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Alecsey [184]
3 years ago
10

Which additional information is needed to solve this story problem? The zoo had 3 baby pandas in 2006 and gave 2 of them to othe

r zoos. How many baby pandas had the zoo given to other zoos in 2006 and 2007 altogether? How much did the pandas weigh? How many pandas were born in 2006? How many pandas did the zoo give away in 2007?
Mathematics
1 answer:
Likurg_2 [28]3 years ago
7 0

Answer:

How may pandas were born in 2006

Step-by-step explanation:

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0.72(x + 2) = 7.2<br><br><br><br> x = _____
mixas84 [53]

Answer:

x = 8

Step-by-step explanation:

Hope that helps

5 0
4 years ago
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Pete can buy 4 cokes for $5 or 10 cokes for $12.50. How much will he would pay for 7 cokes?
frutty [35]

Answer:

$8.75

Step-by-step explanation:

Pete would pay 8.75 because if you divide 5/4 and 12.50/10, you find that each Coke costs 1.25. 1.25 times 7 is 8.75

5 0
3 years ago
the sign of the product of two integers with the same sign is positive what is the sign of the product of 3 integers with the sa
hammer [34]
The answer would be positive if all 3 signs were the same. Yes, the product of two integers with the same sign will be positive, but that goes for any number of integers as long as they still have the same sign.
6 0
3 years ago
Help would be greatly appreciated ​
vodka [1.7K]

Answer:

Step-by-step explanation:

5 0
4 years ago
Which statement is true of normally distributed data ??
balandron [24]

Answer:

The correct option is b. approximately 68% of the data falls within 1 standard deviation (+-1) of the mean

Explanation:

According to the empirical rule of normal distribution:

1. Approximately 68% of the data falls within 1 standard deviation \pm 1 of the mean

2. Approximately 95% of the data falls within 2 standard deviations \pm 2 of the mean

3. Approximately 99.7% of the data falls within 3 standard deviations \pm 3 of the mean.

Therefore, among the given options, only option b adheres to the empirical rule of the normal distribution. Therefore, the option b is correct


8 0
3 years ago
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