What is the length of AB?

The diameter of circle Q is 14 cm. Find the arc length of PR.

mario62 [17]

Answer:

4.15, or 1.32pi

Step-by-step explanation:

arclength is circumference(34/360)

so pi x d (34/360)

3.14 x 14 x (34/360) = 4.15, or 1.32pi

7 0

3 years ago

Danny’s dog weighs 45.33 pounds. His cat only weighs 14.89 pounds.

Zinaida [17]

Answer:

His dog weighs 30.44lbs more than his cat.

Step-by-step explanation:

The question asks how much more his dog weighs, so we use subtraction to find the answer.

8 0

3 years ago

National data in the 1960's showed that about 44% of the adult population had never smoked cigarettes. In 1995 a national health

Marta_Voda [28]

Answer:

a) $0.52 - 1.96\sqrt{\frac{0.52(1-0.52)}{881}}=0.487$

$0.52 + 1.96\sqrt{\frac{0.52(1-0.52)}{881}}=0.553$

The 95% confidence interval would be given by (0.487;0.553)

b) After see the confidence interval we see that the lower limits 0.487>0.44 with 0.44 or 44% the estimated proportion for 1960, we can conclude that we have a significant increase in the adult proportion had never smoked cigarettes.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.52 - 1.96\sqrt{\frac{0.52(1-0.52)}{881}}=0.487$

$0.52 + 1.96\sqrt{\frac{0.52(1-0.52)}{881}}=0.553$

The 95% confidence interval would be given by (0.487;0.553)

Part b

After see the confidence interval we see that the lower limits 0.487>0.44 with 0.44 or 44% the estimated proportion for 1960, we can conclude that we have a significant increase in the adult proportion had never smoked cigarettes at 5% of significance.

4 0

3 years ago

Sodas in a can are supposed to contain an average 12 oz. This particular brand has a standard deviation of 0.1 oz, with an avera

hoa [83]

Answer:

The probability is $P(X < 12) = 0.99286$

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 12 \ oz$

The standard deviation is $\sigma = 0.1 \ oz$

The sample mean is $\= x = 12.1 \ oz$

The sample size is n = 6 packs

The standard error of the mean is mathematically represented as

$\sigma_{\= x } = \frac{\sigma}{\sqrt{n} }$

substituting values

$\sigma_{\= x } = \frac{0.1}{\sqrt{6} }$

$\sigma_{\= x } = 0.0408$

Given that the can’s contents follow a Normal distribution then then the probability that the mean contents of a six-pack are less than 12 oz is mathematically represented as

$P(X < 12) = P ( \frac{X - \mu }{ \sigma_{\= x }} < \frac{\= x - \mu }{ \sigma_{\= x }} )$