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3 years ago

10

The quality control manager at a battery factory picks three batteries at random each day from the production line.

1 answer:

azamat3 years ago

6 0

Answer : 0.001

Step - by - step explanation :

If 90% of the batteries are produced perfectly which equals to . . . 90 / 100. That therefore means that 10 / 100 of the batteries are imperfect.

So it's 0.001 !

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Please expand equation <br> 0.5(5x - 2.2)

Andrews [41]

0.44............is your answer

5 0

3 years ago

Read 2 more answers

Eli invested $ 330 $330 in an account in the year 1999, and the value has been growing exponentially at a constant rate. The val

Cloud [144]

Answer:

The value of the account in the year 2009 will be $682.

Step-by-step explanation:

The acount's balance, in t years after 1999, can be modeled by the following equation.

$A(t) = Pe^{rt}$

In which A(t) is the amount after t years, P is the initial money deposited, and r is the rate of interest.

$330 in an account in the year 1999

This means that $P = 330$

$590 in the year 2007

2007 is 8 years after 1999, so P(8) = 590.

We use this to find r.

$A(t) = Pe^{rt}$

$590 = 330e^{8r}$

$e^{8r} = \frac{590}{330}$

$e^{8r} = 1.79$

Applying ln to both sides:

$\ln{e^{8r}} = \ln{1.79}$

$8r = \ln{1.79}$

$r = \frac{\ln{1.79}}{8}$

$r = 0.0726$

Determine the value of the account, to the nearest dollar, in the year 2009.

2009 is 10 years after 1999, so this is A(10).

$A(t) = 330e^{0.0726t}$

$A(10) = 330e^{0.0726*10} = 682$

The value of the account in the year 2009 will be $682.

4 0

3 years ago

Algebra 2 - Grade 11

saveliy_v [14]

Answer:

for the first equation

f(-3) = 34

f(4) = 6

for the 2nd equation

f(-3) = -56

f(4) ÷ 70

Step-by-step explanation:

my work is attached in a picture.

all you do is substitute each x value into each equation

8 0

2 years ago

What is 0.02% of 40 million pounds?

DanielleElmas [232]

0.02% of 40 million pounds is 8000 (pounds)

5 0

2 years ago

Find all the zeroes of the equation(with simple steps).

uysha [10]

<u>Answer-</u>

<em>The zeros are, $5,\ -5,\ 4i,\ -4i$ </em>

<u>Solution-</u>

$\Rightarrow -3x^4+27x^2+1200=0$

$\Rightarrow -3(x^2)^2+27(x^2)+1200=0$

Here,

a = -3, b = 27, c = 1200

So,

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

$=\dfrac{-27\pm \sqrt{-27^2-4\cdot (-3)\cdot 1200}}{2\cdot (-3)}$

$=\dfrac{-27\pm \sqrt{729+14400}}{-6}$

$=\dfrac{-27\pm 123}{-6}$

$=\dfrac{-27+123}{-6},\ \dfrac{-27- 123}{-6}$

$=\dfrac{96}{-6},\ \dfrac{-150}{-6}$

$=-16,\ 25$

So,

$\Rightarrow x^2=25,\ -16$

$\Rightarrow x=\sqrt{25},\ \sqrt{-16}$

$\Rightarrow x=5,\ -5,\ 4i,\ -4i$

8 0

3 years ago