Please just try and help it’s due in 18 mins. I’ll mark brainliest. :)

Please help? A,b,c,d

Ganezh [65]

Answer:

i did this and it was c

Step-by-step explanation:

8 0

4 years ago

A man and a woman agree to meet at a certain location about 12:30 p.m. if the man arrives at a time uniformly distributed betwee

Svetllana [295]

We define the spaces of the man and the woman's independent arrivals.
In this case,
f(man) = 1 (45-15) = 1/30
f(woman)=1/(60-0)= 1/60
f(man)*f(woman) = 1/1800

Probability = P
P (man-woman)< 5 = 1/1800* integral of (y ) with limits of (x+5) and (x-5) from 45 to 15.
P (man-woman) <5 = 1/6

8 0

3 years ago

Please answer soon (45 Points)

Nataliya [291]

Answer:

Step-by-step explanation:

The answer is the first choice.

The maximum calories per week for the dog is 4900

One week is seven days which means:

maximum calorie/day = $\frac{4900 calorie/day}{7 day}$ =700 calories/day

Thus the dog can have 700 or fewer calories per day

3 0

3 years ago

Use the definition of continuity to determine whether f is continuous at a.

dmitriy555 [2]

$f(x)$ will be continuous at $x=a=7$ if
(i) $\displaystyle\lim_{x\to7}f(x)$ exists,
(ii) $f(7)$ exists, and
(iii) $\displaystyle\lim_{x\to7}f(x)=f(7)$ .

The second condition is immediate, since $f(7)=8918$ has a finite value. The other two conditions can be established by proving that the limit of the function as $x\to7$ is indeed the value of $f(7)$ . That is, we must prove that for any $\varepsilon>0$ , we can find $\delta>0$ such that

$|x-7|$

Now,

$|f(x)-f(7)|=|5x^4-9x^3+x-8925|$

Notice that when $x=7$ , we have $5x^4-9x^3+x-8925=0$ . By the polynomial remainder theorem, we know that $x-7$ is then a factor of this polynomial. Indeed, we can write

$|5x^4-9x^3+x-8925|=|(x-7)(5x^3+26x^2+182x+1275)|=|x-7||5x^3+26x^2+182x+1275|$

This is the quantity that we do not want exceeding $\varepsilon$ . Suppose we focus our attention on small values $\delta$ . For instance, say we restrict $\delta$ to be no larger than 1, i.e. $\delta\le1$ . Under this condition, we have

$|x-7|$

Now, by the triangle inequality,

$|5x^3+26x^2+182x+1275|\le|5x^3|+|26x^2|+|182x|+|1275|=5|x|^3+26|x|^2+182|x|+1275$

If $|x|$ , then this quantity is moreover bounded such that

$|5x^3+26x^2+182x+1275|\le5\cdot8^3+26\cdot8^2+182\cdot8+1275=6955$

To recap, fixing $\delta\le1$ would force $|x|$ , which makes

$|x-7||5x^3+26x^2+182x+1275|$

and we want this quantity to be smaller than $\varepsilon$ , so

$6955|x-7|$

which suggests that we could set $\delta=\dfrac{\varepsilon}{6955}$ . But if $\varepsilon$ is given such that the above inequality fails for $\delta=\dfrac{\varepsilon}{6955}$ , then we can always fall back on $\delta=1$ , for which we know the inequality will hold. Therefore, we should ultimately choose the smaller of the two, i.e. set $\delta=\min\left\{1,\dfrac{\varepsilon}{6955}\right\}$ .

You would just need to formalize this proof to complete it, but you have all the groundwork laid out above. At any rate, you would end up proving the limit above, and ultimately establish that $f(x)$ is indeed continuous at $x=7$ .

5 0

3 years ago

Help please!!!! final exam

LiRa [457]

Answer:

The Rate is 22.5 mph

The distance it traveled was 240 mph

Step-by-step explanation:

6 0

2 years ago