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3 years ago

PLZ HELP!!!!!! It’s due by today .. I will make you the brainliest

Mathematics

1 answer:

Aleks04 [339]3 years ago

8 0

Answer:

draw a line like this / through zero than draw another line like this \ through zero

Step-by-step explanation:

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In a beaded necklace 3/12, of the beads

Alexxandr [17]

Answer: 7/12 of the beads are blue and green

Step-by-step explanation: To make both denominators of fraction 3/12 and 1/3 equal, you have to change both denominators to the lowest common denominator. Since the LCD (lowest common denominator) is 12 you have to multiply 1/3 by 4 to make the two denominators match.

1/3 x 4/4= 4/12

Then to find the total fraction of both green and blue beads on the necklace, you simply have to add the two fractions together.

4/12+3/12= 7/12

hope this helps :)

4 0

3 years ago

laura goes for a cycle from her house to the post office 4km away work out laura's speed cycling to the post office CORBETTMATHS

lys-0071 [83]

ItItItIt takes Laura 15 minutes (0.25 hours) to cycle to the post office.

<h3>Speed and time</h3>

a. It takes Laura 15 minutes (0.25 hours) to cycle to the post office.

b. Speed:

Speed=Distance/Time

Speed=4/0.25

Speed=16 km/hr

c. It takes Laura 20 minutes to cycle to the post office.

d. Time

Time=20 minutes (1/3 hour)

Speed:

Speed=Distance/Time

Speed=4÷1/3

Speed=12 km/hr

Therefore It takes Laura 15 minutes (0.25 hours) to cycle to the post office.

Learn more about speed here:brainly.com/question/13943409

#SPJ1

3 0

2 years ago

Conjunto A = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}

OlgaM077 [116]

hello ✋ my country is Indonesia

6 0

3 years ago

Please help me I don’t understand

DIA [1.3K]

Answer: A)

Explanation: 10-3=7

8 0

3 years ago

There are some blue and red counters in a bag in the ratio 1:3. Two counters are taken at random. The probability of both counte

Vaselesa [24]

Let N be the number of blue counters. This implies that 3N counters are red, and there are 4N counters in total.

Assuming that you don't reinsert the first counter, for the first pick, you have N blue counters over 4N total counters, so you'll pick a blue counter with probability

$\dfrac{N}{4N}=\dfrac{1}{4}$

For the second pick, you're left with N-1 blue counters over 4N-1 counters, so you'll pick a blue counter with probability

$\dfrac{N-1}{4N-1}$

The probability of picking two blue counters with two picks is the product of the two probabilities:

$\dfrac{1}{4}\cdot\dfrac{N-1}{4N-1}$

And we want this to equal 1/20, so we have

$\dfrac{1}{4}\cdot\dfrac{N-1}{4N-1}=\dfrac{1}{20} \iff \dfrac{N-1}{4N-1}=\dfrac{1}{5} \iff 5(N-1)=4N-1$

We can expand the left hand side and solve for N:

$5(N-1)=4N-1 \iff 5N-5=4N-1 \iff N = 4$

So, there are 4 blue counters and 12 red counters, for a total of 16 counters in the bag.

We can indeed verify that the probabilities of picking the two blue counters is

$\dfrac{4}{16}\cdot \dfrac{3}{15} = \dfrac{1}{4}\cdot\dfrac{1}{5}=\dfrac{1}{20}$

7 0

3 years ago