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3 years ago

One horsepower is equal to how many foot pounds of work per second

Mathematics

2 answers:

gizmo_the_mogwai [7]3 years ago

7 0

I belive its 5 fpow i think

SpyIntel [72]3 years ago

3 0

1 horsepower
hp= 550.00 foot pounds per second ft-lb/sec

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You are running a fuel economy study. One of the cars you find is red. It can travel 24 4/5 miles on 4/5 gallon of gasoline. Wha

Lilit [14]

Answer:

1) 28.4 miles per gallon

2) 34 miles per gallon

3) the red car

Step-by-step explanation:

Blue car:

Travels 35 1/2 miles on 1 1/4 gallons of gasoline

Red car:

Travels 27 1/5 miles on 4/5 gallons of gasoline

We want to know the unit rate for miles per hour, and to do it we have to divide the miles they travel by the gallons each car uses:

Blue car:

(35 1/2) / (1 1/4) = 28.4 miles per gallon

Red car:

(27 1/5) / (4/5) = 34 miles per gallon

The red car travels greater distance with 1 gallon of gasoline {and thats about it!}

3 0

3 years ago

SOMEONE PLS HELP ME I REALLY NEED THIS

creativ13 [48]

I’m thinking it’s A cause the others don’t seem right

3 0

3 years ago

Read 2 more answers

Suppose that bugs are present in 1% of all computer programs. A computer de-bugging program detects an actual bug with probabili

lawyer [7]

Answer:

(i) The probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

Step-by-step explanation:

Denote the events as follows:

B = bugs are present in a computer program.

D = a de-bugging program detects the bug.

The information provided is:

$P(B) =0.01\\P(D|B)=0.99\\P(D|B^{c})=0.02$

(i)

The probability that there is a bug in the program given that the de-bugging program has detected the bug is, P (B | D).

The Bayes' theorem states that the conditional probability of an event E given that another event X has already occurred is:

$P(E|X)=\frac{P(X|E)P(E)}{P(X|E)P(E)+P(X|E^{c})P(E^{c})}$

Use the Bayes' theorem to compute the value of P (B | D) as follows:

$P(B|D)=\frac{P(D|B)P(B)}{P(D|B)P(B)+P(D|B^{c})P(B^{c})}=\frac{(0.99\times 0.01)}{(0.99\times 0.01)+(0.02\times (1-0.01))}=0.3333$

Thus, the probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii)

The probability that a bug is actually present given that the de-bugging program claims that bug is present is:

P (B|D) = 0.3333

Now it is provided that two tests are performed on the program A.

Both the test are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is:

P (Bugs are actually present | Detects on both test) = P (B|D) × P (B|D)

$=0.3333\times 0.3333\\=0.11108889\\\approx 0.1111$

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii)

Now it is provided that three tests are performed on the program A.

All the three tests are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is:

P (Bugs are actually present | Detects on all 3 test)

= P (B|D) × P (B|D) × P (B|D)

$=0.3333\times 0.3333\times 0.3333\\=0.037025927037\\\approx 0.037$

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

4 0

3 years ago

Given: AABC, AC = 5 m C = 90° m A= 22° Find: Perimeter of AABC A C B

Leto [7]

9514 1404 393

Answer:

perimeter ≈ 12.4 units

Step-by-step explanation:

The side adjacent to the angle is given. The relationships useful for the other two sides are ...

Tan = Opposite/Adjacent

Cos = Adjacent/Hypotenuse

From these, we have ...

opposite = 5·tan(22°) ≈ 2.02

hypotenuse = 5/cos(22°) ≈ 5.39

Then the perimeter is ...

P = a + b + c = 2.02 + 5 + 5.39 = 12.41

The perimeter of ∆ABC is about 12.4 units.

7 0

3 years ago

6th grade math help me pleaseeee

Ilya [14]

Answer:

D :3 -Your friend, Bill Cipher

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers