How many significant digits should be re-

jacob bought a place mat. The place mat has four sides and only one pair of parallel sides. Which could be the shape of Jacob's

Gwar [14]

A rectangle is the answer

8 0

3 years ago

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HELPPP pleaseeeeeeee

ladessa [460]

Answer:

I think the answer is 200 yards good luck

6 0

3 years ago

Help please!

Stella [2.4K]

Answer:

1.What is the y intercept of the function?

y-coordinate when x = 0, take from the table.

(0, 5)

2.What is the first difference?

Difference of y-values. This is the slope

7-5 = 2

3.Write an equation to represent the function given in the table

Use the first difference and the y-intercept: b = 5, m = 2

y = 2x + 5

4 0

3 years ago

based on the simulation, what is the probability that at most 2 of the next 10 callers will have to wait more than 8 minutes to

Triss [41]

Using the binomial distribution, supposing that 0.3 of the callers have to wait more than 8 minutes to have their calls answered, it is found that there is a 0.3828 = 38.28% probability that at most 2 of the next 10 callers will have to wait more than 8 minutes to have their calls answered.

For each caller, there are only two possible outcomes, either they have to wait more than 8 minutes to have their calls answered, or they do not. The probability of a caller having to wait more than 8 minutes is independent of any other caller, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

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

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

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:

10 callers, hence $n = 10$
Suppose that 0.3 of them have to wait more than 8 minutes, hence $p = 0.3$

The probability that <u>at most 2</u> of the next 10 callers will have to wait more than 8 minutes is:

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

Then

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

$P(X = 0) = C_{10,0}.(0.3)^{0}.(0.7)^{10} = 0.0282$

$P(X = 1) = C_{10,1}.(0.3)^{1}.(0.7)^{9} = 0.1211$

$P(X = 2) = C_{10,2}.(0.3)^{2}.(0.7)^{8} = 0.2335$

Then:

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0282 + 0.1211 + 0.2335 = 0.3828$

0.3828 = 38.28% probability that at most 2 of the next 10 callers will have to wait more than 8 minutes to have their calls answered.

A similar problem is given at brainly.com/question/25537909

3 0

3 years ago

Three sixth-grade classes raised $25.50, $49.75,

USPshnik [31]

Answer:

$37.50

Step-by-step explanation:

They are asking you to find the average and to do that you have to add all the numbers together in a given set then divide that total by the amount of numbers in that set to get your average.

6 0

4 years ago

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