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Eduardwww [97]
3 years ago
8

Math im soooooooooooooo baaaaaaaaaaaadddddddddddddd

Mathematics
2 answers:
storchak [24]3 years ago
8 0

Answer:

The correct answer is A, 73

Step-by-step explanation:

First we have to arrange the whole numbers in ascending order. i-e from smallest to largest. Showing it as follows:

53 54 59 62 64 65 66 68 70 71 75 78 79 79 83 83 86 90 91 94                

Now we would find the median by taking the middle number. But this series consists of even numbers. There are 20 numbers. So we will take the middle two numbers and find their average like follows:

71 + 75 = 146/2  = 73

The answer is 73

lozanna [386]3 years ago
7 0

Answer:

median =73

Step-by-step explanation:

Given list of data is

{94,79,83,78,70,66,68,75,53,54.79,59,83,91,64,65,71,62,86,90}.

Now we need to find about what is the median for the data listed above.

So first we need to rearrange the given data in increasing order. then we can easily find the middle number as median.

Sorted list of data is:

{53, 54, 59, 62, 64, 65, 66, 68, 70, 71, 75, 78, 79, 79, 83, 83, 86, 90, 91, 94}

there are two numbers at the middle 71 and 75

then median = (71+75)/2=146/2=73

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Step-by-step explanation:

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3 years ago
The FDA regulates that a fish that is consumed is allowed to contain at most 1 mg/kg of mercury. In Florida, bass fish were coll
levacccp [35]

Answer:

Yes. At a significance level of 0.05, there is enough evidence to support the claim that the fish in all Florida lakes have different mercury than the allowable amount.

The random variable is the sample mean amount of mercury in the bass fish from the lakes of Florida.

The population parameter is the mean amount of mercury in the bass fish of Florida lakes.

The alternative hypothesis (Ha) states that the amount of mercury significantly differs from 1 mg/kg.

The null hypothesis (H0) states that the amount of mercury is not significantly different from 1 mg/kg.

H_0: \mu=1\\\\H_a:\mu\neq 1

Step-by-step explanation:

<em>The question is incomplete.</em>

<em>There is no data provided.</em>

<em>We will work with a sample mean of 0.95 mg/kg and sample standard deviation of 0.15 mg/kg to show the procedure.</em>

<em />

This is a hypothesis test for the population mean.

The claim is that the fish in all Florida lakes have different mercury than the allowable amount (1 mg of mercury per kg of fish).

Then, the null and alternative hypothesis are:

H_0: \mu=1\\\\H_a:\mu\neq 1

The significance level is assumed to be 0.05.

The sample has a size n=53.

The sample mean is M=0.95.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.15.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.15}{\sqrt{53}}=0.0206

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{0.95-1}{0.0206}=\dfrac{-0.05}{0.0206}=-2.427

The degrees of freedom for this sample size are:

df=n-1=53-1=52

This test is a two-tailed test, with 52 degrees of freedom and t=-2.427, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=2\cdot P(t

As the P-value (0.019) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that the fish in all Florida lakes have different mercury than the allowable amount.

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Answer:

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Step-by-step explanation:

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