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2 years ago

Indicate in standard form the equation of the line passing through the given points.

Mathematics

1 answer:

irina1246 [14]2 years ago

5 0

Answer:

6,2

Step-by-step explanation:

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Linda had 84 fliers to post around town. Last week, she posted 1/3 of them. This week, she posted 2/7 of the remaining fliers. H

Jobisdone [24]

The remaining number of flires that Linda still not posted is 40.

According to the given question.

The total number of fliers Linda have = 84

Fraction of fliers Linda posted = 1/3

And, fraction of fliers Linda posted second time from the remaning fliers = 2/7

Now,

1/3 of 84 fliers = 1/3 × 84 = 28

So, the number of fliers are left with Linda after first posting = 84 - 28 = 56

Again she posted 2/7 fliers. The number of flires Linda posted second time is given by

⇒ 2/7 × 56 = 16

Thereore, the number of fliers that Lindna still not posted = 56 - 16 = 40

Hence, Linda still not posted 40 fliers.

Find out more information about fraction and remaining numbers here:

brainly.com/question/2254581

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4 0

11 months ago

Efficient manufacturing: Efficiency experts study the processes used to manufacture items in order to make them as efficient as

Delicious77 [7]

Answer:

I'm not sure sorry good luck though

3 0

3 years ago

Someone help me with this

viva [34]

Answer and work is shown in picture :)

i hope you have a lovely rest of your day!

6 0

2 years ago

A bag contains 4 red balls, 2 green balls, 3 yellow balls, and 5 blue balls. Find each probability for randomly removing balls w

AfilCa [17]

Answer:

$\frac{15}{4802}$ , $\frac{15}{9604}$ , $\frac{9}{2401}$ , $\frac{9}{4802}$

Step-by-step explanation:

The bag has a total of (4+2+3+5) = 14 balls. Set up the proportions:

Red: $\frac{4}{14}$

green: $\frac{2}{14}$

yellow: $\frac{3}{14}$

blue: $\frac{5}{14}$

Now solve!

Removing 1 yellow, 1 red, 1 green, and 1 blue = $\frac{3}{14} \cdot \frac{4}{14} \cdot \frac{2}{14} \cdot \frac{5}{14} =\frac{15}{4802}$

Removing 1 blue, 1 green, 1 green, and 1 yellow = $\frac{5}{14} \cdot \frac{2}{14} \cdot \frac{2}{14} \cdot \frac{3}{14} =\frac{15}{9604}$

Removing 1 red, 1 red, 1 yellow, and 1 yellow = $\frac{4}{14} \cdot \frac{4}{14} \cdot \frac{3}{14} \cdot \frac{3}{14} =\frac{9}{2401}$

Removing 1 green, 1 yellow, 1 yellow, and 1 red = $\frac{2}{14} \cdot \frac{3}{14} \cdot \frac{3}{14} \cdot \frac{4}{14} =\frac{9}{4802}$

4 0

2 years ago

Hilda draws a card from a deck of cards and it is a seven. The probability that it is the 7 of spades is Select one: O a. 1/3 O

Schach [20]

Answer:b

Step-by-step explanation:

Given

Hilda draws a card from a deck of cards and it is seven.

It is certain that card is 7 so there a 4 cards which can be 7

i.e. 7 of hearts,7 of spades,7 of club,7 of diamond

So probability that it is the 7 of spades is

$\frac{1}{4}=0.25$

8 0

3 years ago