All categories

3 years ago

2x^4–5x^3+x^2+3x+2=? X=5

Mathematics

1 answer:

DanielleElmas [232]3 years ago

3 0

Answer:

$2x^4–5x^3+x^2+3x+2=$

$2 \times 5^4–5 \times 5^3+5^2+3 \times 5+2$

= 2 × 625 – 5 × 125 + 25 + 3 × 5 + 2

=1250 – 625 + 25 + 15 +2

= 1292 – 625

= 667

You might be interested in

In a small town, electricity prices have been rising 2% each year since 2008. Which

mihalych1998 [28]

<h3>Answer: p = 0.12*(1.02)^t</h3>

Explanation:

The general exponential growth equation is

p = A*B^t

where t is the number of years that have gone by after 2008, A is the starting amount, B is the growth multiplier, and p is the price t years after 2008

We know that A = 0.12 is the starting price

The value of B is B = 1.02 which is in the form 1+r since 1.02 = 1 + 0.02 = 1+r

The r value is r = 0.02 and it is positive to represent growth. Keep in mind that 2% = 2/100 = 0.02

So we go from

p = A*B^t

p = 0.12*(1.02)^t

4 0

3 years ago

Michelle draws a card from a standard deck of 52 cards. She replaces the card and draws a second card. What is the probability t

erma4kov [3.2K]

To draw a heart, one would be choosing 1 card of 13 possible hearts, and 0 from the remaining 39 non-hearts. With respect to the entire deck, one would be choosing 1 card from 52 total cards. So the probability of drawing a heart is

$\dfrac{\dbinom{13}1\cdot\dbinom{39}0}{\dbinom{52}1}=\dfrac{13\cdot1}{52}=\dfrac14$

When Michelle replaces the card, the deck returns the normal, so the probability of drawing any card from a given suit is the same, $\dfrac14$ . In other words, drawing a spade is independent of having drawn the heart first.

So the probability of drawing a heart, replacing it, then drawing a spade is $\dfrac14\cdot\dfrac14=\dfrac1{16}$ .

4 0

3 years ago

Read 2 more answers

If we sample from a small finite population without replacement, the binomial distribution should not be used because the event

seropon [69]

Answer:

5/4324 = 0.001156337

Step-by-step explanation:

To better understand the hyper-geometric distribution consider the following example:

There are 100 senators in the US Congress, and suppose 60 of them are republicans so 100 - 60 = 40 are democrats).

We extract a random sample of 30 senators and we want to answer this question:

What is the probability that 10 senators in the sample are republicans (and of course, 30 - 10 = 20 democrats)?

The answer using the h-g distribution is:

$\large \frac{\binom{60}{10}\binom{100-60}{30-10}}{\binom{100}{30}}=\frac{\binom{60}{10}\binom{40}{20}}{\binom{100}{30}}$

Now, imagine there are 56 senators (56 lottery numbers), 6 are republicans (6 winning numbers and 50 losers), we extract a sample of 6 senators (the bettor selects 6 numbers). What is the probability that 4 senators are republicans? (What is the probability that 4 numbers are winners?).

<em>As we see, the situation is exactly the same,</em> but changing the numbers. So the answer would be

$\large \frac{\binom{6}{4}\binom{56-6}{6-4}}{\binom{56}{6}}=\frac{\binom{6}{4}\binom{50}{2}}{\binom{56}{6}}$

Now compute each combination separately:

$\large \binom{6}{4}=\frac{6!}{4!2!}=15\\\\\binom{50}{2}=\frac{50!}{2!48!}=1225\\\\\binom{50}{6}=\frac{50!}{6!44!}=15890700$

and now replace the values:

$\large \frac{\binom{6}{4}\binom{50}{2}}{\binom{56}{6}}=\frac{15*1225}{15890700}=\frac{18375}{15890700}=\frac{5}{4324}$

and that is it.

If the decimal expression is preferred then divide the fractions to get 0.001156337

6 0

3 years ago

please help 20 points if do

NISA [10]

Answer:

2,34x

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Find derivative problem<br> Find B’(6)

dalvyx [7]

Answer:

$B^\prime(6) \approx -28.17$

Step-by-step explanation:

We have:

$\displaystyle B(t)=24.6\sin(\frac{\pi t}{10})(8-t)$

And we want to find B’(6).

So, we will need to find B(t) first. To do so, we will take the derivative of both sides with respect to x. Hence:

$\displaystyle B^\prime(t)=\frac{d}{dt}[24.6\sin(\frac{\pi t}{10})(8-t)]$

We can move the constant outside:

$\displaystyle B^\prime(t)=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)]$

Now, we will utilize the product rule. The product rule is:

$(uv)^\prime=u^\prime v+u v^\prime$

We will let:

$\displaystyle u=\sin(\frac{\pi t}{10})\text{ and } \\ \\ v=8-t$

Then:

$\displaystyle u^\prime=\frac{\pi}{10}\cos(\frac{\pi t}{10})\text{ and } \\ \\ v^\prime= -1$

(The derivative of u was determined using the chain rule.)

Then it follows that:

$\displaystyle \begin{aligned} B^\prime(t)&=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)] \\ \\ &=24.6[(\frac{\pi}{10}\cos(\frac{\pi t}{10}))(8-t) - \sin(\frac{\pi t}{10})] \end{aligned}$

Therefore:

$\displaystyle B^\prime(6) =24.6[(\frac{\pi}{10}\cos(\frac{\pi (6)}{10}))(8-(6))- \sin(\frac{\pi (6)}{10})]$

By simplification:

$\displaystyle B^\prime(6)=24.6 [\frac{\pi}{10}\cos(\frac{3\pi}{5})(2)-\sin(\frac{3\pi}{5})] \approx -28.17$

So, the slope of the tangent line to the point (6, B(6)) is -28.17.

5 0

3 years ago