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2 years ago

List the common factor 30 and 40

1 answer:

3 0

The factors of 30 are 1, 2, 3, 5, 6, 10, 15, and 30. The factors of 40 are 1, 2, 4, 5, 8, 10, 20, and 40. The common factors of 30 and 40 are 1, 2, 5, and 10, and the GCF is 10.

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A total of 276 tickets were sold for the school play. They were either adult tickets or student tickets. The number of student t

NISA [10]

Consider the number of the adult tickets is x
X+2x=276
3x=276
X=92
So the student tickets is =184

4 0

2 years ago

Women's heights average 65 inches with a standard deviation of 2.2 inches in a certain study. Use the 68-95-99.7 rule to determi

vodomira [7]

• We are told that the average height : Mean () = 65

• Standard deviation : SD () = 2.2

• 95 % 0f women height range = +- 3

= 65 +-3( 2.2)

= {65 + 6.6 ; 65 -6.6}

= {71.6 ; 58.4}

• This means that 95 % 0f women height will range between ( 58.4 ) and (71.6 )

3 0

1 year ago

Jerry has taken a random sample of students and determined the number of electives that each student in his sample took last yea

loris [4]

The sample proportion of the the students that took fewer than the mean number of electives is given by the number of people who took less that the mean number of electives (i.e. 7.63) divided by the total number of students in the sample (i.e. 19). The total number of people who took less than 7.63 electives is 10. Therefore, the required sample proportion is 10 / 19.

Read more on Brainly.com - brainly.com/question/1687069#readmore

6 0

3 years ago

X ſocm Jicm A) 2V5 cm B) V13 cm D) V19 cm C) 1 cm ASAP

bulgar [2K]

Answer:

using pythagoras theorem

hyp 2=opp 2+adj 2

=( √6)(√6)+(√7)(√7)

hyp 2=13

hyp=√13

3 0

3 years ago

Read 2 more answers

A researcher records the repair cost for 8 randomly selected washers. A sample mean of $60.46 and standard deviation of $18.36 a

Anuta_ua [19.1K]

Answer:

The critical value is T = 1.895.

The 90% confidence interval for the mean repair cost for the washers is between $48.159 and $72.761

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution.

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 = 8 - 1 = 6

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 6 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.9}{2} = 0.95$ . So we have T = 1.895, which is the critical value.

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.895\frac{18.36}{\sqrt{8}} = 12.301$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 60.46 - 12.301 = $48.159

The upper end of the interval is the sample mean added to M. So it is 60.46 + 12.301 = $72.761

The 90% confidence interval for the mean repair cost for the washers is between $48.159 and $72.761

8 0

3 years ago