Ask question

Ask question

All categories

Dennis_Churaev [7]

3 years ago

8

When asked to determine if a set of points on a graph represents a function or a relation, explain what influences your decision

.

1 answer:

ladessa [460]3 years ago

7 0

You see if the graphs passes the vertical line test.
if any vertical line passes through more than one part of the graph then its not a function - just a relation.

You might be interested in

A yogurt company charges $.72 for a 6 ounce container of yogurt how much does it charge for an 8 ounce container

liraira [26]

Each ounce of yogurt costs $0.12 so 8 ounces would be $0.96

5 0

3 years ago

The length of time a person takes to decide which shoes to purchase is normally distributed with a mean of 8.21 minutes and a st

madam [21]

Answer:

4.55% probability that a randomly selected individual will take less than 5 minutes to select a shoe purchase.

Since Z > -2 and Z < 2, this outcome is not considered unusual.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

If $Z \leq 2$ or $Z \geq 2$ , the outcome X is considered to be unusual.

In this question:

$\mu = 8.21, \sigma = 1.9$

Find the probability that a randomly selected individual will take less than 5 minutes to select a shoe purchase.

This is the pvalue of Z when X = 5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5 - 8.21}{1.9}$

$Z = -1.69$

$Z = -1.69$ has a pvalue of 0.0455.

4.55% probability that a randomly selected individual will take less than 5 minutes to select a shoe purchase.

Since Z > -2 and Z < 2, this outcome is not considered unusual.

8 0

3 years ago

Thanks sm need an answer asap

Novosadov [1.4K]

Answer:

0.8 s

Step-by-step explanation:

Horizontal velocity

= 61cos(35) = 49.968 f/s

To reach 40 feet, it takes

40/49.968 = 0.801 = 0.8 sec

7 0

3 years ago

NEED HELP FAST PLEASE<br> Evaluate the limit.<br><br> a.4<br> c.1<br> b.2<br> d.limit does not exist

BlackZzzverrR [31]

We have the following limit:
(8n2 + 5n + 2) / (3 + 2n)
Evaluating for n = inf we have:
(8 (inf) 2 + 5 (inf) + 2) / (3 + 2 (inf))
(inf) / (inf)
We observe that we have an indetermination, which we must resolve.
Applying L'hopital we have:
(8n2 + 5n + 2) '/ (3 + 2n)'
(16n + 5) / (2)
Evaluating again for n = inf:
(16 (inf) + 5) / (2) = inf
Therefore, the limit tends to infinity.
Answer:
d.limit does not exist

7 0

3 years ago

An urn contains n white balls andm black balls. (m and n are both positive numbers.) (a) If two balls are drawn without replacem

Genrish500 [490]

DISCLAIMER: Please let me rename b and w the number of black and white balls, for the sake of readability. You can switch the variable names at any time and the ideas won't change a bit!

<h2>(a)</h2>

Case 1: both balls are white.

At the beginning we have $b+w$ balls. We want to pick a white one, so we have a probability of $\frac{w}{b+w}$ of picking a white one.

If this happens, we're left with $w-1$ white balls and still $b$ black balls, for a total of $b+w-1$ balls. So, now, the probability of picking a white ball is

$\dfrac{w-1}{b+w-1}$

The probability of the two events happening one after the other is the product of the probabilities, so you pick two whites with probability

$\dfrac{w}{b+w}\cdot \dfrac{w-1}{b+w-1}=\dfrac{w(w-1)}{(b+w)(b+w-1)}$

Case 2: both balls are black

The exact same logic leads to a probability of

$\dfrac{b}{b+w}\cdot \dfrac{b-1}{b+w-1}=\dfrac{b(b-1)}{(b+w)(b+w-1)}$

These two events are mutually exclusive (we either pick two whites or two blacks!), so the total probability of picking two balls of the same colour is

$\dfrac{w(w-1)}{(b+w)(b+w-1)}+\dfrac{b(b-1)}{(b+w)(b+w-1)}=\dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}$

<h2>(b)</h2>

Case 1: both balls are white.

In this case, nothing changes between the two picks. So, you have a probability of $\frac{w}{b+w}$ of picking a white ball with the first pick, and the same probability of picking a white ball with the second pick. Similarly, you have a probability $\frac{b}{b+w}$ of picking a black ball with both picks.

This leads to an overall probability of

$\left(\dfrac{w}{b+w}\right)^2+\left(\dfrac{b}{b+w}\right)^2 = \dfrac{w^2+b^2}{(b+w)^2}$

Of picking two balls of the same colour.

<h2>(c)</h2>

We want to prove that

$\dfrac{w^2+b^2}{(b+w)^2}\geq \dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}$

Expading all squares and products, this translates to

$\dfrac{w^2+b^2}{b^2+2bw+w^2}\geq \dfrac{w^2+b^2-b-w}{b^2+2bw+w^2-b-w}$

As you can see, this inequality comes in the form

$\dfrac{x}{y}\geq \dfrac{x-k}{y-k}$

With x and y greater than k. This inequality is true whenever the numerator is smaller than the denominator:

$\dfrac{x}{y}\geq \dfrac{x-k}{y-k} \iff xy-kx \geq xy-ky \iff -kx\geq -ky \iff x\leq y$

And this is our case, because in our case we have

$x=b^2+w^2$
$y=b^2+w^2+2bw$ so, y has an extra piece and it is larger
$k=b+w$ which ensures that k<x (and thus k<y), because b and w are integers, and so b<b^2 and w<w^2

4 0

3 years ago