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fomenos
3 years ago
13

PLEASE HELP!!!!-0.5 1/2 0.7 -4/4 from least to greatest​

Mathematics
1 answer:
yawa3891 [41]3 years ago
3 0

Step-by-step explanation:

Converting all the numbers to decimals,

1) -0.5

2) 1/2 = 0.5

3) 0.7

4) -4/4 = -1

We see that order of numbers from least to greatest is -1 < -0.5 < 0.5 < 0.7

Please mark Brainliest if this helps!

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The coach recorded how fast his runners ran 1 mile. Emmett ran a mile in 9.73 minutes Kate ran a mile in 10.87 minutes how many
igor_vitrenko [27]

Complete Question

The coach recorded how fast his runners ran 1 mile. Emmett ran a mile in 9.73 minutes, Kate ran a mile in 10.87 minutes, and Rubin ran a mile in 8.46 minutes. How many more minutes did Emmett run than Robin?

a. 1.37 minutes

b. 1.27 minutes

c. 2.27 minutes

d. 1.27 seconds

Answer:

1.27 minutes

Step-by-step explanation:

Emmett ran a mile in 9.73 minutes

Kate ran a mile in 10.87 minutes

Rubin ran a mile in 8.46 minutes.

How many more minutes did Emmett run than Robin?

This is calculated as:

Number of minutes Emmett ran - Number of minutes Rubin ran

= 9.73 minute - 8.46 minutes

= 1.27 minutes

Option b is correct

5 0
3 years ago
The length of a spring varies directly with the mass of an object that is attached to it. When a 30-gram object is attached, the
Yuri [45]
Hi.
your answer is s=3/10m
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3 years ago
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1. One morning, Roshan and Varun were talking to each other face to face at a crossing. If Varun's shadow was exactly to the lef
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Answer: Roshan was facing forward and to the right
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13. The least common multiple of two non-zero integers a and b is the unique positive integer m such that (i) m is a common mult
Vlad [161]

Answer:

[a,b] divides n

Step-by-step explanation:

Let us denote the least common multiple of a and b [a,b]=m.

We want to prove that m divides n, where n is a multiple of a and b.

We suppose m does not divide n, then by the Division Theorem, there exists q and r integers such that:

(1) ... n=mq+r, where 0<r<m

As n is a multiple of a and b, there exists s and t integers such that:

sa=n and tb=n

Same thing happens to m as it is the least common multiple, there exists u and v such that:

ua=m and vb=m

So (1) has the following form:

n=mq+r ⇒ sa=uaq+r ⇒sa-uaq=r⇒(s-uq)a=r and

n=mq+r ⇒ tb=vbq+r ⇒ tb-vbq=r⇒ (t-vq)b=r

So r is a multiple of a and b, but r<m which is a contradiction as, m is the least common multiple of a and b. So this concludes the proof.

So this means that \frac{ab}{m} is and integer.

As m= vb, then \frac{m}{b} is an integer, lets say \frac{m}{b}=v; and as m=ua, then \frac{m}{a}=u.

So \frac{ab}{m}v=\frac{ab}{m}\frac{m}{b}=a, so \frac{ab}{m} divides a; on the other hand, \frac{ab}{m}u=\frac{ab}{m}\frac{m}{a}=b, so \frac{ab}{m} divides b. From this we can conclude that \frac{ab}{m} is a common divisor of a and b.

4 0
3 years ago
In southern California, a growing number of individuals pursuing teaching credentials are choosing paid internships over traditi
Irina18 [472]

Answer:

Candidates are chosen without replacement, thus trials are dependent, meaning that Y has a hypergeometric distribution.

Step-by-step explanation:

Difference between hypergeometric and binomial distributions:

In the hypergeometric distribution, trials are dependent of other trials, while in the binomial distribution, they are independent. This can be examplified that in the hypergeometric distribution, the candidates are chosen from a set without replacement, while in the binomial distribution they are chosen with replacement.

In this question:

Candidates are chosen without replacement, thus trials are dependent, meaning that Y has a hypergeometric distribution.

6 0
2 years ago
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