Rewrite without parenthesis and simplify. Box method (w-6)^2<br><br> WILL MARK BRAINLEST

Let the function f be continuous and differentiable for all x. Suppose you are given that f(−1)=−3, and that f'(x) for all value

Monica [59]

Answer:

Step-by-step explanation:

By MVT,

$f'(c) = \frac{f(5)-f(-1)}{5-1}$

$4f'(c)= f(5) + 3\rightarrow f(5) = 4f'(c) - 3$

Moreover, $f'(x)\leq 4~~~~~~ \forall x,$

$f(5) \leq 16 - 3 =13$

Hence the largest possible value of f(5) is 13

8 0

3 years ago

What is the solution set of the inequality?

jolli1 [7]

Answer:

In the pic

Step-by-step explanation:

If you have any questions about the way I solved it, don't hesitate to ask me in the comments below =)

5 0

3 years ago

Write the next number in this sentence.57,64,71......

Levart [38]

and 7 each time answer Is 78

8 0

3 years ago

An online news source reports that the proportion of smartphone owners who use a certain operating system is 0.216. Two simulati

kaheart [24]

Answer:

The center/ mean will almost be equal, and the variability of simulation B will be higher than the variability of simulation A.

Step-by-step explanation:

Solution

Normally, a distribution sample is mostly affected by sample size.

As a rule, sampling error decreases by half by increasing the sample size four times.

In this case, B sample is 2 times higher the A sample size.

Now, the Mean sampling error is affected and is not higher for A.

But it's sample is huge for this, Thus, they are almost equal

Variability of simulation decreases with increase in number of trials. A has less variability.

With increase number of trials, variability of simulation decreases, so A has less variability.

4 0

3 years ago

Question Help Suppose that the lifetimes of light bulbs are approximately normally distributed, with a mean of 5656 hours and a

koban [17]

Answer:

a)3.438% of the light bulbs will last more than 6262 hours.

b)11.31% of the light bulbs will last 5252 hours or less.

c) 23.655% of the light bulbs are going to last between 5858 and 6262 hours.

d) 0.12% of the light bulbs will last 4646 hours or less.

Step-by-step explanation:

Normally distributed problems can be solved by the z-score formula:

On a normaly distributed set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a value X is given by:

$Z = \frac{X - \mu}{\sigma}$

After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.

In this problem, we have that:

The lifetimes of light bulbs are approximately normally distributed, with a mean of 5656 hours and a standard deviation of 333.3 hours.

So $\mu = 5656, \sigma = 333.3$

(a) What proportion of light bulbs will last more than 6262 hours?

The pvalue of the z-score of $X = 6262$ is the proportion of light bulbs that will last less than 6262. Subtracting 100% by this value, we find the proportion of light bulbs that will last more than 6262 hours.