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3 years ago

129 times 6+1 pronto fast

Mathematics

2 answers:

melamori03 [73]3 years ago

7 0

Answer:

129 x 6=774

774+1=775

Step-by-step explanation:

Hope this helps

erastovalidia [21]3 years ago

6 0

Answer:

903

Step-by-step explanation:

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The circle below is centered at the origin and has a radius of 7. What is its

Rashid [163]

Answer:

x² + y² = 49

Step-by-step explanation:

The equation for a circle is: x² + y² = r²

here, we know our circle's radius (r) to be 7

So, the equation for our circle is: {replace "r" with value}

x² + y² = 7²

which can be simplified to:

x² + y² = 49

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3 years ago

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Write the equation in slope form intercept m=-2(3,-1)

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Answer:

y = -2x + 5

Step-by-step explanation:

Find b using the m and (3, -1) given by plugging them into y = mx + b

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3 years ago

Please solve for me!!

Allushta [10]

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A-base (s); three congruent rectangles.

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3 years ago

Suppose that on the average, 7 students enrolled in a small liberal arts college have their automobiles stolen during the semest

sineoko [7]

Answer:

0.91824 = 91.824% probability that more than 3 students will have their automobiles stolen during the current semester.

Step-by-step explanation:

We have only the mean, which means that the Poisson distribution is used to solve this question.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

Suppose that on the average, 7 students enrolled in a small liberal arts college have their automobiles stolen during the semester.

This means that $\mu = 7$

What is the probability that more than 3 students will have their automobiles stolen during the current semester?

This is:

$P(X > 3) = 1 - P(X \leq 3)$

In which

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-7}*7^{0}}{(0)!} = 0.00091$

$P(X = 1) = \frac{e^{-7}*7^{1}}{(1)!} = 0.00638$

$P(X = 2) = \frac{e^{-7}*7^{2}}{(2)!} = 0.02234$

$P(X = 3) = \frac{e^{-7}*7^{3}}{(3)!} = 0.05213$

Then

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.00091 + 0.00638 + 0.02234 + 0.05213 = 0.08176
$

$P(X > 3) = 1 - P(X \leq 3) = 1 - 0.08176 = 0.91824$

0.91824 = 91.824% probability that more than 3 students will have their automobiles stolen during the current semester.

3 0

3 years ago