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3 years ago

7

The probability that a battery you buy lasts at most 100 hours is about DONE

1 answer:

belka [17]3 years ago

6 0

Answer:

4 percent

Step-by-step explanation:

just took the test :)

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LITERAL EQUATIONS: A literal equation is one consisting of all letters (or at least mostly letters). The Latin word literalis me

Crank

Answer:

D. b = r+z

Step-by-step explanation:

Given the expression b-r = z, we are to solve for b. To do this, we will add 'r' to both sides of the equation as shown;

b-r+r = z+r

Since -r+r = 0, substitute:

b+0 = z+r

b = z+r

Hence the resulting equation when r is added to both sides of the equation is b = z+r

Hence option D is correct

6 0

3 years ago

A box contains 5 plain pencils and 5 pens. A second box contains 3 color pencils and 5 crayon. One item from each box is chosen

Dmitry_Shevchenko [17]

Hello! :)

What is the probability of choosing a pen from the first box? It is 5/10 = 1/2.

What is the probability of selecting a crayon from the second box? It is 5/8.

The probability of picking a pen from the first box and a crayon from the second box is 5/8 x 1/2 = 5/16.

Hope this helped and I hope I answered in time!

Good luck!

~ Destiny ^_^

7 0

3 years ago

Estimate the quotient for the following problem 645 divided by 69

vampirchik [111]

9.285714285714286 645=650 69=70

6 0

3 years ago

Use integration by parts to find the integrals in Exercise.<br> ∫(4x-12)e-8x dx.

stepladder [879]

Answer:

$e(2x^{2} -12x)-4x^{2}+C$

Step-by-step explanation:

We have been given an indefinite integral as $\int \left(4x-12\right)e-8x\:dx$ . We are asked to find the given integral.

Let us solve our given problem.

$\int \left(4x-12\right)e\:dx-\int 8x\:dx$

Take out constant:

$e\int \left(4x-12\right)\:dx-8\int x\:dx$

$e(\int 4x\:dx -\int 12\right\:dx)-8\int x\:dx$

$e(\frac{4x^{1+1}}{2} -12x)-8*\frac{x^{1+1}}{1+1}+C$

$e(\frac{4x^{2}}{2} -12x)-8*\frac{x^{2}}{2}+C$

$e(2x^{2} -12x)-4x^{2}+C$

Therefore, our required integral would be $e(2x^{2} -12x)-4x^{2}+C$ .

5 0

3 years ago

Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th

grigory [225]

Answer:

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

The sample space is:

$S = \{1,1,2,2,2,3\}$

And the probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{6}$

$P(1) = \frac{1}{3}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{6}$

$P(2) = \frac{1}{2}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{6}$

P(Odd Number) is then calculated as:

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{6}$

$P(Odd\ Number) = \frac{3}{6}$

$P(Odd\ Number) = \frac{1}{2}$

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

$S = \{1,1,2,2,2,3,3\}$

The probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{7}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{7}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{7}$

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{7}$

$P(Odd\ Number) = \frac{3}{7}$

Comparing P(Odd Number) before and after

$P(Odd\ Number) = \frac{1}{2}$ --- Before

$P(Odd\ Number) = \frac{3}{7}$ --- After

<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>

7 0

3 years ago