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3 years ago

How do I find out how much fabric she'll need??

Mathematics

2 answers:

Oksana_A [137]3 years ago

8 0

I can't see the whole thing, but first you have to muliply the height and the width of the front face. Then you have to mutiply that by 4. After that, you have to multiply the width and height of the side face. Then multiply that by 2.

Fantom [35]3 years ago

5 0

Read the answer above me, its better.

You might be interested in

Please help me find the area of shaded region and step by step

Bumek [7]

Answer:

Part 1) $A=60\ ft^2$

Part 2) $A=80\ cm^2$

Part 3) $A=96\ m^2$

Part 4) $A=144\ cm^2$

Part 5) $A=9\ m^2$

Part 6) $A=(49\pi -33)\ in^2$

Step-by-step explanation:

Part 1) we know that

The shaded region is equal to the area of the complete rectangle minus the area of the interior rectangle

The area of rectangle is equal to

$A=bh$

where

b is the base of rectangle

h is the height of rectangle

$A=(12)(7)-(8)(3)$

$A=84-24$

$A=60\ ft^2$

Part 2) we know that

The shaded region is equal to the area of the complete rectangle minus the area of the interior square

The area of square is equal to

$A=b^2$

where

b is the length side of the square

$A=(12)(8)-(4^2)$

$A=96-16$

$A=80\ cm^2$

Part 3) we know that

The area of the shaded region is equal to the area of four rectangles plus the area of one square

$A=4(4)(5)+(4^2)$

$A=80+16$

$A=96\ m^2$

Part 4) we know that

The shaded region is equal to the area of the complete square minus the area of the interior square

$A=(15^2)-(9^2)$

$A=225-81$

$A=144\ cm^2$

Part 5) we know that

The area of the shaded region is equal to the area of triangle minus the area of rectangle

The area of triangle is equal to

$A=\frac{1}{2}(b)(h)$

where

b is the base of triangle

h is the height of triangle

$A=\frac{1}{2}(6)(7)-(6)(2)$

$A=21-12$

$A=9\ m^2$

Part 6) we know that

The area of the shaded region is equal to the area of the circle minus the area of rectangle

The area of the circle is equal to

$A=\pi r^{2}$

where

r is the radius of the circle

$A=\pi (7^2)-(3)(11)$

$A=(49\pi -33)\ in^2$

7 0

3 years ago

If x*y =<br>1, then 7 * 28 =<br>

gladu [14]

196 I think hope this helps

8 0

3 years ago

Read 2 more answers

Suppose that one person in 10,000 people has a rare genetic disease. There is an excellent test for the disease; 98.8% of the pe

nirvana33 [79]

Answer:

A)The probability that someone who tests positive has the disease is 0.9995

B)The probability that someone who tests negative does not have the disease is 0.99999

Step-by-step explanation:

Let D be the event that a person has a disease

Let $D^c$ be the event that a person don't have a disease

Let A be the event that a person is tested positive for that disease.

P(D|A) = Probability that someone has a disease given that he tests positive.

We are given that There is an excellent test for the disease; 98.8% of the people with the disease test positive

So, P(A|D)=probability that a person is tested positive given he has a disease = 0.988

We are also given that one person in 10,000 people has a rare genetic disease.

So, $P(D)=\frac{1}{10000}$

Only 0.4% of the people who don't have it test positive.

$P(A|D^c)$ = probability that a person is tested positive given he don't have a disease = 0.004

$P(D^c)=1-\frac{1}{10000}$

Formula: $P(D|A)=\frac{P(A|D)P(D)}{P(A|D)P(D^c)+P(A|D^c)P(D^c)}$

$P(D|A)=\frac{0.988 \times \frac{1}{10000}}{0.988 \times (1-\frac{1}{10000}))+0.004 \times (1-\frac{1}{10000})}$

P(D|A)= $\frac{2470}{2471}$ =0.9995

P(D|A)= $0.9995$

A)The probability that someone who tests positive has the disease is 0.9995

(B)

$P(D^c|A^c)$ =probability that someone does not have disease given that he tests negative

$P(A^c|D^c)$ =probability that a person tests negative given that he does not have disease =1-0.004

=0.996

$P(A^c|D)$ =probability that a person tests negative given that he has a disease =1-0.988=0.012

Formula: $P(D^c|A^c)=\frac{P(A^c|D^c)P(D^c)}{P(A^c|D^c)P(D^c)+P(A^c|D)P(D)}$

$P(D^c|A^c)=\frac{0.996 \times (1-\frac{1}{10000})}{0.996 \times (1-\frac{1}{10000})+0.012 \times \frac{1}{1000}}$

$P(D^c|A^c)=0.99999$

B)The probability that someone who tests negative does not have the disease is 0.99999

8 0

3 years ago

) Set up a double integral for calculating the flux of F=3xi+yj+zk through the part of the surface z=−5x−2y+2 above the triangle

Fynjy0 [20]

The surface (call it $S$ ) is a triangle with vertices at the points

$x=0,y=0\implies z=2\implies(0,0,2)$

$x=0,y=2\implies z=-2\implies(0,2,-2)$

$x=2,y=0\implies z=-8\implies(2,0,-8)$

Parameterize $S$ by

$\vec s(u,v)=(1-v)(2,0,-8)+v\bigg((1-u)(0,2,-2)+u(0,0,2)\bigg)=(2-2v,2v-2uv,-8+6v+4uv)$

with $0\le u\le1$ and $0\le v\le1$ . Take the normal vector to $S$ to be

$\vec s_v\times\vec s_u=(20v,8v,4v)$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^1\int_0^1(6-6v,2v-2uv,-8+6v+4uv)\cdot(20v,8v,4v)\,\mathrm du\,\mathrm dv$

$=\displaystyle8\int_0^1\int_0^1(11v-10v^2)\,\mathrm du\,\mathrm dv=\boxed{\frac{52}3}$

8 0

4 years ago

A distribution with µ = 55 and σ = 6 is being standardized so that the new mean and standard deviation will be µ = 50 and σ = 10

Alekssandra [29.7K]

Solution:

$\hbox{Let the distribution standarized is z}\\ \hbox{given:-}\mu =50\hbox{ and} \sigma =6\hbox{will become } \mu=50 \hbox{ and} \sigma =10\\\hbox{distribution standarized with value x=52 }\\\rm{so}\\\hbox{the value of score is:}\\z=\frac{x-\mu }{\sigma}\\\rm{if}\\p(x< 52)=p(z$

3 0

3 years ago