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3 years ago

This is the difference between the estimated value and the true value

Mathematics

1 answer:

evablogger [386]3 years ago

6 0

Answer:

Estimated is close but not completely correct, and true value has the highest accuracy.

Step-by-step explanation:

Hopefully that helped! :)

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Ryan bought some candy for Valentine's Day. Each piece of candy that Ryan buys costs $0.75 and then table below shows the cost o

s2008m [1.1K]

Answer:

where is the rest of the question?

Step-by-step explanation:

3 0

2 years ago

10 points thanks to those who help

nignag [31]

A) a 45 degree angle

3 0

3 years ago

Kaden works 5 days. The median number of hours he works is 3. The mean number of hours he works is 4.

alisha [4.7K]

Answer:

(a) 1, 2, 3, 6, 8

Step-by-step explanation:

Given

$Median = 3$

$Mean = 4$

Required

The sequence of hours worked each day

<em>See attachment for options</em>

From the question, we understand that:

$Median = 3$

This means that, the middle number is 3 (when sorted)

So, we can conclude that (b) and (c) cannot be true because their middle numbers are 6 and 5 respectively

Next, is to determine the mean of (a) and (d)

The mean of a data is calculated as:

$\bar x = \frac{\sum x}{n}$

So, we have:

(a)

$\bar x = \frac{1+ 2+ 3+ 6+ 8}{5}$

$\bar x = \frac{20}{5}$

$\bar x = 4$

(d)

$\bar x = \frac{1+ 2+ 3+ 4+ 5}{5}$

$\bar x = \frac{15}{5}$

$\bar x = 3$

Option (a) is true, because it has:

$Median = 3$

$Mean = 4$

7 0

3 years ago

Suppose that the pages per book in a library are normally distributed with an unknown mean and standard deviation. The book leng

Elanso [62]

Answer:

A t-score of 2.0244 should be used to find the 99% confidence interval for the population mean

Step-by-step explanation:

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 39 - 1 = 38

Now, we have to find a value of T, which is found looking at the t table, with 38 degrees of freedom(y-axis) and a confidence level of 0.99( $t_{99}$ ). So we have T = 2.0244.

A t-score of 2.0244 should be used to find the 99% confidence interval for the population mean

6 0

3 years ago

Given the following definition, compute Q(5). Q(n) = 0 if n = 0 2 if n = 1 4 if n = 2 Q(n − 1) + Q(n − 2) + Q(n − 3) if n > 2

aalyn [17]

We have
$Q(0)=0\\
Q(1)=2\\
Q(2)=4$
and $Q(n)=Q(n-1)+Q(n-2)+Q(n-3)\mbox{ when } n\ \textgreater \ 2.$
So
$Q(3)=Q(2)+Q(1)+Q(0)=4+2+0=6\\
Q(4)=Q(3)+Q(2)+Q(1)=6+4+2=12\\
Q(5)=Q(4)+Q(3)+Q(2)=12+6+4=22.
$

3 0

3 years ago