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3 years ago

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Four-leaf clovers are a St. Patrick’s Day tradition. My class needs to cut out 220 four-leaf clovers to help decorate our school

. We cut out 46 clovers on Monday, 44 clovers on Tuesday, 46 clovers on Wednesday, and 44 clovers on Thursday. How many clovers do we need to cut out on Friday to reach our goal?

1 answer:

Anarel [89]3 years ago

6 0

Answer:

40

Step-by-step explanation:

46+46+44+44= 180

220-180=40

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If Log 20 = 1.301 what is the value of Logsub100 20?

Charra [1.4K]

Answer:

$log_{100}20 = 0.6505$

Step-by-step explanation:

log20 = 1.301 (when you see only "log" with no base, it is taken as a base 10)

This basically means:

$10^{1.301} = 20$ (This is the log in exponential form)

Since we don't know what $log_{100}20$ is equal to, we will say $log_{100}20$ = x

So to solve for $log_{100}20$ = x, you do the same thing. (convert to exponential form)

$100^{x} =20$

Now you will notice that both of these equations are equal to 20.

Since 20 = 20,

we can say $100^{x}$ = $10^{1.301}$

Another way of saying 100, is $10^{2}$ (make the bases the same)

Now we get $10^{2x} = 10^{1.301}$

(we get 2x because an exponent to the power of an exponent ( $2^{x}$ ) is the same as 2 * x)

Because you have the same base, you can just ignore the 10s and focus on the exponents. So you get:

2x = 1.301

x = 0.6505

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Find an equation of the tangent plane to the given parametric surface at the specified point.

Neko [114]

Answer:

Equation of tangent plane to given parametric equation is:

$\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

Step-by-step explanation:

Given equation

$r(u, v)=u cos (v)\hat{i}+u sin (v)\hat{j}+v\hat{k}$ ---(1)

Normal vector tangent to plane is:

$\hat{n} = \hat{r_{u}} \times \hat{r_{v}}\\r_{u}=\frac{\partial r}{\partial u}\\r_{v}=\frac{\partial r}{\partial v}$

$\frac{\partial r}{\partial u} =cos(v)\hat{i}+sin(v)\hat{j}\\\frac{\partial r}{\partial v}=-usin(v)\hat{i}+u cos(v)\hat{j}+\hat{k}$

Normal vector tangent to plane is given by:

$r_{u} \times r_{v} =det\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\cos(v)&sin(v)&0\\-usin(v)&ucos(v)&1\end{array}\right]$

Expanding with first row

$\hat{n} = \hat{i} \begin{vmatrix} sin(v)&0\\ucos(v) &1\end{vmatrix}- \hat{j} \begin{vmatrix} cos(v)&0\\-usin(v) &1\end{vmatrix}+\hat{k} \begin{vmatrix} cos(v)&sin(v)\\-usin(v) &ucos(v)\end{vmatrix}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u(cos^{2}v+sin^{2}v)\hat{k}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u\hat{k}\\$

at u=5, v =π/3

$=\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k}$ ---(2)

at u=5, v =π/3 (1) becomes,

$r(5, \frac{\pi}{3})=5 cos (\frac{\pi}{3})\hat{i}+5sin (\frac{\pi}{3})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=5(\frac{1}{2})\hat{i}+5 (\frac{\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=\frac{5}{2}\hat{i}+(\frac{5\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

From above eq coordinates of r₀ can be found as:

$r_{o}=(\frac{5}{2},\frac{5\sqrt{3}}{2},\frac{\pi}{3})$

From (2) coordinates of normal vector can be found as

$n=(\frac{\sqrt{3} }{2},-\frac{1}{2},1)$

Equation of tangent line can be found as:

$(\hat{r}-\hat{r_{o}}).\hat{n}=0\\((x-\frac{5}{2})\hat{i}+(y-\frac{5\sqrt{3}}{2})\hat{j}+(z-\frac{\pi}{3})\hat{k})(\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k})=0\\\frac{\sqrt{3}}{2}x-\frac{5\sqrt{3}}{4}-\frac{1}{2}y+\frac{5\sqrt{3}}{4}+z-\frac{\pi}{3}=0\\\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

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3 years ago

Find angle measures and arc measures

Vlad1618 [11]

Answer:

Step-by-step explanation:

a. We know that KH is a diameter of the circle and that along angles along it on either side adds up to 180° as it's a straight line.

-Therefore, we have that:

$m\angle a+ \angle GaH=180\textdegree\\\\\\m\angle a=180-35\\\\=145\textdegree$

Hence, the angle a is equal to 180°

b.We know that GJ is a diameter of the circle and that along angles along it on either side adds up to 180° as it's a straight line.

-Therefore, we have that:

$m\angle b+\angle GaK=180\textdegree\\\\m\angle b=180\textdegree -145\textdegree\\\\\\\\=35\textdegree$

Hence, the angle b is equal to 35°

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3 years ago

There are 48 students in an elementary statistics class. On the basis of years of experience, the instructor knows that the time

Stella [2.4K]

Answer:

a) 64.06% probability that he is through grading before the 11:00 P.M. TV news begins.

b) The hardness distribution is not given. But you would have to find s when n = 39, then the probability would be 1 subtracted by the pvalue of Z when X = 51.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For the sum of n trials, the mean is $\mu*n$ and the standard deviation is $s = \sigma\sqrt{n}$

In this question:

$n = 48, \mu = 48*5 = 240, s = 4\sqrt{48} = 27.71$

These values are in minutes.

(a) If grading times are independent and the instructor begins grading at 6:50 P.M. and grades continuously, what is the (approximate) probability that he is through grading before the 11:00 P.M. TV news begins?

From 6:50 PM to 11 PM there are 4 hours and 10 minutes, so 4*60 + 10 = 250 minutes. This probability is the pvalue of Z when X = 250. So

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

By the Central Limit Theorem

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

$Z = \frac{250 - 240}{27.71}$

$Z = 0.36$

$Z = 0.36$ has a pvalue of 0.6406

64.06% probability that he is through grading before the 11:00 P.M. TV news begins.

(b) What is the (approximate) probability that the sample mean hardness for a random sample of 39 pins is at least 51?

The hardness distribution is not given. But you would have to find s when n = 39(using the standard deviation of the population divided by the square root of 39, since it is not a sum here), then the probability would be 1 subtracted by the pvalue of Z when X = 51.

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