I can't figure out how to make the equation.

Solve for 30 points and brainliest<br> 7³(8)÷3(2)-1800.3 repeating

Arturiano [62]

Answer:

29.033333

Step-by-step explanation:

6 0

3 years ago

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g The completion times for a job task range from 11.8 minutes to 19.4 minutes and are thought to be uniformly distributed. What

nevsk [136]

Answer: 0.5263

Step-by-step explanation:

Given : The completion times for a job task range from 11.8 minutes to 19.4 minutes .

The density function for uniform distribution function for interval [a,b] :-

$f(x)=\dfrac{1}{b-a}$

Then the density function for the given situation:-

$f(x)=\dfrac{1}{19.4-11.8}=\dfrac{1}{7.6}$

The required interval : $16.8-12.8=4$

Now, the probability that it will require between 12.8 and 16.8 minutes to perform the task will be :-

$\dfrac{4}{7.6}=0.526315789474\approx0.5263$

5 0

3 years ago

Why would a person who only needs a car for the daily commute most likely prefer leasing a car over buying?

prohojiy [21]

Answer:

anyone would prefer leasing a car just because it has some benefits such as; Lower monthly payment than buying a car, much smaller down payment is required and one get to drive a new car every few years.

Step-by-step explanation:

When leasing a car, you are paying to use the car during its first few years.

-Paying for the depreciation of the vehicle while you use it

-Price the car sells for new less the price the car will sell for at the end of the lease=depreciation

So anyone would prefer leasing a car just because it has some benefits such as; Lower monthly payment than buying a car, much smaller down payment is required and one get to drive a new car every few years.

6 0

3 years ago

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Which expression is equivalent to 2x + x2 + 3x + 5x2 - 10x?

masha68 [24]

Add similar terms:
(2x + 3x - 10x) + (5x² + x<span>²) </span>= -5x + 6x<span>² (ANSWER: C)</span>

6 0

4 years ago

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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