Greatest common factor

How do i write the number 8.517 in two other forms

marta [7]

Improper fraction 8517/1000 and as the mixed number 8 and 517/1000

6 0

4 years ago

an institution has 500 full time students and part time students. the institution charges each full time student a $20 activity

earnstyle [38]

350 of the 20 dollar student and 150 of the 15 dollar ones

8 0

4 years ago

Find mZLKM. м. M K/(3x – 70) 60° L mZLKM= l

Bingel [31]

Answer:

125

Step-by-step explanation:

( 3x - 70 ) is exterior angle.

60 and x are interior angles.

Formula : -

Exterior angle of a triangle

= Sum of two interior angles in a triangle

3x - 70 = x + 60

3x - x = 60 + 70

2x = 130

x = 130 / 2

x = 65

m∠LKM = 3x - 70

= 3 ( 65 ) - 70

= 195 - 70

m∠LKM = 125

8 0

3 years ago

In 1998, as an advertising campaign, the Nabisco Company announced a "1000 Chips Challenge," claiming that every 18-ounce bag of

Phoenix [80]

Answer:

A 95% confidence interval for the mean number of chips in a bag of Chips Ahoy Cookies is [1187.96, 1288.44].

Step-by-step explanation:

We are given that statistics students at the Air Force Academy (no kidding) purchased some randomly selected bags of cookies and counted the chocolate chips.

Some of their data are given below; 1219, 1214, 1087, 1200, 1419, 1121, 1325, 1345, 1244, 1258, 1356, 1132, 1191, 1270, 1295, 1135.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean number of chocolate chips = $\frac{\sum X}{n}$ = 1238.2

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 94.3

n = sample of car drivers = 16

$\mu$ = population mean number of chips in a bag

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.131 < $t_1_5$ < 2.131) = 0.95 {As the critical value of t at 15 degrees of

freedom are -2.131 & 2.131 with P = 2.5%}

P(-2.131 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.131) = 0.95

P( $-2.131 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.131 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.131 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.131 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $1238.2-2.131 \times {\frac{94.3}{\sqrt{16} } }$ , $1238.2+2.131 \times {\frac{94.3}{\sqrt{16} } }$ ]

= [1187.96, 1288.44]

Therefore, a 95% confidence interval for the mean number of chips in a bag of Chips Ahoy Cookies is [1187.96, 1288.44].

7 0

3 years ago

Somebody please help me. I cant figure this out once so ever.

WITCHER [35]

Answer:

A

Step-by-step explanation:

The Fibonacci sequence is the sum of the two terms before it. It looks like this:

1, 1, 2, 3, 5, 8, 13, 21, 34, ...

As you can see, 13 and 21 are indeed consecutive terms.

7 0

3 years ago