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3 years ago

What is 89,659rounding to the nearest hundred thousand

Mathematics

1 answer:

algol [13]3 years ago

3 0

89,659 rounded to the nearest hundred thousand = 100,000

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Step-by-step explanation:

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2 years ago

The frequency (F) and length (L) of a wave are related by the following formula: F x L = 3.0 x 10^10. Consider that the waveleng

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Which statement is true? Lisa has one and four twelves cups of grapes. Dawn has one and eleven twelves cups of grapes. Together

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Answer:

The fourth one: Gavin has one and thirteen fourteenths buckets of water. Daven has one and three fourteenths buckets of water. Together they have three and two fourteenths buckets of water.

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lozanna [386]

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2 years ago

g 78% of all Millennials drink Starbucks coffee at least once a week. Suppose a random sample of 50 Millennials will be selected

Tatiana [17]

Answer:

We know that n = 50 and p =0.78.

We need to check the conditions in order to use the normal approximation.

$np=50*0.78=39 \geq 10$

$n(1-p)=50*(1-0.78)=11 \geq 10$

Since both conditions are satisfied we can use the normal approximation and the distribution for the proportion is given by:

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

With the following parameters:

$\mu_ p = 0.78$

$\sigma_p = \sqrt{\frac{0.78*(1-0.78)}{50}}= 0.0586$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

We know that n = 50 and p =0.78.

We need to check the conditions in order to use the normal approximation.

$np=50*0.78=39 \geq 10$

$n(1-p)=50*(1-0.78)=11 \geq 10$

Since both conditions are satisfied we can use the normal approximation and the distribution for the proportion is given by:

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

With the following parameters:

$\mu_ p = 0.78$

$\sigma_p = \sqrt{\frac{0.78*(1-0.78)}{50}}= 0.0586$

6 0

3 years ago