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Komok [63]
3 years ago
12

The teachers had to transport 175 students by bus on a field trip. Each bus holds 64 students. How many buses do they need to or

der so no one is left behind
Mathematics
1 answer:
strojnjashka [21]3 years ago
6 0
At least 3 buses. Two buses can be filled completely and the third will be filled with 47 students.
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BRAINLIESTTT ASAP! MATH EXPERT PLEASE HELP ME :)
kirza4 [7]

Answer:

It's true.

Explanation:

2(segment ST) = RT

divide both sides by two

ST = RT/2

You would have two equal halves of RT. One half equals RS. The other half equals ST.

Here's a rough picture of the line.

R _____________S_____________ T

S is in the middle of the line which makes it the midpoint.

It's true. S is the midpoint of RT.

3 0
3 years ago
Read 2 more answers
100 points !!!!!!!!!!!!! please help
m_a_m_a [10]

We have to,

write a word problem for this equation,

→ 2/3 × 5

Let's write a word problem,

<u>For one student 2/3 part of one cake is needed. Then how </u><u>many</u><u> </u><u>cakes will they have to buy for 5 students?</u>

5 0
3 years ago
Read 2 more answers
Graph the system by hand.
weqwewe [10]

Answer:

dont khan academy

Step-by-step explanation:

idk

why

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cheet

7 0
3 years ago
For the following exercise, determine the range (possible values) of the random variable, X. A batch of 300 machined parts conta
solmaris [256]

Answer:

{0,1,2,3,4,5}

Step-by-step explanation:

We are given that

Total number of  machine parts=300

Number of defective machine parts=10

Total number of good machine parts=300-10=290

Sample contain parts that do not conform to customer requirement=5

X  is a random variable which is the number of parts in a sample of 5 parts that do not conform to customer requirements.

We have to find the correct answer.

The sample contain 5 parts

Therefore, the possible values of random variable X

0,1,2,3,4,5

Hence, the range of X is given by

{0,1,2,3,4,5}

5 0
3 years ago
Question 3 (Multiples and Factors] Three numbers are given below. Use prime factorisation to determine the HCF and LCM 1848 132
Ilia_Sergeevich [38]

Prime factorization involves rewriting numbers as products

The HCF and the LCM of 1848, 132 and 462​ are 66 and 1848 respectively

<h3>How to determine the HCF</h3>

The numbers are given as: 1848, 132 and 462

Using prime factorization, the numbers can be rewritten as:

1848 = 2^3 * 3 * 7 * 11

132 =  2^2 * 3 * 11

462 = 2 * 3 * 7 * 11

The HCF is the product of the highest factors

So, the HCF is:

HCF = 2 * 3 * 11

HCF = 66

<h3>How to determine the LCM</h3>

In (a), we have:

1848 = 2^3 * 3 * 7 * 11

132 =  2^2 * 3 * 11

462 = 2 * 3 * 7 * 11

So, the LCM is:

LCM = 2^3 * 3 * 7 * 11

LCM  = 1848

Hence, the HCF and the LCM of 1848, 132 and 462​ are 66 and 1848 respectively

Read more about prime factorization at:

brainly.com/question/9523814

4 0
2 years ago
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