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3 years ago

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A play director set up four rows of eight chairs in front of the stage as shown. A seat is randomly chosen. Which expression giv

es the probability that it is a front row seat or an aisle seat?

1 answer:

faust18 [17]3 years ago

3 0

The answer is a,,,,,,,

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the letter "T" makes up an estimated 8% of a certain language. Assume this still correct. A random sample of 1000 letters is tak

rusak2 [61]

Answer:

0.242

Step-by-step explanation:

p = 0.08, q = 1 − p = 0.92, and n = 1000.

The mean and standard deviation are:

μ = p = 0.08

σ = √(pq/n) = 0.00858

The z score is:

z = (x − μ) / σ = -0.70

Using a calculator or z score table, the probability is:

P(Z < -0.70) = 0.242

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3 years ago

Dante plans to put 80% of his paycheck into a savings ccount and spend the other 20%. His paycheck this week was $295. About how

Anni [7]

The 20% statement is useless

80% of $295 is what?
80%×295=??
.80 × 295 = $236.00 in savings

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3 years ago

The length of a pool is 3 feet more than twice its width. If the perimeter of the pool is 72 feet, find the dimensions of the po

photoshop1234 [79]

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3 years ago

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Find the sum of the first 12 terms of the sequence. Show all work for full credit. 1, -4, -9, -14, . . . (2 points)

My name is Ann [436]

<u>Answer:</u>

$S_n=$ -348

<u>Step-by-step explanation:</u>

We are given the following arithmetic sequence and we are to find the sum of its first 12 terms:

1, -4, -9, -14, . . .

For that, we will use the formula for the sum of the arithmetic mean:

$S_n=\frac{n}{2} (a_1+a_n)$

We know the value of the first term ( $a_n$ ) but we need to find the value of $a_{12}$ . So we will use the following formula:

$a_{12}=a_1+(n-1)d$

$a_{12}=(-4)+(12-1)(5)$

$a_{12}=-59$

Substituting these values in the sum formula to get:

$S_n=\frac{12}{2} (1+(-59))$

$S_n=$ -348

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3 years ago

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Identify the number as real, complex, pure imaginary, or nonreal complex.<br> - 169

barxatty [35]

Answer:

real

Step-by-step explanation:

-169 is a negative integer and integers are real numbers

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3 years ago

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