Mr. Stanford divides the students in his class into 4 equal teams. There

What is the least positive integer divisible by the numbers 2, 4, and 7?

Viefleur [7K]

Answer:

Least positive integer divisible by the numbers 2, 4, and 7 is 28

Step-by-step explanation:

We can find the least positive integer divisible by the numbers 2, 4, and 7 by taking the LCM

First lets List all prime factors for each number.

Prime Factorization of 2

2 is prime => $2^1$

Prime Factorization of 4 is:

2 x 2 => $2^2$

Prime Factorization of 7 is:

7 is prime => $7^1$

For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.

The new superset list is

2, 2, 7

Multiply these factors together to find the LCM.

LCM = 2 x 2 x 7 = 28

7 0

4 years ago

Sarah and amanda each have 2 bags with 4 marbles in each .how many marbles do they have altogether?

OLEGan [10]

16 marbles...................

7 0

3 years ago

A university wants to compare out-of-state applicants' mean SAT math scores (?1) to in-state applicants' mean SAT math scores (?

nordsb [41]

Answer:

d. Yes, because the confidence interval does not contain zero.

Step-by-step explanation:

We are given that the university looks at 35 in-state applicants and 35 out-of-state applicants. The mean SAT math score for in-state applicants was 540, with a standard deviation of 20.

The mean SAT math score for out-of-state applicants was 555, with a standard deviation of 25.

Firstly, the Pivotal quantity for 95% confidence interval for the difference between the population means is given by;

P.Q. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t__n__1-_n__2-2$

where, $\bar X_1$ = sample mean SAT math score for in-state applicants = 540

$\bar X_2$ = sample mean SAT math score for out-of-state applicants = 555

$s_1$ = sample standard deviation for in-state applicants = 20

$s_2$ = sample standard deviation for out-of-state applicants = 25

$n_1$ = sample of in-state applicants = 35

$n_2$ = sample of out-of-state applicants = 35

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(35-1)\times 20^{2} +(35-1)\times 25^{2} }{35+35-2} }$ = 22.64

<em>Here for constructing 95% confidence interval we have used Two-sample t test statistics.</em>

So, 95% confidence interval for the difference between population means ( $\mu_1-\mu_2$ ) is ;

P(-1.997 < $t_6_8$ < 1.997) = 0.95 {As the critical value of t at 68 degree

of freedom are -1.997 & 1.997 with P = 2.5%}

P(-1.997 < $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < 1.997) = 0.95

P( $-1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ${(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}$ < $1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.95

P( $(\bar X_1-\bar X_2)-1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ( $\mu_1-\mu_2$ ) < $(\bar X_1-\bar X_2)+1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.95

<u>95% confidence interval for</u> ( $\mu_1-\mu_2$ ) =

[ $(\bar X_1-\bar X_2)-1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ , $(\bar X_1-\bar X_2)+1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ]

=[ $(540-555)-1.997 \times {22.64 \times \sqrt{\frac{1}{35} +\frac{1}{35} } }$ , $(540-555)+1.997 \times {22.64 \times \sqrt{\frac{1}{35} +\frac{1}{35} } }$ ]

= [-25.81 , -4.19]

Therefore, 95% confidence interval for the difference between population means SAT math score for in-state and out-of-state applicants is [-25.81 , -4.19].

This means that the mean SAT math scores for in-state students and out-of-state students differ because the confidence interval does not contain zero.

So, option d is correct as Yes, because the confidence interval does not contain zero.

6 0

4 years ago

What is the next term in the sequence 1/2 1/6 1/18 1/54

ddd [48]

1/162
2*3=6
6*3=18
18*3=54
54*3=162

7 0

3 years ago

Can anykne solve for any of these? If yiu can put what letter you did.

Aneli [31]

A) y - 8= -2/3 (x + 3)
B) y + 1 = 4 (x - 5)
C) y - 8 = -3/7 (x - 4)
D) y - 7 = 2/3 (x - 5)
E) y + 5 = -5/3 (x + 3)

4 0

3 years ago