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2 years ago

Help please!!!!!!!!!!!!

Mathematics

1 answer:

Nady [450]2 years ago

8 0

Answer:

Step-by-step explanation:

a. 1 & 2

b. 3 & 5

c. 2/15

d. 2/5 x 5/3

e. 10/15= 2/3

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1. You roll a standard number cube. Find P(number greater than 2).

ra1l [238]

Answer:

P={3,4,5,6}

Step-by-step explanation:

Cube=6 sides={1,2,3,4,5,6}

P>2

P={3,4,5,6}

7 0

3 years ago

What is the least common multiple (LCM) of 3 and 5? O 30 O 2 O 15 O 8 pls help

Ksivusya [100]

Answer:

Step-by-step explanation:

magical it is 15 because it is what it is

8 0

2 years ago

Simplify the expression by arranging the steps in sequence based on the order of operations 1/6(30x+24)-1/7(63-7×)

alex41 [277]

Answer:

6x-5

Step-by-step explanation:

We apply distributive property to get rid of the first parenthesis, by multiplying each term inside by the factor 1/6:

$\frac{1}{6} (30x+24)-\frac{1}{7} (63-7x)\\\frac{30}{6} x+\frac{24}{6} -\frac{1}{7} (63-7x)\\5x+4-\frac{1}{7} (63-7x)$

now we get rid of the second parenthesis by multiplying the factor " $-\frac{1}{7}$ " times each of the terms inside the parenthesis (notice here that this factor is negative , so it will flip the signs of the terms it multiplies):

$5x+4-\frac{1}{7} (63-7x)\\5x+4-\frac{63}{7} +\frac{7}{7} x\\5x+4-9+x$

Next we combine like terms:

$5x+4-9+x\\5x+x+4-9\\6x-5$

7 0

2 years ago

Find the probability that a randomly selected dog is Male given it is a boxer. Enter your answer in decimal form rounded to two

RoseWind [281]

Answer:

0.63

Step-by-step explanation:

M: The selected dog is a male. P(M) = 24/40 = 0.6

B: The selected dog is a boxer. P(B) = 22/40 = 0.55

M and B: 14/40 = 0.35

P(M/B) = 0.35 / 0.55 = 0.63

3 0

3 years ago

Use Cramer’s rule to solve for x: x + 4y − z = −14 5x + 6y + 3z = 4 −2x + 7y + 2z = −17

V125BC [204]

Looks like the system is

x + 4y - z = -14

5x + 6y + 3z = 4

-2x + 7y + 2z = -17

or in matrix form,

$\mathbf{Ax} = \mathbf b \iff \begin{bmatrix} 1 & 4 & -1 \\ 5 & 6 & 3 \\ -2 & 7 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -14 \\ 4 \\ -17 \end{bmatrix}$

Cramer's rule says that

$x_i = \dfrac{\det \mathbf A_i}{\det \mathbf A}$

where $x_i$ is the solution for i-th variable, and $\mathbf A_i$ is a modified version of $\mathbf A$ with its i-th column replaced by $\mathbf b$ .

We have 4 determinants to compute. I'll show the work for det(A) using a cofactor expansion along the first row.

$\det \mathbf A = \begin{vmatrix} 1 & 4 & -1 \\ 5 & 6 & 3 \\ -2 & 7 & 2 \end{vmatrix}$

$\det \mathbf A = \begin{vmatrix} 6 & 3 \\ 7 & 2 \end{vmatrix} - 4 \begin{vmatrix} 5 & 3 \\ -2 & 2 \end{vmatrix} - \begin{vmatrix} 5 & 6 \\ -2 & 7 \end{vmatrix}$

$\det \mathbf A = ((6\times2)-(3\times7)) - 4((5\times2)-(3\times(-2)) - ((5\times7)-(6\times(-2)))$

$\det\mathbf A = 12 - 21 - 40 - 24 - 35 - 12 = -120$

The modified matrices and their determinants are

$\mathbf A_1 = \begin{bmatrix} -14 & 4 & -1 \\ 4 & 6 & 3 \\ -17 & 7 & 2\end{bmatrix} \implies \det\mathbf A_1 = -240$

$\mathbf A_2 = \begin{bmatrix} 1 & -14 & -1 \\ 5 & 4 & 3 \\ -2 & -17 & 2 \end{bmatrix} \implies \det\mathbf A_2 = 360$

$\mathbf A_3 = \begin{bmatrix} 1 & 4 & -14 \\ 5 & 6 & 4 \\ -2 & 7 & -17 \end{bmatrix} \implies \det\mathbf A_3 = -480$

Then by Cramer's rule, the solution to the system is

$x = \dfrac{-240}{-120} \implies \boxed{x = 2}$

$y = \dfrac{360}{-120} \implies \boxed{y = -3}$

$z = \dfrac{-480}{-120} \implies \boxed{z = 4}$

5 0

1 year ago