 12/5 + 4k

Answer:

The likelihood of a battery being produced by Factory A and being defective is 0.012, which corresponds to 1.2%.

Step-by-step explanation:

We have two manufacturers of batteries for mobile phones: Factory A and Factory B. Factory A is responsible for 60% of the total battery production, while Factory B accounts for the remaining 40%.

Out of the batteries made by Factory A, 2% are found to be defective, whereas 4% of those from Factory B are defective.

Let P(A) denote the probability that a battery comes from Factory A, therefore P(A) = 0.60.

The probability that a battery originates from Factory B can be denoted as P(B) = 0.40.

Furthermore, define D as the event indicating that a battery is defective.

Thus, the probability of a defect given that a battery is from Factory A is P(D/A) = 0.02.

For Factory B, the probability of a defect given it is from there is P(D/B) = 0.04.

The probability for a battery being made at Factory A and also being defective can be calculated as follows:

Probability that Factory A produces batteries Probability that a battery from Factory A is defective

                       =  P(A) P(D/A)

                       =  0.60 0.02 = 0.012  or  1.2%

Conclusively, the probability that a battery is both from Factory A and defective is 1.2%.