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Lelechka [254]
3 years ago
11

In the video at the beginning of the lesson, how many 2-topping pizza combinations were there if there were 7 toppings to choose

from?
Mathematics
2 answers:
elena-s [515]3 years ago
8 0

Answer:

3 or 3 and 1/2

Step-by-step explanation:

2 toppings per pizza

2, 4, 6 and the half pizza with one topping

Greeley [361]3 years ago
3 0
The answer is 3 or 3 1/2
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Pedro has 3 boxes of 6 markers. Rafael has 5 boxes of 5 markers. One of the friends arranges all his markers into 2 equal groups
Komok [63]

Answer:

see below

Step-by-step explanation:

Pedro has 3 boxes of 6 markers.

3*6 = 18 markers

We can divide this by 2 because this is even

18/2 = 9

There are 9 in each group

Rafael has 5 boxes of 5 markers

5*5 = 25

We cannot divide this by 2 because this is odd

3 0
3 years ago
I really need this answer quick
agasfer [191]
I got 168 because if you multiply 8x12 then you would get 96 and on the sides u need to multiply 3x12
3 0
3 years ago
Read 2 more answers
What expression is equivalent to (2^3)^-5
aliya0001 [1]

Answer:

2^-15 or 1/2^15

Step-by-step explanation:

3*-5 is -15 so that makes it 2^-15

Apply the exponent rule a^-b = 1/a^b

That changes it to 1/2^15

Hope this helped! :)

7 0
3 years ago
Pay to Park
iragen [17]

Answer:

5 minutes = 3

1 hour = 3

1 hour 50 minutes = 6

2 hours = 6

2 hours 1 minute = 9

3 and a half = 12

Step-by-step explanation:

7 0
3 years ago
According to a study conducted in one city, 39% of adults in the city have credit card debts of more than $2000. A simple random
Butoxors [25]

Answer:

The sampling distribution of the sample proportion of adults who have credit card debts of more than $2000 is approximately normally distributed with mean \mu = 0.39 and standard deviation s = 0.0488

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

In this question:

p = 0.39, n = 100

Then

s = \sqrt{\frac{0.39*0.61}{100}} = 0.0488

By the Central Limit Theorem:

The sampling distribution of the sample proportion of adults who have credit card debts of more than $2000 is approximately normally distributed with mean \mu = 0.39 and standard deviation s = 0.0488

5 0
3 years ago
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