Confidence intervals for the means provide an estimate for where the true mean lies.

Mathematics

1 answer:

stepladder [879]1 year ago

3 0

It is true that the confidence intervals for the mean provide an estimate for where the true mean lies.

In statistics, a confidence interval denotes the likelihood that a population parameter will fall between a set of values for a given proportion of the time. A confidence interval depicts the likelihood that a parameter will fall between two values near the mean. Confidence intervals quantify the degree of uncertainty or certainty in a sampling procedure.

The mean is a basic mathematical average of two or more values. There are two sorts of means that may be calculated: the arithmetic mean and the geometric mean. A mean tells you the average of a bunch of values, which helps you contextualize each data point.

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Find the right quotient. 36m 5 n 5 ÷ (12m 3)

polet [3.4K]

ANSWER

The right quotient is

$3 {m}^{2} {n}^{5}$

EXPLANATION

The given expression is :

$\frac{36 {m}^{5} {n}^{5} }{12 {m}^{3} }$

$\frac{36 {m}^{5} {n}^{5} }{12 {m}^{3} } = \frac{36}{12} \times \frac{ {m}^{5} }{ {m}^{3}} \times {n}^{5}$

$\frac{36 {m}^{5} {n}^{5} }{12 {m}^{3} } = 3\times \frac{ {m}^{5} }{ {m}^{3}} \times {n}^{5}$

Recall that,

$\frac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n}$
We apply this property to obtain;

$\frac{36 {m}^{5} {n}^{5} }{12 {m}^{3} } = 3 \times {m}^{5 - 3} \times {n}^{5}$

$\frac{36 {m}^{5} {n}^{5} }{12 {m}^{3} } = 3 {m}^{2} {n}^{5}$

5 0

3 years ago

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Sam invests £1800 in the bank for four

KIM [24]

Answer:

£2088

US$ 2,536.48 dollars (2021)

Step-by-step explanation:

4% of 1800 = 72

72 x 4 = 288

1800 + 288 = 2088