All categories

3 years ago

3:9 in simpliest form

Mathematics

2 answers:

s344n2d4d5 [400]3 years ago

8 0

Answer:

1:3

Step-by-step explanation:

$3 \div 3 \: \: \: \: \: \: \: \: \: 9 \div 3 \\ \: \: \: \: \: 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 3$

Softa [21]3 years ago

5 0

Answer:

1:3

Step-by-step explanation:

3 divide 3 1

9 divide 3 3

You might be interested in

"Tongue Piercing May Speed Tooth Loss, Researchers Say" is the headline of an article. The article describes a study of 51 young

Paul [167]

Answer:

The 95% confidence interval for the proportion of young adults with pierced tongues who have receding gums is (0.24, 0.506).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$n = 51, \pi = \frac{19}{51} = 0.373$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.373 - 1.96\sqrt{\frac{0.373*0.627}{51}} = 0.24$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.373 + 1.96\sqrt{\frac{0.373*0.627}{51}} = 0.506$

The 95% confidence interval for the proportion of young adults with pierced tongues who have receding gums is (0.24, 0.506).

3 0

3 years ago

Battery packs in electric go-carts need to last a fairly long time. The run-times (time until it needs to be recharged) of the b

pav-90 [236]

Answer:

The minimum level for which the battery pack will be classified as highly sought-after class is 2.42 hours

Step-by-step explanation:

Problems of normally distributed samples are 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.

In this problem, we have that:

$\mu = 2, \sigma = 0.33$