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3 years ago

(-6-4) divided by (-5)

2 answers:

6 0

The answer is 2. first you would simplify -6 and -4 which would equal -10. once you divide it by -5 you’ll get positive 2

fenix001 [56]3 years ago

3 0

Answer:

Step-by-step explanation:

Simplify the following:

(-6 - 4)/(-5)

Hint: | Group the negative terms in -6 - 4 together and factor out the minus sign.

-6 - 4 = -(6 + 4):

(-(6 + 4))/(-5)

Hint: | Evaluate 6 + 4.

6 + 4 = 10:

(-10)/(-5)

Hint: | In (-10)/(-5), the numbers 10 in the numerator and -5 in the denominator have gcd greater than one.

The gcd of 10 and -5 is 5, so (-10)/(-5) = (-(5×2))/(5 (-1)) = 5/5×(-2)/(-1) = (-2)/(-1):

(-2)/(-1)

Hint: | Cancel common terms in the numerator and denominator of (-2)/(-1).

(-2)/(-1) = (-1)/(-1)×2 = 2:

Answer: 2

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Two percent of the customers of a store buy cigars. Half of the customers who buy cigars buy beer. 25 percent who buy beer buy c

enyata [817]

Answer:

0.95

Step-by-step explanation:

The computation of the probability that a customer neither buys beer nor buys cigars is given below;

Given that, the probabilities are

The customers who purchased cigars be 0.02

The customers who purchased cigars + beer 0.50

And, the customers who purchased beer + cigars be 0.25

Now the probabilities where the customer purchased both

= 0.05 × 0.02

= 0.10

The probability where the customer purchased beer is

= 0.01 ÷ 0.25

= 0.04

Now the probability where a customer neither buys beer nor buys cigars is

= 1 - 0.02 + 0.04 - 0.01

= 0.95

4 0

3 years ago

HELP! <br> Plot at least three points using the equation:<br> y = 1/2 x - 2

frutty [35]

Answer:

First point - (0,-2)

second - (2,-1)

Third - ( 4, 0)

Step-by-step explanation:

5 0

3 years ago

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:

kondaur [170]

$\text{Proof by induction:}$
$\text{Test that the statement holds or n = 1}$

$LHS = (3 - 2)^{2} = 1$
$RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS$
$\text{Thus, the statement holds for the base case.}$

$\text{Assume the statement holds for some arbitrary term, n= k}$
$1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}$

$\text{Prove it is true for n = k + 1}$
$RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$

$LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}$
$= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}$
$= \frac{k(6k^{2} + 15k + 11) + 2}{}$
$= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$
$= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}$
$= RHS$

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.

3 0

3 years ago

*NEED HELP QUICKLY* son needs a math assignment done within the hour so he can submit it and get a decent grade. i tried helping

ahrayia [7]

time would be the independent variable as it doesnt depend on the distance. distance is dependant as you need a certain amount of time to get somewhere..