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3 years ago

Multiply please and thank u

Mathematics

1 answer:

WARRIOR [948]3 years ago

7 0

First of all, you need to distribute the -2 to the numbers in the parenthesis
so -2 (12k^2 - 3)
-24k^2 +6
back into the problem,
$\frac{k+3}{-24k^2+6}$

Hope this helps :)

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3n + 9 = 27 n = ? solve for n

Debora [2.8K]

Answer:

Step-by-step explanation:

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3 years ago

Please help Explanation will be appreciated thanks,

Genrish500 [490]

Answer: A; 6.3%

Step-by-step explanation:

Problem 1

For the first problem, we first want to find y so that we can plug it into the expression.

We can use elimination method for the system of equations to solve.

3x+3y=21

3x-y=5

We subtract both equations to eliminate x.

4y=16 [divide both sides by 4]

y=4

Now that we know y, we can plug it into the expression.

$\frac{4}{2} -3$ [divide]

$2-3$ [subtract]

$-1$

We know that the answer is A.

--------------------------------------------------------------------------------------------------------

Problem 2

For the second problem, we need to know how to calculate percent error. The formula for precent error is $\frac{|measured-exact|}{exact} *100%$ . We know that the exact value is 80 because the buyer was supposed to given 80. 75 is the measured value because that was what the buyer was given.

$\frac{|75-80|}{80} *100%$ [subtract]

$\frac{|-5|}{80} *100%$ [solve absolue value]

$\frac{5}{80} *100%$ [divide]

$0.0625*100%$ [multiply]

$6.25$

Since the problem said to round to one decimal place, we know that the answer is 6.3%.

4 0

2 years ago

A car dealership sells 0, 1, or 2 luxury cars on any day. When selling a car, the dealer also tries to persuade the customer to

melisa1 [442]

Answer:

Mean = 1.42

Variance = 0.58

Step-by-step explanation:

Given: X denote the number of luxury cars sold in a given day, and Y denote the number of extended warranties sold.

Also, joint probability function of X and Y are given.

To find:

mean and variance of X

Solution:

From the given joint probability function of X and Y,

$P(X=0)=\frac{1}{6}\\P(X=1)=\frac{1}{12}+\frac{1}{6}=\frac{1+2}{12}=\frac{3}{12}\\P(X=2)=\frac{1}{12}+\frac{1}{3}+\frac{1}{6}=\frac{1+4+2}{12}=\frac{7}{12}$

Mean of X:

$E(X)=\sum XP(X)\\=0\left ( \frac{1}{6} \right )+1\left ( \frac{3}{12} \right )+2\left ( \frac{7}{12} \right )\\=0+\frac{3}{12}+\frac{14}{12}\\=\frac{17}{12}=1.42$

Variance of X:

$E(X^2)=\sum X^2P(X)\\=0^2\left ( \frac{1}{6} \right )+1^2\left ( \frac{3}{12} \right )+2^2\left ( \frac{7}{12} \right )\\=0+\frac{3}{12}+\frac{28}{12}\\=\frac{31}{12}$

$var(X)=E\left [ X^2 \right ]-\left ( E\left [ X \right ] \right )^2\\=\frac{31}{12}-\left ( \frac{17}{12} \right )^2\\=\frac{31}{12}-\frac{289}{144}\\=\frac{372-289}{144}\\=\frac{83}{144}\\=0.58$

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3 years ago

(2 + 3i) + (3 – 4i) (3 – 5i) – (–2 – i) (2 − 4i)(1+ 3i) 2i(−1+ 3i)

grandymaker [24]

Answer:

See below ~

Step-by-step explanation:

Question 1

(2 + 3i) + (3 - 4i)
5 - i

Question 2

(3 – 5i) – (–2 – i)
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5 - 4i

Question 3

(2 − 4i)(1+ 3i)
2 - 4i + 6i - 12i²
14 + 2i

Question 4

2i(-1 + 3i)
-2i + 6i²
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1 year ago

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Wittaler [7]

Answer:

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3 years ago