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3 years ago

0.8 -0.50 3 with solution

Mathematics

2 answers:

kolezko [41]3 years ago

5 0

Answer:

0.8-0.503=0.297

0.8-0.50= 0.3

Step-by-step explanation:

kaheart [24]3 years ago

4 0

Answer:

0.3

Step-by-step explanation:

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I don't know but i need points so thanks

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3 years ago

"The counting rule that is used for counting the number of experimental outcomes when n objects are selected from a set of N obj

Liula [17]

<u>Options</u>

Counting rule for permutations
Counting rule for multiple-step experiments
Counting rule for combinations
Counting rule for independent events

Answer:

(C)Counting rule for combinations

Step-by-step explanation:

When selecting n objects from a set of N objects, we can determine the number of experimental outcomes using permutation or combination.

When the order of selection is important, we use permutation.

However, whenever the order of selection is not important, we use combination.

Therefore, The counting rule that is used for counting the number of experimental outcomes when n objects are selected from a set of N objects where order of selection is not important is called the counting rule for combinations.

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3 years ago

Suppose that administrators at a large urban high school want to gain a better understanding of the prevalence of bullying withi

boyakko [2]

Answer:

lower limit =0.261

upper limit =0.369

Step-by-step explanation:

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".

Description in words of the parameter p

$p$ represent the real population proportion of people that have experienced bullying

X= 63 people in the random sample that have experienced bullying

n=200 is the sample size required

$\hat p=\frac{63}{200}=0.315$ represent the estimated proportion of people that have experienced bullying

$z_{\alpha/2}$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Numerical estimate for p

In order to estimate a proportion we use this formula:

$\hat p =\frac{X}{n}$ where X represent the number of people with a characteristic and n the total sample size selected.

$\hat p=\frac{63}{200}=0.315$ represent the estimated proportion of people that they were planning to pursue a graduate degree

Confidence interval

The confidence interval for a proportion is given by this formula

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

For the 90% confidence interval the value of $\alpha=1-0.90=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution.