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3 years ago

Which steps could be used to solve this story problem?

Mathematics

1 answer:

Airida [17]3 years ago

8 0

C. Multiply $ 5 by 3, multiply $ 3 by 4, then multiply $ 13 by 3....then add the products together

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1. A vector starts at point ( -2, 1) and ends at point (-6, -2) what is the horizontal component of the vector?

OLga [1]

Answer:

ok I guess we should I am not good I am mad so yeah

6 0

2 years ago

Read 2 more answers

Can someone plsss help me

balu736 [363]

Answer:

y = -5/3 x + 40

Step-by-step explanation:

The general form is y = ax + b, where a is the slope and b is the y intercept.

So immediately you can see that b must be 40, since the line crosses the y axis at 40.

The slope can be calculated by finding two "nice" points on the line that are also grid lines.

How about (0,40) and (6, 30).

Then, calculate the slope by dividing the y difference by the x difference:

(30-40)/(6-0) = -10/6 = -5/3

So C must be the answer.

6 0

3 years ago

A 4ft piece of pipe needs to be divided into pieces that are 2/3ft long. How many pieces are made? How many remaining?

Katen [24]

4ft divided by 2/3ft

4/1 divided by 2/3

4/1 times 3/2 is 12/2 or 6 pieces, 0 remaining.

3 0

3 years ago

Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th

grigory [225]

Answer:

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

The sample space is:

$S = \{1,1,2,2,2,3\}$

And the probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{6}$

$P(1) = \frac{1}{3}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{6}$

$P(2) = \frac{1}{2}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{6}$

P(Odd Number) is then calculated as:

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{6}$

$P(Odd\ Number) = \frac{3}{6}$

$P(Odd\ Number) = \frac{1}{2}$

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

$S = \{1,1,2,2,2,3,3\}$

The probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{7}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{7}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{7}$

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{7}$

$P(Odd\ Number) = \frac{3}{7}$

Comparing P(Odd Number) before and after

$P(Odd\ Number) = \frac{1}{2}$ --- Before

$P(Odd\ Number) = \frac{3}{7}$ --- After

<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>

7 0

3 years ago

What's number 10 please answer

Alik [6]

Sorry what grade are you in

3 0

3 years ago

Read 2 more answers