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2 years ago

Help me with this please I will give you extra points

Mathematics

1 answer:

Ierofanga [76]2 years ago

6 0

$\qquad\qquad\huge\underline{{\sf Answer}}♨$

Total outcomes in which Billy can get 5 or more : 2 <u>outcomes</u>

Total outcomes = 6

So, the probability of getting 5 or higher is ~

$\qquad \sf \dashrightarrow \: \dfrac{favorable \: outcomes}{total \: outcomes}$

$\qquad \sf \dashrightarrow \: \dfrac{2}{6}$

$\qquad \sf \dashrightarrow \: \dfrac{1}{3}$

Therefore, the correct choice is C

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Find vertex for y=x^2 - 10x + 2

NARA [144]

The vertex would be (5,-23)

8 0

3 years ago

Find the number of ways to distribute six different toys to three different children such that each child gets at least one toy.

Lisa [10]

Answer:

Step-by-step explanation:

Given that there are six different toys and they are to be distributed to three different children.

The restraint here is each child gets atleast one toy.

Let us consider the situation as this.

Since each child has to get atleast one toy no of ways to distribute

any 3 toys to the three children each. This can be done by selecting 3 toys from 6 in 6C3 ways and distributing in 3! ways

So 3 toys to each one in 6x5x4 =120 ways

Now remaining 3 toys can be given to any child.

Hence remaining 3 toys can be distributed in 3x3x3 =27 ways

Total no of ways

= 120(27)

= 3240

8 0

3 years ago

Standard deviation of 300, 785, 670, 187, 760, 724

White raven [17]

STEP 1

Mean = $\frac{300+785+670+187+760+724}{6} = \frac{3426}{6}=571$

STEP 2

Subtract the mean from each data value then square the answer
$300-571=-271$ ⇒ $(-271)^{2} =73441$
$785-571=214$ ⇒ $(214)^{2}=45796$
$670-571=99$ ⇒ $(99)^{2}=9801$
$187-571=-384$ ⇒ $(-384)^{2} =147456$
$760-571=189$ ⇒ $189^{2}=35721$
$724-571=153$ ⇒ $153^{2} =23409$

STEP 3

Add the squared answers in STEP 2 then find the mean

$\frac{73441+45796+9801+147456+35721+23409}{6}= \frac{335624}{6}=55937.3$

STEP 4

Square root the answer in STEP 3
$\sqrt{55937.3} =236.5$

Standard deviation is 236.5

7 0

3 years ago

Help fast please geometry

frozen [14]

Answer:

3√6

Explanation:

explanation is in the image above

HOPE THIS HELPS PLEASE MARK ME BRAINIEST

6 0

2 years ago

Read 2 more answers

interpret r(t) as the position of a moving object at time t. Find the curvature of the path and determine thetangential and norm

Igoryamba

Answer:

The curvature is $\kappa=1$

The tangential component of acceleration is $a_{\boldsymbol{T}}=0$

The normal component of acceleration is $a_{\boldsymbol{N}}=1 (2)^2=4$

Step-by-step explanation:

To find the curvature of the path we are going to use this formula:

$\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}$

where

$\boldsymbol{T}}$ is the unit tangent vector.

$\frac{ds}{dt}=|| \boldsymbol{r}'(t)}||$ is the speed of the object

We need to find $\boldsymbol{r}'(t)$ , we know that $\boldsymbol{r}(t)=cos \:2t \:\boldsymbol{i}+sin \:2t \:\boldsymbol{j}+ \:\boldsymbol{k}$ so

$\boldsymbol{r}'(t)=\frac{d}{dt}\left(cos\left(2t\right)\right)\:\boldsymbol{i}+\frac{d}{dt}\left(sin\left(2t\right)\right)\:\boldsymbol{j}+\frac{d}{dt}\left(1)\right\:\boldsymbol{k}\\\boldsymbol{r}'(t)=-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}$

Next , we find the magnitude of derivative of the position vector

$|| \boldsymbol{r}'(t)}||=\sqrt{(-2\sin \left(2t\right))^2+(2\cos \left(2t\right))^2} \\|| \boldsymbol{r}'(t)}||=\sqrt{2^2\sin ^2\left(2t\right)+2^2\cos ^2\left(2t\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4\left(\sin ^2\left(2t\right)+\cos ^2\left(2t\right)\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4}\sqrt{\sin ^2\left(2t\right)+\cos ^2\left(2t\right)}\\\\\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)+\sin ^2\left(x\right)=1\\\\|| \boldsymbol{r}'(t)}||=2\sqrt{1}=2$

The unit tangent vector is defined by

$\boldsymbol{T}}=\frac{\boldsymbol{r}'(t)}{||\boldsymbol{r}'(t)||}$

$\boldsymbol{T}}=\frac{-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}}{2} =\sin \left(2t\right)+\cos \left(2t\right)$

We need to find the derivative of unit tangent vector

$\boldsymbol{T}'=\frac{d}{dt}(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j}) \\\boldsymbol{T}'=-2\cdot(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j})$