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3 years ago

The density of hydrogen is 0.0000899 .

Mathematics

2 answers:

Cloud [144]3 years ago

8 0

It would be D if I did this right

Westkost [7]3 years ago

3 0

The answer would be D, as it is the closest option.

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Which answer choice below correctly represents the quotient of 846 = 35?

avanturin [10]

Answer:

where image?

Step-by-step explanation:

8 0

3 years ago

What is the parameter of a rectangle with width 10 inches and length 13 inches

svetoff [14.1K]

The parameter of a rectangle can be alculated using the following rule:
parameter = 2(length + width)

from the givens:
length = 13 inches
width = 10 inches

so, just substitute with the givens in the equation to calculate the parameter:
parameter = 2(10+13) = 2*23 = 46 inches

8 0

3 years ago

3x+15-7x=-7 solve for x

xenn [34]

3x+15−7x=−7

Simplify:

3x+15−7x=−7

3x+15+−7x=−7

(3x+−7x)+(15)=−7(Combine Like Terms)

−4x+15=−7

Subtract 15 from both sides.

−4x+15−15=−7−15

−4x=−22

Divide both sides by -4.

−4x/−4=−22/−4

x=11/2

5 0

3 years ago

Read 2 more answers

The value of y is directly proportional to the value of x If y=81 when x=54, what is the value of y when x=38

Arisa [49]

Answer:

y = 57

Step-by-step explanation:

y α x

y = kx ; k = Constant of proportionality

81 = k* 54

k = 81/54

k = 1.5

Value of y when x = 38

y = kx

y = 1.5 * 38

y = 57

Hence, y = 57 when x = 38

5 0

3 years ago

A football team has a probability of .75 of winning when playing any of the other four teams in its conference. If the games are

Alexeev081 [22]

Answer:

0.3164 = 31.64% probability the team wins all its conference games

Step-by-step explanation:

For each conference game, there are only two possible outcomes. Either the team wins it, or they lose. The probability of winning a game is independent of any other game. This means that the binomial probability distribution is used 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.

A football team has a probability of .75 of winning when playing any of the other four teams in its conference.

The probability means that $p = 0.75$ , and four games means that $n = 4$

If the games are independent, what is the probability the team wins all its conference games?

This is P(X = 4). So

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

$P(X = 4) = C_{4,4}.(0.75)^{4}.(0.25)^{0} = 0.3164$

0.3164 = 31.64% probability the team wins all its conference games

4 0

3 years ago