The ratio of girls to students at a middle school is 10 to 18. If there are 270 students at the school, how many are girls?

Please help me with this question. Thank you!

mr_godi [17]

A-1,6

B-(-1,-5)

c-(-8,2)

D.4,-9

E.3,8

7 0

4 years ago

what is an equation of the line that passes through the point (-3,-5) and is parallel to the line 2x+3y=15

Mumz [18]

Answer:y=-2x/3+10

Step-by-step explanation:

7 0

3 years ago

The probability of flu symptoms for a person not receiving any treatment is 0.038. In a clinical trial of a common drug used to

alexgriva [62]

Answer:

36.32% probability that at least 47 people experience flu symptoms. This is not an unlikely event, so this suggests that flu symptoms are not an adverse reaction to the drug.

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:

$n = 1164, p = 0.038$

So

$\mu = E(X) = np = 1164*0.038 = 44.232$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = 6.5231$

Estimate the probability that at least 47 people experience flu symptoms.

Using continuity correction, this is $P(X \geq 47 - 0.5) = P(X \geq 46.5)$ , which is 1 subtracted by the pvalue of Z when X = 46.5.

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

$Z = \frac{46.5 - 44.232}{6.5231}$

$Z = 0.35$

$Z = 0.35$ has a 0.6368

1 - 0.6368 = 0.3632

36.32% probability that at least 47 people experience flu symptoms. This is not an unlikely event, so this suggests that flu symptoms are not an adverse reaction to the drug.

6 0

3 years ago

Solving multi-step<br>1.2+1.25g=8.62<br>g=

shutvik [7]

Here are the answers

4 0

4 years ago

Office occupancy in a city is an indication of the economic health of the region in which it is located. A random sample of offi

Pani-rosa [81]

Answer:

$(0.13-0.1) - 2.58 \sqrt{\frac{0.13(1-0.13)}{165} +\frac{0.10(1-0.10)}{140}}=-0.0664$

$(0.13-0.1) + 2.58 \sqrt{\frac{0.13(1-0.13)}{165} +\frac{0.10(1-0.10)}{140}}=0.124$

We are confident at 99% that the difference between the two proportions is between $-0.0664 \leq p_1 -p_2 \leq 0.124$ . And since the confidence interval cotains the value 0 we don't have enough evidence to conclude that we have significant differences between the to proportions in these two cities

Step-by-step explanation:

$p_1$ represent the real population proportion for 1

$\hat p_1 =\frac{22}{165}=0.13$ represent the estimated proportion for 1

$n_1=165$ is the sample size required for 1

$p_2$ represent the real population proportion for 2

$\hat p_2 =\frac{14}{140}=0.10$ represent the estimated proportion for 2

$n_2=140$ is the sample size required for 2

$z$ represent the critical value for the margin of error

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_1 -\hat p_2) \pm z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}$

For the 99% confidence interval the significance is $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , and the critical value using the normal standard distribution.

$z_{\alpha/2}=2.58$

Replacing we got:

$(0.13-0.1) - 2.58 \sqrt{\frac{0.13(1-0.13)}{165} +\frac{0.10(1-0.10)}{140}}=-0.0664$

$(0.13-0.1) + 2.58 \sqrt{\frac{0.13(1-0.13)}{165} +\frac{0.10(1-0.10)}{140}}=0.124$

We are confident at 99% that the difference between the two proportions is between $-0.0664 \leq p_1 -p_2 \leq 0.124$ . And since the confidence interval cotains the value 0 we don't have enough evidence to conclude that we have significant differences between the to proportions in these two cities

8 0

3 years ago