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Snowcat [4.5K]
2 years ago
7

In a class of 25 students, 15 of them have a cat,

Mathematics
1 answer:
Tju [1.3M]2 years ago
5 0

Answer:

9/25

Step-by-step explanation:

I used to struggle on these too, but I'll exaplain the best I can. Firstly, we know that 3 people don't have a cat or a dog, so what we do is we minus that from 25. This becomes 22. Now, we have to add the amount of people who have a cat to the amount of people who have a dog, and then minus it from 22, because that is the amount of people without double ups in the class with at least either one of the pets given. So, 16+15=31. Now, we do 31-22, which is equal to 9. Therefore, 9/25 people, (because the question asked out of random in the class), have a cat and a dog. This is because the double up are the extra amount of people who had more than one pet. Hopefully this was helpful. I'm sorry if it is not as probability is a hard topic to explain.

:))))))))))))))))))))))))))

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Find the percent increase or decrease: 43 to 78
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Answer:

81.39% increase.

Step-by-step explanation:

Given: There is change in number from 43 to 78.

First lets find the amount of change or difference in number.

Difference in number= 78-43= 35

∴ Difference in number show that there is an increase of 35 number.

Now, finding the percent change in the number.

Percent= \frac{Difference\ in\ number}{base\ number} \times 100

⇒ Percent change in number= \frac{35}{43} \times 100= 81.39\%

∴ 81.39% increase in the number.

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The most recent public health statistics available indicate that 23.6​% of American adults smoke cigarettes. Using the​ 68-95-99
artcher [175]

Answer:

There is a​ 68% chance that between 17​% and 30​% are​ smokers.

There is a​ 95% chance that between 10​% and 37​% are​ smokers.

There is a​ 99.7% chance that between 4​% and 44​% are​ smokers.

Step-by-step explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes <em>n</em> > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of the sampling distribution of sample proportion is:

 \mu_{\hat p}=p\\

The standard deviation of the sampling distribution of sample proportion is:

 \sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}

Given:

<em>n</em> = 40

<em>p</em> = 0.236

Compute the mean and standard deviation of this sampling distribution of sample proportion as follows:

\mu_{\hat p}=p=0.236

\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.236(1-0.236)}{40}}=0.067

The Empirical Rule states that in a normal distribution with mean <em>µ</em> and standard deviation <em>σ</em>, nearly all the data will fall within 3 standard deviations of the mean. The empirical rule can be divided into three parts:

  • 68% data falls within 1 standard-deviation of the mean.  

        That is P (µ - σ ≤ X ≤ µ + σ) = 0.68.

  • 95% data falls within 2 standard-deviations of the mean.

        That is P (µ - 2σ ≤ X ≤ µ + 2σ) = 0.95.

  • 99.7% data falls within 3 standard-deviations of the mean.

        That is P (µ - 3σ ≤ X ≤ µ + 3σ) = 0.997.

Compute the range of values that has a probability of 68% as follows:

P (\mu_{\hat p} - \sigma_{\hat p} \leq  \hat p \leq  \mu_{\hat p} + \sigma_{\hat p}) = 0.68\\P(0.236-0.067\leq  \hat p \leq 0.236+0.067)=0.68\\P(0.169\leq  \hat p \leq0.303)=0.68\\P(0.17\leq  \hat p \leq0.30)=0.68

Thus, there is a​ 68% chance that between 17​% and 30​% are​ smokers.

Compute the range of values that has a probability of 95% as follows:

P (\mu_{\hat p} - 2\sigma_{\hat p} \leq  \hat p \leq  \mu_{\hat p} + 2\sigma_{\hat p}) = 0.95\\P(0.236-2\times 0.067\leq  \hat p \leq 0.236+2\times0.067)=0.95\\P(0.102\leq  \hat p \leq 0.370)=0.95\\P(0.10\leq  \hat p \leq0.37)=0.95

Thus, there is a​ 95% chance that between 10​% and 37​% are​ smokers.

Compute the range of values that has a probability of 99.7% as follows:

P (\mu_{\hat p} - 3\sigma_{\hat p} \leq  \hat p \leq  \mu_{\hat p} + 3\sigma_{\hat p}) = 0.997\\P(0.236-3\times 0.067\leq  \hat p \leq 0.236+3\times0.067)=0.997\\P(0.035\leq  \hat p \leq 0.437)=0.997\\P(0.04\leq  \hat p \leq0.44)=0.997

Thus, there is a​ 99.7% chance that between 4​% and 44​% are​ smokers.

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