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marysya [2.9K]
3 years ago
12

What is 43 in expanded form? A. 3 × 3 × 3 × 3 B. 4 × 4 × 4 C. 12 × 12 × 12 D. 8 × 8

Mathematics
1 answer:
densk [106]3 years ago
7 0
4^3 = 4 \times 4 \times 4
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one more than six times that number is 25. select the solution and graph that represents the original number
Brums [2.3K]

6n+1 = 25

subtract 1 from each side

6n+1-1 = 25-1

6n = 24

divide each side by 6

6n/6 = 24/6

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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]

Answer:

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.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

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

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

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

4 0
3 years ago
Find the distance between each pair of points.round your awnser to the nearest tenths,if nessesary (-3,-4),(0,0) problem 2 (-8,0
Arte-miy333 [17]

Answer:

1st one: -3,-4 away

2nd one: Lets just say I got kinda confused on that one so sorry for any inconvenience on my behalf from this question.

Step-by-step explanation:

I will call (-3,-4) Point A and I'll call (0,0) Point B. Point A is at (-3,-4) and Point B is obviously at the origin. That makes it easy, because you just subtract Point A by Point B to find how far away they are.

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3 years ago
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