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2 years ago

11

A factory produces 90 pairs of designer jeans. An inspector randomly chooses 6 pairs of jeans and discovers that 1 of the pairs

of jeans is defective.

2 answers:

natita [175]2 years ago

7 0

For the first one the answer is 1 in 6 jeans or 16.67%

ExtremeBDS [4]2 years ago

6 0

Answer:

For The Second One

The Answer Is 15

Step-by-step explanation:

You might be interested in

while shopping at a clearence sale samantha finds a $60 dress on sale for 25% off. samantha also finds a 50% off coupon.

tatuchka [14]

She would pay 15 dollars for the dress because it would be 75 percent off.

6 0

3 years ago

Write 2/3 as a percent. Round your answer to the nearest tenth of a percent.

sweet [91]

divide 2 b 3

2 divided by 3 = 0.625

move decimal point 2 places to the right to turn to a percentage

0.625 = 62.5%

8 0

2 years ago

The minute hand on a clock is 9 centimeters long and travels through an arc of 252 degrees every 42 minutes. To the nearest tent

kkurt [141]

Answer:

39.6 cm

Step-by-step explanation:

Applying

s = 2πrθ/360................ Equation 1

Where s = length of an arc or distance traveled by the minutes hand of the clock during the 42 munites, r = length of the minutes hand of the clock, θ = Angle traveled by the minute hand of the clock for every 42 minutes

From the question,

Given: r = 9 cm, θ = 252°

Constant: π = 22/7 = 3.14

Substitute these values into equation 1

s = (2×3.14×9×252)/360

s = 39.564

s = 39.6 cm

4 0

3 years ago

Add or subtract each expression.Simplify your final answer show all work.

Elan Coil [88]

1) 2x-1 or x=1/2
2) 5a-16 or a=16/5

8 0

3 years ago

Consider the following region R and the vector field F. a. Compute the two-dimensional curl of the vector field. b. Evaluate bo

Shalnov [3]

Looks like we're given

$\vec F(x,y)=\langle-x,-y\rangle$

which in three dimensions could be expressed as

$\vec F(x,y)=\langle-x,-y,0\rangle$

and this has curl

$\mathrm{curl}\vec F=\langle0_y-(-y)_z,-(0_x-(-x)_z),(-y)_x-(-x)_y\rangle=\langle0,0,0\rangle$

which confirms the two-dimensional curl is 0.

It also looks like the region $R$ is the disk $x^2+y^2\le5$ . Green's theorem says the integral of $\vec F$ along the boundary of $R$ is equal to the integral of the two-dimensional curl of $\vec F$ over the interior of $R$ :

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA$

which we know to be 0, since the curl itself is 0. To verify this, we can parameterize the boundary of $R$ by

$\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle$

$\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt$

with $0\le t\le2\pi$ . Then

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt$

$=\displaystyle5\int_0^{2\pi}(\sin t\cos t-\sin t\cos t)\,\mathrm dt=0$

7 0

3 years ago