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Stella [2.4K]
3 years ago
9

The cost of a jacket increased from $65.00 to $72.15. What is the percentage increase of the cost of the jacket?

Mathematics
1 answer:
Karo-lina-s [1.5K]3 years ago
6 0

Answer:

11%

Step-by-step explanation:

$72.15 - $65.00=$7.15 /$65.00=.11 or 11%

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What is 3/4 + 3.3 - 1 1/2?
guapka [62]

Answer:

2.55

Step-by-step explanation:1

Convert 1121\frac{1}{2}1​2​​1​​ to improper fraction. Use this rule: abc=ac+bca \frac{b}{c}=\frac{ac+b}{c}a​c​​b​​=​c​​ac+b​​.

34+3.3−1×2+12\frac{3}{4}+3.3-\frac{1\times 2+1}{2}​4​​3​​+3.3−​2​​1×2+1​​

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Simplify 1×21\times 21×2 to 222.

34+3.3−2+12\frac{3}{4}+3.3-\frac{2+1}{2}​4​​3​​+3.3−​2​​2+1​​

3

Simplify 2+12+12+1 to 333.

34+3.3−32\frac{3}{4}+3.3-\frac{3}{2}​4​​3​​+3.3−​2​​3​​

4

Simplify.

5.12\frac{5.1}{2}​2​​5.1​​

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2.552.552.55

6 0
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

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3 years ago
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Answer:

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Step-by-step explanation:

Angle QPR + Angle PQR = Angle HRQ

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