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4 years ago

How many meters are there in 21 feet?

Mathematics

2 answers:

Kobotan [32]4 years ago

7 0

There are 6.4008 meters in 21 feet.

Orlov [11]4 years ago

4 0

Hey there!:

1 meter --------------- 3.28084 feet
? meters -------------- 21 feet

therefore:

( 21 * 1 ) / ( 3.28084 ) =

21 / 3.28084 => 6.4008 meters

hope this helps!

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PLSS HELP ASAP TYSM <33 SORRY I COULDN'T FIND SCIENCE LOLL

dimaraw [331]

Answer:

Rock D.

Step-by-step explanation:

We can assume that the force that the catapult does is always the same.

So, here we need to remember Newton's second law:

F = m*a

force equals mass times acceleration.

Where acceleration is the rate of change of the velocity.

So, if we want the rock to hit closer to the catapult, the rock must be less accelerated than rock B.

So, we can rewrite:

a = F/m

So, as larger is the mass of the rock, smaller will be the acceleration of the rock after it leaves the catapult (because the mass is in the denominator). So if we want to have a smaller acceleration, we need to choose a rock with a larger mass than rock B.

Assuming that the mass depends on the size, the only one that has a mass larger than rock B is rock D.

So we can assume that rock D is the correct option.

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3 years ago

14=h/5 Help me please

ValentinkaMS [17]

Answer:

14=h/5

h=14*5

h=70

Check:

70/5=14

Hope this helps

6 0

3 years ago

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Kitty [74]

Answer:

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5 0

3 years ago

Find the radius of the circle whose equation is (x + 5)2 + (y + 8)2 = 16

Oxana [17]

The radius is 4.

Find the radius of the circle whose equation is (x + 5)2 + (y + 8)2 = 16

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3 years ago

Richard has just been given an l0-question multiple-choice quiz in his history class. Each question has five answers, of which o

myrzilka [38]

Answer:

a) 0.0000001024 probability that he will answer all questions correctly.

b) 0.1074 = 10.74% probability that he will answer all questions incorrectly

c) 0.8926 = 89.26% probability that he will answer at least one of the questions correctly.

d) 0.0328 = 3.28% probability that Richard will answer at least half the questions correctly

Step-by-step explanation:

For each question, there are only two possible outcomes. Either he answers it correctly, or he does not. The probability of answering a question correctly is independent of any other question. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Each question has five answers, of which only one is correct

This means that the probability of correctly answering a question guessing is $p = \frac{1}{5} = 0.2$

10 questions.

This means that $n = 10$

A) What is the probability that he will answer all questions correctly?

This is $P(X = 10)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{10,10}.(0.2)^{10}.(0.8)^{0} = 0.0000001024$

0.0000001024 probability that he will answer all questions correctly.

B) What is the probability that he will answer all questions incorrectly?

None correctly, so $P(X = 0)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.2)^{0}.(0.8)^{10} = 0.1074$

0.1074 = 10.74% probability that he will answer all questions incorrectly

C) What is the probability that he will answer at least one of the questions correctly?

This is

$P(X \geq 1) = 1 - P(X = 0)$

Since $P(X = 0) = 0.1074$ , from item b.

$P(X \geq 1) = 1 - 0.1074 = 0.8926$

0.8926 = 89.26% probability that he will answer at least one of the questions correctly.

D) What is the probability that Richard will answer at least half the questions correctly?

This is

$P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{10,5}.(0.2)^{5}.(0.8)^{5} = 0.0264$

$P(X = 6) = C_{10,6}.(0.2)^{6}.(0.8)^{4} = 0.0055$

$P(X = 7) = C_{10,7}.(0.2)^{7}.(0.8)^{3} = 0.0008$

$P(X = 8) = C_{10,8}.(0.2)^{8}.(0.8)^{2} = 0.0001$

$P(X = 9) = C_{10,9}.(0.2)^{9}.(0.8)^{1} \approx 0$

$P(X = 10) = C_{10,10}.(0.2)^{10}.(0.8)^{0} \approx 0$

$P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.0264 + 0.0055 + 0.0008 + 0.0001 + 0 + 0 = 0.0328$

0.0328 = 3.28% probability that Richard will answer at least half the questions correctly

8 0

3 years ago