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2 years ago

Does the graph represent a function? just a lil help for khan academy students

Mathematics

1 answer:

cupoosta [38]2 years ago

4 0

Yeah also you got it correct, congrats.

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For a food truck event, you sold 750 chicken tacos and 520 beef tacos.

Nesterboy [21]

Answer: 750x+520y=total cost

Step-by-step explanation:

750x+520y=total cost

7 0

2 years ago

Read 2 more answers

Find two numbers such that the sum of their squares is 100. The square of the larger number is equal to the sum of the product o

Shtirlitz [24]

Answer with Step-by-step explanation:

Let x and y are two numbers

According to question

$x^2+y^2=100$ ....(1)

Suppose x is larger than y

$x^2=x(4y+8x)$

$x=4y+8x$

$x-8x=4y$

$-7x=4y$

$y=-\frac{7}{4}x$

Substitute the value of y

$x^2+(-\frac{7}{4}x)^2=100$

$x^2+\frac{49}{16}x^2=100$

$\frac{16x^2+49x^2}{16}=100$

$\frac{65}{16}x^2=100$

$x^2=\frac{16\times 100}{65}$

$x=\sqrt{\frac{1600}{65}}=\frac{40}{\sqrt{65}}=\frac{40}{65}\sqrt{65}=\frac{8}{13}\sqrt{65}$

$y=-\frac{7}{4}\times \frac{8}{13}\sqrt{65}=-\frac{14}{13}\sqrt{65}$

3 0

2 years ago

Is a good economist always objective or

Mashutka [201]

anwer

yes good economist is always objective.

3 0

2 years ago

What is the value of 2/7+-3/4+-2/7+4/3

vodka [1.7K]

Answer:

7/12

Step-by-step explanation:

5 0

2 years ago

Show that if X is a geometric random variable with parameter p, then

Lubov Fominskaja [6]

Answer:

$\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}=-\frac{p ln p}{1-p}$

Step-by-step explanation:

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

$P(X=x)=(1-p)^{x-1} p$

Let X the random variable that measures the number os trials until the first success, we know that X follows this distribution:

$X\sim Geo (1-p)$

In order to find the expected value E(1/X) we need to find this sum:

$E(X)=\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}$

Lets consider the following series:

$\sum_{k=1}^{\infty} b^{k-1}$

And let's assume that this series is a power series with b a number between (0,1). If we apply integration of this series we have this:

$\int_{0}^b \sum_{k=1}^{\infty} r^{k-1}=\sum_{k=1}^{\infty} \int_{0}^b r^{k-1} dt=\sum_{k=1}^{\infty} \frac{b^k}{k}$ (a)

On the last step we assume that $0\leq r\leq b$ and $\sum_{k=1}^{\infty} r^{k-1}=\frac{1}{1-r}$ , then the integral on the left part of equation (a) would be 1. And we have:

$\int_{0}^b \frac{1}{1-r}dr=-ln(1-b)$

And for the next step we have:

$\sum_{k=1}^{\infty} \frac{b^{k-1}}{k}=\frac{1}{b}\sum_{k=1}^{\infty}\frac{b^k}{k}=-\frac{ln(1-b)}{b}$

And with this we have the requiered proof.

And since $b=1-p$ we have that:

$\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}=-\frac{p ln p}{1-p}$

4 0

3 years ago