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3 years ago

Is the following relation a function ?

Mathematics

2 answers:

Dafna11 [192]3 years ago

6 0

No, it is not a function. It fails the vertical line test (VLT). VLT state that if you draw a vertical line, it must only intersect the graph one time. If it intersections more than once, then it is not a function.

Hoochie [10]3 years ago

5 0

No because it fails the vertical line test. Note that a vertical line can be drawn through the graph which passes through 2 points. If it was a function any vertical line would pass through only one point on the graph.

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The mean waiting time at the drive-through of a fast-food restaurant from the time the food is ordered to when it is received is

alexira [117]

Answer:

a) And if we replace we got: $\bar X= 78$

$s = 15.391$

b) $78-3.25\frac{15.391}{\sqrt{10}}=62.182$

$78-3.25\frac{15.391}{\sqrt{10}}=93.818$

So on this case the 99% confidence interval would be given by (62.182;93.818)

Step-by-step explanation:

Dataset given: 109 67 58 76 65 80 96 86 71 72

Part a

For this case we can calculate the sample mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And if we replace we got: $\bar X= 78$

And the deviation is given by:

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And if we replace we got $s = 15.391$

Part b

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=10-1=9$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,9)".And we see that $t_{\alpha/2}=3.25$

Now we have everything in order to replace into formula (1):

$78-3.25\frac{15.391}{\sqrt{10}}=62.182$

$78-3.25\frac{15.391}{\sqrt{10}}=93.818$

So on this case the 99% confidence interval would be given by (62.182;93.818)

8 0

2 years ago

In the United States, voters who are neither Democrat nor Republican are called Independent. It is believed that 11% of voters a

Radda [10]

Answer:

a) 0.0214 = 2.14% probability that none of the people are Independent.

b) 0.8516 = 85.16% probability that fewer than 6 are Independent.

c) 0.8914 = 89.14% probability that more than 2 people are Independent.

Step-by-step explanation:

For each people, there are only two possible outcomes. Either they are independent, or they are not. For each person asked, the probability of them being Independent voters is the same. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

It is believed that 11% of voters are Independent.

This means that $p = 0.11$

A survey asked 33 people to identify themselves as Democrat, Republican, or Independent.

This means that $n = 33$

A. What is the probability that none of the people are Independent?

This is P(X = 0). So

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

$P(X = 0) = C_{33,0}.(0.11)^{0}.(0.89)^{33} = 0.0214$

0.0214 = 2.14% probability that none of the people are Independent.

B. What is the probability that fewer than 6 are Independent?

This is

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

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

$P(X = 0) = C_{33,0}.(0.11)^{0}.(0.89)^{33} = 0.0214$

$P(X = 1) = C_{33,1}.(0.11)^{1}.(0.89)^{32} = 0.0872$

$P(X = 2) = C_{33,2}.(0.11)^{2}.(0.89)^{31} = 0.1724$

$P(X = 3) = C_{33,3}.(0.11)^{3}.(0.89)^{30} = 0.2202$

$P(X = 4) = C_{33,4}.(0.11)^{4}.(0.89)^{29} = 0.2041$

$P(X = 5) = C_{33,5}.(0.11)^{5}.(0.89)^{28} = 0.1463$

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0214 + 0.0872 + 0.1724 + 0.2202 + 0.2041 + 0.1463 = 0.8516$

0.8516 = 85.16% probability that fewer than 6 are Independent.

C. What is the probability that more than 2 people are Independent?

This is:

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

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

$P(X = 0) = C_{33,0}.(0.11)^{0}.(0.89)^{33} = 0.0214$

$P(X = 1) = C_{33,1}.(0.11)^{1}.(0.89)^{32} = 0.0872$

$P(X < 2) = 0.0214 + 0.0872 = 0.1086$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.1086 = 0.8914$

0.8914 = 89.14% probability that more than 2 people are Independent.

8 0

2 years ago

Solve using the quadratic formula. Show all work. Write each solution in simplest form. No decimals.

dimaraw [331]

Answer:

u can download the gauthmath app it's very helpful i PROMISE :)

5 0

2 years ago

Read 2 more answers

The product of 7 and p

ICE Princess25 [194]

The product means to multiple

7 0

3 years ago

Read 2 more answers

In the regular pentagon below, if AP = 4 m and BC = 10 m, find it's area.

12345 [234]

First symmetrically cut the pentagon to get 5 pieces. (So that you will get the idea that this polygon is divided into 5 triangles)

Then find the area of one of the triangles:

Area of Triangle = $\frac{1}{2} bh$

A = $\frac{1}{2}$ × 4 × 10

A = 20 m²

To find the area of the whole pentagon shape:

A = 20 × 5 = 100 m²

<em>Hope it helps!</em>

4 0

2 years ago