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Vedmedyk [2.9K]

3 years ago

10

What is the perimeter of Justin's garden in inches? I can't find the right answer 6 yards and 8 yards its a rectangle garden pls

help me

1 answer:

madreJ [45]3 years ago

3 0

Answer:

28 yards

Step-by-step explanation:

It's perimeter so you add (6x2)+(8x2)

The x2 is because there's 2 of each side

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Edward wants to determine which skateboarder won a race. The actual time taken by four skateboarders to skate the same distance

iogann1982 [59]

Skateboarder B. 4.24 will show as a different time since the stopwatch rounds to the nearest tenth and to round up it has to be 0.5+ and to round down you need 0.4 or below and in this case Skateboarder B. has a time of 4.24 thus it has a different time than the other times.

I Hope this helped you!

7 0

3 years ago

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use Taylor's Theorem with integral remainder and the mean-value theorem for integrals to deduce Taylor's Theorem with lagrange r

Vadim26 [7]

Answer:

As consequence of the Taylor theorem with integral remainder we have that

$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \int^a_x f^{(n+1)}(t)\frac{(x-t)^n}{n!}dt$

If we ask that $f$ has continuous $(n+1)$ th derivative we can apply the mean value theorem for integrals. Then, there exists $c$ between $a$ and $x$ such that

$\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}dt = \frac{f^{(n+1)}(c)}{n!} \int^a_x (x-t)^n d t = \frac{f^{(n+1)}(c)}{n!} \frac{(x-t)^{n+1}}{n+1}\Big|_a^x$

Hence,

$\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}d t = \frac{f^{(n+1)}(c)}{n!} \frac{(x-t)^{(n+1)}}{n+1} = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1} .$

Thus,

$\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}d t = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$

and the Taylor theorem with Lagrange remainder is

$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$ .

Step-by-step explanation:

5 0

3 years ago

The score on an exam from a certain MAT 112 class, X, is normally distributed with μ=78.1 and σ=10.8.

salantis [7]

a) $X$

b) 0.1539

c) 0.1539

d) 0.6922

Step-by-step explanation:

a)

In this problem, the score on the exam is normally distributed with the following parameters:

$\mu=78.1$ (mean)

$\sigma = 10.8$ (standard deviation)

We call X the name of the variable (the score obtained in the exam).

Therefore, the event "a student obtains a score less than 67.1) means that the variable X has a value less than 67.1. Mathematically, this means that we are asking for:

$X$

And the probability for this to occur can be written as:

$p(X$

b)

To find the probability of X to be less than 67.1, we have to calculate the area under the standardized normal distribution (so, with mean 0 and standard deviation 1) between $z=-\infty$ and $z=Z$ , where Z is the z-score corresponding to X = 67.1 on the s tandardized normal distribution.

The z-score corresponding to 67.1 is:

$Z=\frac{67.1-\mu}{\sigma}=\frac{67.1-78.1}{10.8}=-1.02$

Therefore, the probability that X < 67.1 is equal to the probability that z < -1.02 on the standardized normal distribution:

$p(X$

And by looking at the z-score tables, we find that this probability is:

$p(z$

And so,

$p(X$

c)

Here we want to find the probability that a randomly chosen score is greater than 89.1, so

$p(X>89.1)$

First of all, we have to calculate the z-score corresponding to this value of X, which is:

$Z=\frac{89.1-\mu}{\sigma}=\frac{89.1-78.1}{10.8}=1.02$

Then we notice that the z-score tables give only the area on the left of the values on the left of the mean (0), so we have to use the following symmetry property:

$p(z>1.02) =p(z$

Because the normal distribution is symmetric.

But from part b) we know that

$p(z$

Therefore:

$p(X>89.1)=p(z>1.02)=0.1539$

d)

Here we want to find the probability that the randomly chosen score is between 67.1 and 89.1, which can be written as

$p(67.1$

Or also as

$p(67.1$

Since the overall probability under the whole distribution must be 1.

From part b) and c) we know that:

$p(X$

$p(X>89.1)=0.1539$

Therefore, here we find immediately than:

$p(67.1$

7 0

3 years ago

Let f(x)=x2−2x−4 . What is the average rate of change for the quadratic function from x=−1 to x = 4?

goblinko [34]

Answer:

The average rate of change for f(x) from x=−1 to x = 4 is, 1

Step-by-step explanation:

Average rate A(x) of change for a function f(x) over [a, b] is given by:

$A(x) = \frac{f(b)-f(a)}{b-a}$

As per the statement:

$f(x) = x^2-2x-4$

we have to find the average rate of change from x = -1 to x = 4

At x = -1

$f(-1) = (-1)^2-2(-1)-4 = 1+2-4 = -1$

and

at x = 4

$f(4) = (4)^2-2(4)-4 = 16-8-4 = 4$

Substitute these in [1] we have;

$A(x) = \frac{f(4)-f(-1)}{4-(-1)}$

⇒ $A(x) = \frac{4-(-1)}{4+1}$

⇒ $A(x) = \frac{5}{5}$

Simplify:

A(x) = 1

Therefore, the average rate of change for f(x) from x=−1 to x = 4 is, 1

4 0

3 years ago

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What is the length of the diagonal (d) in the solid below?

Cerrena [4.2K]

The length would be 12

8 0

2 years ago

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