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Alona [7]
2 years ago
5

A manufacturer tests a certain number of the headsets that are produced each day. If 280 units of that model are tested, and the

manufacturer finds 273 headsets without any defects, what is the chance of finding a defective headset? Round your answer to the nearest tenth.
There is a
% chance of finding a defective headset.
Mathematics
2 answers:
Natali5045456 [20]2 years ago
8 0

Answer: 1%

Step-by-step explanation:

goblinko [34]2 years ago
8 0

Answer:

2.5

Step-by-step explanation:

273/280 = 97.5%

2.5% are defective

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Use repeated addition to find the solution to each multiplication problem. Change any improper fractions to mixed numbers. 4x1/3
Nata [24]

Answer:

1\frac{1}{3}

Step-by-step explanation:

Given the following question:

4\times\frac{1}{3}

To find the answer simply multiply the numerators and the denominators by each other.

4\times\frac{1}{3}
4=\frac{4}{1}
\frac{4}{1} \times\frac{1}{3}
4\times1=4
1\times3=3
=\frac{4}{3}
\frac{4}{3} =4\div3=1\frac{1}{3}
1\frac{1}{3}

Hope this helps.

7 0
2 years ago
Enter the numerator and the denominator of the fraction in simplest form that is equal to 0.998¯¯¯¯¯¯¯¯.
Nookie1986 [14]

Answer:

998/1000

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499/500 simplified

6 0
3 years ago
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Step-by-step explanation:

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8 0
2 years ago
Which absolute value function defines this graph?
bixtya [17]
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6 0
3 years ago
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Solve the initial value problem y'=2 cos 2x/(3+2y),y(0)=−1 and determine where the solution attains its maximum value.
zloy xaker [14]

Answer:

y=\frac{-3\pm\sqrt{4sin2x+1}}{2}

x={\pi}{4}

Step-by-step explanation:

We are given that

y'=\frac{2cos2x}{3+2y}

y(0)=-1

\frac{dy}{dx}=\frac{2cos2x}{3+2y}

(3+2y)dy=2cos2x dx

Taking integration on both sides then we get

\int (3+2y)dy=2\int cos 2xdx

3y+y^2=sin2x+C

Using formula

\int x^n=\frac{x^{n+1}}{n+1}+C

\int cosx dx=sinx

Substitute x=0 and y=-1

-3+1=sin0+C

-2=C

sin0=0

Substitute the value of C

y^2+3y=sin2x-2

y^2+3y-sin 2x+2=0

y=\frac{-3\pm\sqrt{(3)^2-4(1)(-sin2x+2)}}{2}

By using quadratic formula

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

y=\frac{-3\pm\sqrt{9+4sin2x-8}}{2}=\frac{-3\pm\sqrt{4sin2x+1}}{2}

Hence, the solution y=\frac{-3\pm\sqrt{4sin2x+1}}{2}

When the solution is maximum then y'=0

\frac{2cos2x}{3+2y}=0

2cos2x=0

cos2x=0

cos2x=cos\frac{\pi}{2}

cos\frac{\pi}{2}=0

2x=\frac{\pi}{2}

x=\frac{\pi}{4}

3 0
3 years ago
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