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3 years ago

SOMEONE HELP ME !

Mathematics

1 answer:

Ivahew [28]3 years ago

3 0

Answer:

10. 11 2/5

11. 6 1/2

2. 15 9/10

3. 10

Step-by-step explanation:

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Pls help this is due in soon!!

Rasek [7]

Answer:

Step-by-step explanation:

6 0

3 years ago

Select the three equations that pass through the points (-4, -16) and (5,2).

dezoksy [38]

Answer:

The equation of the line is $y-2 = 2(x - 5)$ , that passes through points $(-4, -16)\ and\ (5,2)$

Step-by-step explanation:

Given points are $(-4, -16)\ and\ (5,2)$

We need to write the equation of the line that passes through these points

First, we will find slope of this line

$m=\frac{y_2-y_1}{x_2-x_1}\\\\m=\frac{2-(-16)}{5-(-4)}\\\\m=\frac{2+16}{5+4}\\\\m=\frac{18}{9}=2$

So, our slope is $2$ . Now, we will plug the point $(5,2)$ in the equation below.

$(y-y_1)=m(x-x_1)\\y-2=2(x-5)\\$

So, the equation of the line is $y-2 = 2(x - 5)$ .

5 0

3 years ago

Read 2 more answers

Solve for the value of x. 1.x+8=-10 2.X-5=7 3.5x=30 4.X over 5 =-8

BabaBlast [244]

-18 negative eighteen
6 positive six
12 positive twelve
-40 negative forty

8 0

3 years ago

The manager of a music store has kept records of

Marysya12 [62]

Answer:

(a) $P(x\le 3) = 0.75$

(b) $P(x\le 3) = 0.75$

(b) is the same as (a)

(d) $P(x = 1\ or\ 2) = 0.55$

(e) $P(x > 2) = 0.45$

Step-by-step explanation:

Given

$\begin{array}{ccccccc}{CDs} & {1} & {2} & {3} & {4} & {5} & {6\ or\ more}\ \\ {Prob} & {0.30} & {0.25} & {0.20} & {0.15} & {0.05} & {0.05}\ \ \end{array}$

Solving (a): Probability of 3 or fewer CDs

Here, we consider:

$\begin{array}{cccc}{CDs} & {1} & {2} & {3} \ \\ {Prob} & {0.30} & {0.25} & {0.20} \ \ \end{array}$

This probability is calculated as:

$P(x\le 3) = P(1) + P(2) + P(3)$

This gives:

$P(x\le 3) = 0.30 + 0.25 + 0.20$

$P(x\le 3) = 0.75$

Solving (b): Probability of at most 3 CDs

Here, we consider:

$\begin{array}{cccc}{CDs} & {1} & {2} & {3} \ \\ {Prob} & {0.30} & {0.25} & {0.20} \ \ \end{array}$

This probability is calculated as:

$P(x\le 3) = P(1) + P(2) + P(3)$

This gives:

$P(x\le 3) = 0.30 + 0.25 + 0.20$

$P(x\le 3) = 0.75$

(b) is the same as (a)

Solving (c): Probability of 5 or more CDs

Here, we consider:

$\begin{array}{ccc}{CDs} & {5} & {6\ or\ more}\ \\ {Prob} & {0.05} & {0.05}\ \ \end{array}$

This probability is calculated as:

$P(x \ge 5) = P(5) + P(6\ or\ more)$

This gives:

$P(x\ge 5) = 0.05 + 0.05$

$P(x \ge 5) = 0.10$

Solving (d): Probability of 1 or 2 CDs

Here, we consider:

$\begin{array}{ccc}{CDs} & {1} & {2} \ \\ {Prob} & {0.30} & {0.25} \ \ \end{array}$

This probability is calculated as:

$P(x = 1\ or\ 2) = P(1) + P(2)$

This gives:

$P(x = 1\ or\ 2) = 0.30 + 0.25$

$P(x = 1\ or\ 2) = 0.55$

Solving (e): Probability of more than 2 CDs

Here, we consider:

$\begin{array}{ccccc}{CDs} & {3} & {4} & {5} & {6\ or\ more}\ \\ {Prob} & {0.20} & {0.15} & {0.05} & {0.05}\ \ \end{array}$

This probability is calculated as:

$P(x > 2) = P(3) + P(4) + P(5) + P(6\ or\ more)$

This gives:

$P(x > 2) = 0.20+ 0.15 + 0.05 + 0.05$

$P(x > 2) = 0.45$

3 0

3 years ago

State the range of the following function

Basile [38]

Answer:

Y=8x

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers