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3 years ago

83/63 as a mixed number?

2 answers:

4 0

Since 63 goes into 83 once, with a remainder of 20, write the mixed fraction like so:

1 20/63

Also, because you cannot simplify this answer anymore, this is the final answer.

Hope this helps!

Fofino [41]3 years ago

4 0

83/63 as a mixed number is 1 20/63

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Calculus 3 chapter 16

o-na [289]

Evaluate $\vec F$ at $\vec r$ :

$\vec F(x,y,z) = x\,\vec\imath + y\,\vec\jmath + xy\,\vec k \\\\ \implies \vec F(\vec r(t)) = \vec F(\cos(t), \sin(t), t) = \cos(t)\,\vec\imath + \sin(t)\,\vec\jmath + \sin(t)\cos(t)\,\vec k$

Compute the line element $d\vec r$ :

$d\vec r = \dfrac{d\vec r}{dt} dt = \left(-\sin(t)\,\vec\imath+\cos(t)\,\vec\jmath+\vec k\bigg) \, dt$

Simplifying the integrand, we have

$\vec F\cdot d\vec r = \bigg(-\cos(t)\sin(t) + \sin(t)\cos(t) + \sin(t)\cos(t)\bigg) \, dt \\ ~~~~~~~~= \sin(t)\cos(t) \, dt \\\\ ~~~~~~~~= \dfrac12 \sin(2t) \, dt$

Then the line integral evaluates to

$\displaystyle \int_C \vec F\cdot d\vec r = \int_0^\pi \frac12\sin(2t)\,dt \\\\ ~~~~~~~~ = -\frac14\cos(2t) \bigg|_{t=0}^{t=\pi} \\\\ ~~~~~~~~ = -\frac14(\cos(2\pi)-\cos(0)) = \boxed{0}$

3 0

2 years ago

A stack of 6 gift boxes is 12.36 inches high. What is<br> the height of 32 gift boxes?

Rom4ik [11]

Answer:

Step-by-step explanation:

Discussion

6 gift boxes are 12.36 inches high

32 gift box are x inches high

This is a proportion question. A proportion is made of 2 equal ratios. If you know 3 parts of the proportion, you can solve for the 4th,

6/32 = 12.36/x Cross multiply

6x = 12.36 * 32 Combine the right.

6x = 395.52 Divide by 6

6x/6 = 395.52/6

x = 65.92

Answer: 65.92

Note

Using a proportion saves you the trouble of finding 1 box and multiplying whatever number you get by 32

3 0

2 years ago

PLEASE HELP !!!!!!!!

Aloiza [94]

275 is the answer to the question

8 0

4 years ago

Suppose a certain study reported that 27.7% of high school students smoke.

Norma-Jean [14]

<u>Given</u>:

Probability of student smoke,

P = 27.7%

= 0.277

Number of students,

n = 632

$q = 1-p$

$=1-0.277$

$=0.723$

(i)

Here,

Number of students (n) = 60

then,

⇒ $n_P=60\times 0.277$

$=16.62$

⇒ $n_q=60\times 0.723$

$=43.38$

We can see that $n_P > 10$ and $n_q>10$ so the normal approximation condition are met.

Now,

$\mu = n_P= 16.62$

$\sigma = \sqrt{n_{Pq}}$

$= \sqrt{60\times 0.277\times 0.723}$

$=3.9664$

Now,

⇒ $P(X$

$=P(Z_{18.5})$

The Z-score is:

= $\frac{18.5-16.62}{3.4664}$

= $0.5423$

hence,

The probability will be:

⇒ $P(Z_{18.5}) = 0.7062$

or,

⇒ $P(Z$

(ii)

Here,

Number of students (n) = 75

$\mu = n_P = 75\times 0.277$

$=20.775$

$\sigma = \sqrt{n_{Pq}}$

$=\sqrt{75\times 0.277\times 0.723}$

$=3.8756$

Now,

⇒ $P(X>17) = P(X> 17.5)$

$=1-P(X \leq 17.5)$

$=1-P(Z_{17.5})$

The Z-score is:

= $\frac{17.5-20.775}{3.8756}$

= $-0.9740$

then, $P(Z_{17.5}) = 0.165$

hence,

The probability will be:

⇒ $P(X>17) = 1-0.165$

$=0.835$

Learn more about probability here:

brainly.com/question/9825651

4 0

3 years ago

Q4H: The question is in the picture below (please show work) :

Ipatiy [6.2K]

-3x-5=16
-3x-5(+5)=16+5
-3x=21
-3x/-3=21/-3
x=-7

-4(y-2)=12
-4y+8=12
-4y=4
y=-1

8.88=4.44(x-7)
8.88=4.44x-31.08
39.96=4.44x
9=x

5(y+2/5)=-13
5y+2=-13
5y=-15
y=-3

4 0

3 years ago