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satela [25.4K]
3 years ago
15

6. Jessie sorted the coins in her bank. Shemade 7 stacks of 6 dimes and 8 stacksof 5 nickels. She then found 1 dime and1 nickel.

How many dimes and nickelsdoes Jessie have in all? (Lesson 2.12)

Mathematics
2 answers:
Salsk061 [2.6K]3 years ago
6 0
7 stacks of 6 dimes = 6•7 which is 42
8 stacks of 5 nickles = 8•5 which is 40
1 dime + 1 nickle
Add all of them together .......42+40+1+1=84

Hope this helped!❤️
Leokris [45]3 years ago
3 0
Lets write down the information that we know.

Jessie has 7 stacks of 6 dimes, so Jessie has a total of 6*7 dimes

Dimes=7*6=42dimes  

Jessie has 42 dimes, but then she found another one so lets add one more dime.

42+1=43 dimes

Jessie has 8 stacks of 5 nickels, so Jessie has 8*5 nickels

Nickels=8*5=40nickels

Jessie has 40 nickels, but then she found another one so lets add 1 to the total number of nickels.

40+1=41nickels

Now we can answer the question.

How many dimes and nickels does jessie have in all?

nickels+dimes=
41+43=84 coins

Jessie has 84 coins in total
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3 years ago
20% of US High School teens vape. A local High School has implemented campaigns to reduce vaping among students and believes tha
zaharov [31]

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10.93% probability of observing 51 or fewer vapers in a random sample of 300

Step-by-step explanation:

I am going to use the normal approximation to the binomial to solve this question.

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Normal probability distribution

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In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

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In this problem, we have that:

n = 300, p = 0.2

So

\mu = E(X) = np = 300*0.2 = 60

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{300*0.2*0.8} = 6.9282

What is the approximate probability of observing 51 or fewer vapers in a random sample of 300?

Using continuity corrections, this is P(X \leq 51 + 0.5) = P(X \leq 51.5), which is the pvalue of Z when X = 51.5 So

Z = \frac{X - \mu}{\sigma}

Z = \frac{51.5 - 60}{6.9282}

Z = -1.23

Z = -1.23 has a pvalue of 0.1093.

10.93% probability of observing 51 or fewer vapers in a random sample of 300

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bezimeni [28]

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