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3 years ago

What is 7/12 foot long

Mathematics

1 answer:

elena-s [515]3 years ago

4 0

7.8333333 will be the answer

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FIRST CORRECT ANSWER WILL GET BRAINLIEST

dezoksy [38]

Answer:

Step-by-step explanation:

vertical line through x at one

vertical lines have undefined slopes

can take two points to verify (1,1) (1,2) m= 2-1

1-1

1

0 undefined zero in denominator

8 0

3 years ago

if a car averages 22 miles per gallon how many gallons of gas are needed to drive the car for 6 hours at the rate of 55 miles pe

m_a_m_a [10]

Answer:

15 gallons PLEASE GIVE BRAINLIEST

Step-by-step explanation:

55 x 6 = 330 total miles

330 ÷ 22 = 15 gallons of gas are needed

6 0

3 years ago

On a middle school basketball court, there is a large circle painted on the

igor_vitrenko [27]

Answer:

113.04 feet²

Step-by-step explanation:

Use the circle area formula, A = $\pi$ r²

Sine the diameter is 12 feet, the radius is 6 feet

Plug in 6 as r into the formula, and 3.14 as pi, then solve:

A = $\pi$ r²

A = (3.14)(6²)

A = (3.14)(36)

A = 113.04

So, the area of the circle is approximately 113.04 feet²

4 0

3 years ago

Read 2 more answers

Each day, Leila records the number of news articles she reads. Here are her results for the last nine days.

Daniel [21]

The mode is whatever number that repeats itself the most, which would be 3 in this case. To find the range you would find the difference between the highest number and the lowest number. The highest number is 8 and the lowest is 1 so 8-1=7. The range is 7.

4 0

3 years ago

Astronomers treat the number of stars X in a given volume of space as Poisson random variable. The density of stars in the Milky

cupoosta [38]

Answer:

Explained below.

Step-by-step explanation:

The random variable X is defined as the number of stars in a given volume of space.

$X\sim \text{Poisson}\ (\lambda=1)$

The probability mass function of X is:

$p_{X}(x)=\frac{e^{-\lambda}\lambda^{x}}{x!}$

(7)

Compute the probability of exactly two stars in 16 cubic light-years as follows:

$P(X=2)=\frac{e^{-1}\times 1^{2}}{2!}=\frac{e^{-1}}{2}=\frac{0.36788}{2}=0.18394\approx 0.184$

(8)

Compute the probability of three or more stars in 16 cubic light-years as follows:

$P(X\geq 3)=1-P(X$

(9)

In 16 cubic light years there is only 1 star.

Then in 1 cubic light years there will be, (1/16) stars.

Then in 4 cubic light years there will be, 4 × (1/16) = (1/4) stars.

(10)

In 16 cubic light years there is only 1 star.

Then in 1 cubic light years there will be, (1/16) stars.

Then in t cubic light years there will be, [t × (1/16)] stars.

7 0

3 years ago