What is the simplified value of the exponential expression 27 1/3?<br> 1/3<br> 1/9<br> 3<br> 9

Savatey [412]

I hope this helps you

5 0

3 years ago

A test is worth 50 points. Multiple- choice questions are worth 1 point, and short answer questions are worth 3 points. if the t

STatiana [176]

15 Short Answer Questions
( =45 points)

5 Multiple Choice Questions
( = 5 points)

4 0

3 years ago

Read 2 more answers

Which of the following is equal to 393,500? 0.3935 × 10 -6 39.35 × 10 -4 3.935 × 10 -5 3.935 × 10 5

SCORPION-xisa [38]

Answer:

No answer is equal to 393,500

Step-by-step explanation:

You can try to solve each question.

0.3935 × 10 -6 = -2.065

39.35 × 10 -4 = 389.5

3.935 × 10 -5 = 34.35

3.935 × 10 +5 = 44.35

None of these equal to 393,500

(Maybe I am getting things wrong I am not in a smart mood today)

4 0

3 years ago

Solve for x in -14x⁴ - 8x³ = 32

gayaneshka [121]

Answer:

No solutions

Step-by-step explanation:

There can be no solutions to your question

7 0

2 years ago

It is estimated that 0.54 percent of the callers to the Customer Service department of Dell Inc. will receive a busy signal. Wha

stira [4]

Using the binomial distribution, it is found that there is a 0.8295 = 82.95% probability that at least 5 received a busy signal.

<h3>What is the binomial distribution formula?</h3>

The formula is:

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

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

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

In this problem:

0.54% of the calls receive a busy signal, hence p = 0.0054.

A sample of 1300 callers is taken, hence n = 1300.

The probability that at least 5 received a busy signal is given by:

$P(X \geq 5) = 1 - P(X < 5)$

In which:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4).

Then:

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

$P(X = 0) = C_{1300,0}.(0.0054)^{0}.(0.9946)^{1300} = 0.0009$

$P(X = 1) = C_{1300,1}.(0.0054)^{1}.(0.9946)^{1299} = 0.0062$

$P(X = 2) = C_{1300,2}.(0.0054)^{2}.(0.9946)^{1298} = 0.0218$

$P(X = 3) = C_{1300,3}.(0.0054)^{3}.(0.9946)^{1297} = 0.0513$

$P(X = 4) = C_{1300,4}.(0.0054)^{4}.(0.9946)^{1296} = 0.0903$

Then:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0009 + 0.0062 + 0.0218 + 0.0513 + 0.0903 = 0.1705.

$P(X \geq 5) = 1 - P(X < 5) = 1 - 0.1705 = 0.8295$

0.8295 = 82.95% probability that at least 5 received a busy signal.

More can be learned about the binomial distribution at brainly.com/question/24863377

#SPJ1

6 0

1 year ago

What is the simplified value of the exponential expression 27 1/3? 1/3 1/9 3 9