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3 years ago

Write four numbers that round to 700,000 when rounded to the nearest hundred thousand

2 answers:

6 0

699,512
655,456
689,433
701,001 this is the correct answer and btw i copied this from my e-notes and i own the rights to those

miskamm [114]3 years ago

4 0

699,548
655,417
689,414
701,000

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What are the equivalent fractions for 2 1/5 and 1 5/6 using 30 as the common denominator

Anuta_ua [19.1K]

Answer:

2 6/30 and 1 25/30

Step-by-step explanation:

First, you need to find 30÷5 and 30÷6. You do this because you need to know the multiplier when you create an equivalent fraction. 30÷5 is 6 and 30÷6 is 5. 2 1/5. You need to multiply the 1 and 5 by 6, that is 2 6/30. 1 5/6. 5 and 6 multiplied by 5 is 25 and 30. 1 5/6=1 25/30

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2 years ago

Find the Value of Y! Thank u!!!

Virty [35]

Answer:

my bad if im wrong but i got y=35

Step-by-step explanation:

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3 years ago

A grocery store sells tangerines in 2/5 kg bags. A customer bought 4 kg of tangerines for a school party. How many bags did he b

Alex777 [14]

10kg bags I believe

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2 years ago

Read 2 more answers

A school store buys pencils in bulk and sells them to students for a small profit. The school pays $2.00 for 100 pencils and sel

Valentin [98]

Answer:

1,000 pencils

Step-by-step explanation:

First, there are 10,000 cents in $100.

So divide 10,000 by 10.

1,000.

I hope this helps!

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3 years ago

Two athletic teams play a series of games; the first team to win 4 games is declared the overall winner. Suppose that one of the

kodGreya [7K]

Answer:

p=0.7103 (4-game series)
p=0.6480 (2-game series)

Step-by-step explanation:

Let X be the random variable equal the the first 4 straight wins. An overall win for the stronger team implies a negative binomial function with the parameters n=4, p=0.6:

$P(X=4)={{i-1}\choose {4-1}}0.6^40.4^{i-4},\ i=4,5,6,7$

#We find probabilities for the different values of i:

$P(X=4)={3\choose 3}0.6^4=0.1296\\\\P(X=5)={4\choose 3}0.6^40.4^1=0.2074\\\\P(X=6)={5\choose 3}0.6^40.4^2=0.2074\\\\P(X=4)={6\choose 3}0.6^40.4^3=0.1659$

Hence, probability of the stronger team winning overall is:

$=P(X=4)+P(X=5)+P(X=6)+P(X=7)\\\\=0.7103$

#Define Y as the random variable for winning 2/3 games.:

$P(Y=2)={1\choose 1}0.6^2=0.3600\\\\P(Y=3)={2\choose3}0.6^20.4=0.2880\\\\P(win)=0.2880+0.3600=0.6480$

Hence, probability of the stronger team winning in 2 out 3 game series is 0.6480

The stronger team has a higher chance of winning in a 4-game series(0.7103>0.6480)

8 0

3 years ago