Giving brainlyest to the correct answer!!

You measure 25 turtles' weights, and find they have a mean weight of 31 ounces. Assume the population standard deviation is 12.8

Firlakuza [10]

Answer:

Margin of error = 4.21 ounces

Step-by-step explanation:

According to the Question,

Given That, You measure 25 turtles' weights, and find they have a mean weight of 31 ounces. Assume the population standard deviation is 12.8 ounces

Therefore, Sample mean = 31 ounces , Sample size(n) = 25 , Alpha(α) = 0.10 & Population standard deviation(σ) = 12.8 ounces

Thus, Margin of error = $Z_{critical}$ × σ / √n ( $Z_{critical}$ at α=.010 is 1.645)

Putting The Values, We get

1.645 × (12.8 / √25 ) ⇒ 4.2112 ≈ 4.21

Thus, the maximum margin of error associated with a 90% confidence interval for the true population mean turtle weight is 4.21 ounces

8 0

3 years ago

Witch of the following are categorical data

Alisiya [41]

Step-by-step explanation:

Examples of categorical variables are race, genders, ages, and education levels. While the closing two variables may be considered in a numerical manner by using exact values for age and the high grade completed, it is always informative to put such variables into a relatively small number of groups.

6 0

3 years ago

If Whitney wrote the decimal representations for the first 300 positive integer multiples of 5 and did not write any other numbe

fiasKO [112]

Answer:

201 times

Step-by-step explanation:

Since it's the first 300 positive integer multiples of 5, then the last integer will be: 300 × 5 = 1500

While the first is 5.

The sequence is like this;

5, 10, 15, 20, 25, 30, 35, .. 1500

Since for every 2 integers that are multiples of 5, 1 will include the digit 5, then it means that, number of times the unit place will have 5 is; 300/5 = 150 times

Now, between 5 and 100, the only value that has 5 in it's tense place is 50 & 55.

So for every 100 numbers, we have 2 times to write 5 in the tens place. Thus, for 1500 numbers, we will write 5 in the tens place: 1500/100 × 2 = 30 times

Now, for the hundreds place, from 500 and 600, we have 21 multiples of 5 inclusive of 500 and 600 but since we want the one that has 5 in the hundreds place, then it is 20 as they all start with 5 excluding 600.

Also, from 1000 to 1500, the only number that has 5 in its hundreds place is 1500.

Thus,total times 5 is written in the hundreds place = 21 times

Total number of times 5 is written = 150 + 30 + 21 = 201 times

5 0

3 years ago

The drama club is selling tickets to its play. An adult ticket costs $15 and a student ticket costs $11. The auditorium will sea

spin [16.1K]

Answer:

83 adult tickets and 217 student tickets.

Step-by-step explanation:

Let number of adult tickets sold = $x$

Given that total number of tickets = 300

So, number of student tickets = 300 - $x$

Cost of adult ticket = $15

Cost of student ticket = $11

Total collection from adult tickets = $ $15x$

Total collection from student tickets = $(300-x)\times 11 = 3300-11x$

Given that overall collection = $3630

$15x+(3300-11x) = 3630\\\Rightarrow 15x-11x=3630-3300\\\Rightarrow 4x = 330\\\Rightarrow x = 82.5$

So, for atleast $3630 collection, there should be 83 adult tickets and (300-83 = 217 student tickets.

Now , collection = $3632

4 0

3 years ago

The lee's spend $ 31 on movie tickets for 2 adults and 3 children. The Smith's spent $ 26 on movie tickets for 2 adults and 2 ch

Trava [24]

Answer:

Cost of adult ticket is $8 and Cost of children ticket is $5.

Step-by-step explanation:

Given:

Let the cost of each Adult ticket be 'x'.

Let the cost of each child ticket be 'y'.

Now Given:

The lee's spend $ 31 on movie tickets for 2 adults and 3 children.

Number of adult tickets = 2

Number of children ticket = 3

Total money spent = $31

So we can say that;

Total money spent is equal to sum Number of adult tickets multiplied by cost of each Adult ticket and sum Number of children tickets multiplied by cost of each child ticket.

framing in equation form we get;

$2x+3y=31 \ \ \ \ equation \ 1$

Also Given:

The Smith's spent $ 26 on movie tickets for 2 adults and 2 children

Number of adult tickets = 2

Number of children ticket = 2

Total money spent = $26

So we can say that;

Total money spent is equal to sum Number of adult tickets multiplied by cost of each Adult ticket and sum Number of children tickets multiplied by cost of each child ticket.

framing in equation form we get;

$2x+2y=26 \ \ \ \ equation \ 2$

Now Subtracting equation 2 from equation 1 we get;

$2x+3y-(2x+2y)=31-26\\\\2x+3y-2x-2y=5\\\\y=\$5$

Now substituting the value of y in equation 1 we get;

$2x+3y=31\\\\2x+3\times5 =31\\\\2x+15=31\\\\2x=31-15\\\\2x=16x\\\\\frac{2x}{2}\frac{16}{2}\\\\x=\$8$

Hence Cost of adult ticket is $8 and Cost of children ticket is $5.

6 0

3 years ago