Nabuko buys 14 dog biscuits for $1.68 if nabuko pays $4.20 for dog biscuits how many did she buy

Let T be the plane-2x-2y+z =-13. Find the shortest distance d from the point Po=(-5,-5,-3) to T, and the point Q in T that is cl

GaryK [48]

Answer:

d=10u

Q(5/3,5/3,-19/3)

Step-by-step explanation:

The shortest distance between the plane and Po is also the distance between Po and Q. To find that distance and the point Q you need the perpendicular line x to the plane that intersects Po, this line will have the direction of the normal of the plane $n=(-2,-2,1)$ , then r will have the next parametric equations:

$x=-5-2\lambda\\y=-5-2\lambda\\z=-3+\lambda$

To find Q, the intersection between r and the plane T, substitute the parametric equations of r in T

$-2x-2y+z =-13\\-2(-5-2\lambda)-2(-5-2\lambda)+(-3+\lambda) =-13\\10+4\lambda+10+4\lambda-3+\lambda=-13\\9\lambda+17=-13\\9\lambda=-13-17\\\lambda=-30/9=-10/3$

Substitute the value of $\lambda$ in the parametric equations:

$x=-5-2(-10/3)=-5+20/3=5/3\\y=-5-2(-10/3)=5/3\\z=-3+(-10/3)=-19/3\\$

Those values are the coordinates of Q

Q(5/3,5/3,-19/3)

The distance from Po to the plane

$d=\left| {\to} \atop {PoQ}} \right|=\sqrt{(\frac{5}{3}-(-5))^2+(\frac{5}{3}-(-5))^2+(\frac{-19}{3}-(-3))^2} \\d=\sqrt{(\frac{5}{3}+5))^2+(\frac{5}{3}+5)^2+(\frac{-19}{3}+3)^2} \\d=\sqrt{(\frac{20}{3})^2+(\frac{20}{3})^2+(\frac{-10}{3})^2}\\d=\sqrt{\frac{400}{9}+\frac{400}{9}+\frac{100}{9}}\\d=\sqrt{\frac{900}{9}}=\sqrt{100}\\d=10u$

7 0

3 years ago

A group of 22 students participated in a race. Their scores are below: Score (points) 4–5 6–7 8–9 10–11 12–13 Number of Students

Anna71 [15]

A histogram would best represent the data.
Dot plot is a good way to show a certain trend. However, when it comes to the number of a item, in this case the number of students in a certain score range, a histogram will be the most intuitive and concise. Thus, a histogram best represent the data presented here.

6 0

3 years ago

Equation that has solutions x = -2 and x = 8

VLD [36.1K]

If you were to solve this equation you would get answers of x=-2 and x=-8

3 0

4 years ago

Find the missing number 1:2 = 3:_

Angelina_Jolie [31]

The answer is 6, hope this helped!

4 0

3 years ago

Read 2 more answers

A food company sells salmon to various customers. The mean weight of the salmon is 37 lb with a standard deviation of 2 lbs. The

brilliants [131]

Answer:

The standard deviation for the mean weigth of Salmon is 2/3 lbs for restaurants, 2/7 lbs for grocery stores and 1/4 lbs for discount order stores.

Step-by-step explanation:

The mean sample of the sum of n random variables is

$\overline{X} = \frac{X_1+X_2+...+X_n}{n}$

If $X_1, ..., X_n$ are indentically distributed and independent, like in the situation of the problem, then the variance of $X_1 + .... + X_n$ will be the sum of the variances, in other words, it will be n times the variance of $X_1$ .

However if we multiply this mean by 1/n (in other words, divide by n), then we have to divide the variance by 1/n², thus $\overkine{X} = \frac{V(X_1)}{n}$ and as a result, the standard deviation of $\overline{X}$ is the standard deviation of $X_1$ divided by $\sqrt{n}$ .

Since the standard deviation of the weigth of a Salmon is 2 lbs, then the standard deviations for the mean weigth will be:

Restaurants: We have boxes with 9 salmon each, so it will be $\frac{2}{\sqrt{9}} = \frac{2}{3}$
Grocery stores: Each carton has 49 salmon, thus the standard deviation is $\frac{2}{\sqrt{49}} = \frac{2}{7}$
Discount outlet stores: Each pallet has 64 salmon, as a result, the standard deviation is $\frac{2}{\sqrt{64}} = \frac{1}{4}$

We conclude that de standard deivation of the mean weigth of salmon of the types of shipment given is: 2/3 lbs for restaurants, 2/7 lbs for grocery stores and 1/4 lbs for discount outlet stores.

7 0

3 years ago