I do not understand the 5 question.

Choose the correct simplification of (2x3 + x2 − 4x) − (9x3 − 3x2).

Mrrafil [7]

You start off by multiplying the second bracket by -1 to get rid of the bracket.
So,
(2x3 + x2 - 4x) - 9x3 + 3x2 =
2x3 +x2 - 4x - 9x3 + 3x2 =
(2x3 - 9x3) + (x2 + 3x2) - 4x =
-7x3 + 4x2 - 4x

7 0

3 years ago

Read 2 more answers

Factorise <br> a) 4p²-81<br> b) x²-13x+36

taurus [48]

Answer:

A = (2p + 9) (2p - 9)

B = (x - 9) (x - 4)

Step-by-step explanation:

For A : Rewrite 4p^2 as (2p)^2.

(2p)^2−81

Rewrite 81 as 9^2.

(2p)^2−9^2

Since both terms are perfect squares, factor using the difference of squares formula, a^2 − b^2 = ( a + b ) ( a − b ) where a = 2p and b = 9 .

(2p + 9) (2p − 9)

For B : Consider the form x^2 + bx + c . Find a pair of integers whose product is c and whose sum is b . In this case, whose product is 36 and whose sum is − 13 .

-9, -4

(x - 9) (x - 4)

I hope this helps.

7 0

3 years ago

Infections seldom start with a single bacterium. Suppose that you cut yourself on a rusty nail that puts 25

Soloha48 [4]

Answer: There would be 12,000 bacteria 8 hours after the initial infection.

Step-by-step explanation: You start with 25 cells and if they divide every 15 minutes, 25 multiplied by 15 gets you 375 cells every quarter of an hour. Then if you multiply that by the 32 quarter hours, you get 12,000 cells. To check that you can also multiply 25 by 15, still 375, then 375 by 4 for each quarter of the hour, then that gets you 1,500 cells every hour. If you multiply that by 8 for the 8 hours they have time to divide, you still get 12,000 cells.

I hope this helps!

3 0

3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Anettt [7]

Answer:

Choice C: approximately 121 green beans will be 13 centimeters or shorter.

Step-by-step explanation:

What's the probability that a green bean from this sale is shorter than 13 centimeters?

Let the length of a green bean be $X$ centimeters.

$X$ follows a normal distribution with

mean $\mu = 11.2$ and
standard deviation $\sigma = 2.1$ .

In other words,

$X\sim \text{N}(11.2, 2.1^{2})$ ,

and the probability in question is $X \le 13$ .

Z-score table approach:

Find the z-score of this measurement:

$\displaystyle z= \frac{x-\mu}{\sigma} = \frac{13-11.2}{2.1} = 0.857143$ . Closest to 0.86.

Look up the z-score in a table. Keep in mind that entries on a typical z-score table gives the probability of the left tail, which is the chance that $Z$ will be less than or equal to the z-score in question. (In case the question is asking for the probability that $Z$ is greater than the z-score, subtract the value from table from 1.)

$P(X\le 13) = P(Z \le 0.857143) \approx 0.8051$ .

"Technology" Approach

Depending on the manufacturer, the steps generally include:

Locate the cumulative probability function (cdf) for normal distributions.
Enter the lower and upper bound. The lower bound shall be a very negative number such as $-10^{9}$ . For the upper bound, enter $13$
Enter the mean and standard deviation (or variance if required).
Evaluate.

For example, on a Texas Instruments TI-84, evaluating $\text{normalcdf})(-1\text{E}99,\;13,\;11.2,\;2.1 )$ gives $0.804317$ .

As a result,

$P(X\le 13) = 0.804317$ .

Number of green beans that are shorter than 13 centimeters:

Assume that the length of green beans for sale are independent of each other. The probability that each green bean is shorter than 13 centimeters is constant. As a result, the number of green beans out of 150 that are shorter than 13 centimeters follow a binomial distribution.

Number of trials $n$ : 150.
Probability of success $p$ : 0.804317.

Let $Y$ be the number of green beans out of this 150 that are shorter than 13 centimeters. $Y\sim\text{B}(150,0.804317)$ .

The expected value of a binomial random variable is the product of the number of trials and the probability of success on each trial. In other words,

$E(Y) = n\cdot p = 150 \times 0.804317 = 120.648\approx 121$

The expected number of green beans out of this 150 that are shorter than 13 centimeters will thus be approximately 121.

7 0

3 years ago

f the centripetal and thus frictional force between the tires and the roadbed of a car moving in a circular path were reduced, w

LUCKY_DIMON [66]

The frictional force between the tires and the road prevent the car from skidding off the road due to centripetal force.

If the frictional force is less than the centripetal force, the car will skid when it navigates a circular path.

The diagram below shows that when the car travels at tangential velocity, v, on a circular path with radius, r, the centripetal acceleration of v²/ r acts toward the center of the circle.

The resultant centripetal force is (mv²)/r, which should be balanced by the frictional force of μmg, where μ = coefficient of kinetic friction., and mg is the normal reaction on a car with mass, m.

This principle is applied on racing tracks, where the road is inclined away from the circle to give the car an extra restoring force to overcome the centripetal force.

8 0

3 years ago