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4 years ago

What is coefficient tern and constant 6xy-5xy

Mathematics

1 answer:

GREYUIT [131]4 years ago

3 0

Coefficient is 6 and -5
constant is xy

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Step-by-step explanation:

8x6=48

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4x7=28

48+28+8=84

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3 years ago

These are the cost and revenue functions for a line of trumpets sold at a music store: RX) = 76x -0.25x2 C(x) = -7.75x + 5,312.5

Ede4ka [16]

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Step-by-step explanation: I got it wrong lol

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3 years ago

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About A red and a blue die are thrown. Both dice are fair. The events A, B, and C are defined as follows: A: The sum on the two

Flauer [41]

$p(A) = \frac{1}{2}$

$p(B) = \frac{1}{6}$

$p(C) = \frac{1}{6}$

$P(A | C)=\frac{1}{2}$

Solution:

The probability of an event is given as:

$\text { probability of an event }=\frac{\text { number of favorable outcomes }}{\text { total number of outcomes }}$

In throwing one die, the total number of outcomes = 6 { 1, 2, 3, 4, 5 , 6}

First let us calculate p(A):

The event is defined as: The sum on the two dice is even

Sum on two dice is even if and only if either both dice turn up odd or both even.

The odd outcomes in thowing a single die = 3 {1, 3, 5}

The even outcomes in throwing a single die = 3 {2, 4, 6}

The probability that both turn up odd is:

$\text { probability of both die turing up odd }=\frac{3}{6} \times \frac{3}{6}=\frac{1}{4}$

Similarly, the probability that both turn up even is:

$\text { probability of both die turing up even }=\frac{3}{6} \times \frac{3}{6}=\frac{1}{4}$

probability that the sum on two dice is even = probability that both turn up odd + probability that both turn up even

$\text { probability of sum on two dice is even }=\mathrm{p}(\mathrm{A})=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$

Thus $p(A) = \frac{1}{2}$

Let us calculate p(B):

The event B is defined as: The sum on the two dice is at least 10

The total possible outcomes of two die is given as:

{(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6) }

Since each individual die can turn up any of the numbers 1, 2, 3, 4, 5, 6 the event "sum of the two dice will be at least 10" is :

atleast 10 means that sum can be 10 or greater than 10

{(4,6), (6,4), (5,5), (5,6), (6,5), (6,6)}

Here favourable outcomes = 6

Total number of outcomes = 36

Hence, the probability that the sum of the two dice will be at least 10 is:

$\text { probability that the sum of the two dice will be at least } 10=\frac{6}{36}=\frac{1}{6}$

Thus $p(B) = \frac{1}{6}$

Let us calculate p(C):

The event C is defined as: The red die comes up 5

Favourable outcomes = {(1,5),(2,5),(3,5),(4,5),(5,5),(6,5)}

$\text { probability of red die comes up } 5 \text { is the event }=\frac{6}{36}=\frac{1}{6}$

Thus $p(C) = \frac{1}{6}$

B) What is p(A l C)

$P(A | C)=\frac{p(A \cap C)}{P(C)}$

$\mathrm{A} \cap \mathrm{C}=\{(1,5),(3,5),(5,5)\}$

$p(A \cap C)=\frac{3}{36}=\frac{1}{12}$

$P(A | C)=\frac{p(A \cap C)}{P(C)}=\frac{\frac{1}{12}}{\frac{1}{6}}=\frac{1}{2}$

Thus $P(A | C)=\frac{1}{2}$

8 0

4 years ago

Will Mark Brainlest helpppp

Anvisha [2.4K]

Answer:

R = $\left[\begin{array}{ccc}-3&-2\\1&-3\\\end{array}\right]$

Step-by-step explanation:

P - Q + R = I ( I is the identity matrix )

$\left[\begin{array}{ccc}2&4\\3&5\\\end{array}\right]$ - $\left[\begin{array}{ccc}-2&2\\4&1\\\end{array}\right]$ + R = $\left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right]$ ( subtract corresponding elements )

$\left[\begin{array}{ccc}2-(-2)&4-2\\3-4&5-1\\\end{array}\right]$ + R = $\left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right]$

$\left[\begin{array}{ccc}4&2\\-1&4\\\end{array}\right]$ + R = $\left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right]$

R = $\left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right]$ - $\left[\begin{array}{ccc}4&2\\-1&4\\\end{array}\right]$ = $\left[\begin{array}{ccc}-3&-2\\1&-3\\\end{array}\right]$

6 0

3 years ago

Poease help! Thank you

DIA [1.3K]

Answer:

28 and 12t

Step-by-step explanation:

4 x 7

4 x 3t

6 0

3 years ago

Read 2 more answers