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3 years ago

8. The length of an actual road is

Mathematics

2 answers:

lina2011 [118]3 years ago

8 0

Answer:

10 inches

Step-by-step explanation:

1,000 / 100 = 10

Natasha_Volkova [10]3 years ago

3 0

Answer:

It's A.

Step-by-step explanation:

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Dina has a number cube that she uses for a game. The number cube has the shape of a triangular pyramid, with 4 faces numbered 1

yaroslaw [1]

Answer:

A) The relative frequencies in the table are reasonably close; C) The relative frequency of rolling an even number is 0.51.

Step-by-step explanation:

The relative frequencies given are at most 0.02 apart. This means they are reasonably close.

The theoretical probability for each outcome would be 1/4, or 0.25; this means the theoretical probability of rolling an even number would be 0.50, not 0.51.

However, the relative frequency of rolling an even number would be 0.24+0.27 = 0.51.

Since the relative frequencies are reasonably close, the number cube is likely to be fair.

6 0

3 years ago

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I need help with all plzz

garri49 [273]

1. yes, it is a proportional relationship
2. k=3
3. y=3x
4. 3
5.y=35x
6. y=7
hope this helps!

5 0

3 years ago

Read 2 more answers

Summer School 2020 Math 8 NWLSD

Ray Of Light [21]

Answer:

8 granola bars in each box

Step-by-step explanation:

The equation is 4x+5x=72

4x and 5x are like terms, so combine those 9x=72

Use the division property of equality to solve for x, x=8

5 0

3 years ago

A ball is kicked with an inicial velocity of

vesna_86 [32]

Answer:

bbbbbbbbbbbbbbbbbbbbbb

4 0

3 years ago

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