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hram777 [196]
1 year ago
12

How many terms are in the binomial expansion of (3x 5)9? 8 9 10 11

Mathematics
1 answer:
swat321 year ago
5 0

Answer:

10

Step-by-step explanation:

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Taylor is using money from her coin try to go to the fair. She wants to spend no more than $17 there. She has $8 in bills and so
ZanzabumX [31]
There are 4 quarters in dollars. Keep that in mind.

Since Taylor is spending no more than 17 dollars, she probably only has 17 dollars. We already know that she has 8 dollars in bills, so we can subtract 8 dollars from 17 dollars because we need to find the amount of dollars in quarters. 

The answer for that is 9 dollars. Since there are 4 quarters in the dollars, you need to multiply 9 by 4 to find the amount of quarters. 
4 0
2 years ago
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2. Use (<, >, =) to compare 23 and 34.
lara31 [8.8K]

Answer:

Step--step explanation:

2/3 = 0.67

3/4= 0.75

So 2/3 < 3/4

8 0
2 years ago
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To convert 12 yards to inches, you would use the ratio .36inches/1 yard true or false ?
Taya2010 [7]
False. It's 36 not .36 and you would just do 12x36 since 36 inches is 1 yd. But if you want to make it complicated just do 12/1 times 36/1 which still equals 432.
8 0
3 years ago
Solve for x.<br> 13(x-3) = 39<br> X=1<br> X=4<br> x=6<br> x=10
Valentin [98]
13 factored out is 13x - 39 add 39 to both sides it equals 13x=78 and 78 divided by 13 is 6 the anwser is x=6
7 0
2 years ago
A consumer organization estimates that over a​ 1-year period 20​% of cars will need to be repaired​ once, 8​% will need repairs​
Sladkaya [172]

Answer:

Probability that a car need to be repaired​ once = 20% = 0.20

Probability that a car need to be repaired​ twice = 8% = 0.08

Probability that a car need to be repaired​ three or more = 2% = 0.02

a) If you own two​ cars what is the probability that  neither will need​ repair?

Probability that a car need to be repaired​ once , twice and thrice or more= 0.20+0.08+0.02=0.3

Probability that car need no repair = 1-0.3=0.7

Neither car will need repair=0.7 \times 0.7=0.49

​b) both will need​ repair?

Probability both will need​ repair = 0.3 \times 0.3=0.09

c)at least one car will need​ repair

Neither car will need repair=0.7 \times 0.7=0.49

Probability that at least one car will need​ repair= 1-0.49 = 0.51

6 0
2 years ago
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