1. I mark as brainliest

Masja [62]

Answer:

(A) The rate of change in the price of a bushel of corn in the current year is $7.

(B) The price of a bushel of corn in the current year is $2 more than the price of a bushel of corn in the previous year.

Step-by-step explanation:

The graph for the prices of different numbers of bushels of corn at a store in the current year is shown below.

Part A:

The rate of change in the price of a bushel of corn in the current year based upon the number of bushels is known as the slope of the line.

The formula to compute the slope is:

$\text{Slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Consider the ordered pairs: (4, 28) and (10, 70)

Compute the slope of the line as follows:

$\text{Slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$=\frac{70-28}{10-4}\\\\=\frac{42}{6}\\\\=7$

Thus, the rate of change in the price of a bushel of corn in the current year is $7.

Part B:

The data for the price of bushels in the previous year is as follows:

Number of Bushels Price

2 10

4 20

6 30

8 40

Compute the rate of change in the price of a bushel of corn in the previous year based upon the number of bushels as follows:

Consider the ordered pairs: (2, 10) and (6, 30)

$\text{Slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$=\frac{30-10}{6-2}\\\\=\frac{20}{4}\\\\=5$

The rate of change in the price of a bushel of corn in the previous year is $5.

Thus, the price of a bushel of corn in the current year is $2 more than the price of a bushel of corn in the previous year.

5 0

3 years ago

Convert to scientific notation

andreev551 [17]

Answer:

8.245*10^6

Step-by-step explanation:

scientific notation is known as c*10^n

c : any number from 1-10

n : the power of base 10

the number in our case between 1 and 10 is 8.245

but then we want to move the decimal 6 spaces to the right, so we multiply that by 10^6

8 0

3 years ago

You roll a 6-sided die.

Lerok [7]

Answer:

$\huge\boxed{\text{P(not factor of 35) = } \frac{2}{3}}$

Step-by-step explanation:

We can use basic probability to find the probability that this roll is not a factor of 35.

First off, we know that with a six sided die there are 6 possible things we can roll.

1, 2, 3, 4, 5, or 6

Now, what are the factors of 35? The factors of 35 will be any whole number that can be multiplied by another whole number to get 35.

We know $1 \cdot 35 = 35$ , so two factors are 1 and 35.
We know $5 \cdot 7 = 35$ , so two factors are 5 and 7.

Therefore, the factors of 35 are 1, 5, 7, 35.

Both 5 and 7 are inside the range of 1-6. So the probability of rolling a side that's a factor of 35 will be $\frac{2}{6} = \frac{1}{3}$ since there are two factors and 6 possible options.

This means, logically, there is a $\frac{4}{6} = \frac{2}{3}$ chance of not rolling a factor of 35.

Hope this helped!

5 0

2 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!

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

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

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.

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

What is the measure of angle TRV? 20° 50° 60° 130°

Svetllana [295]

Answer: 130

Step-by-step explanation:

8 0

3 years ago

1. I mark as brainliest​

1. I mark as brainliest