eight out of 10 dentist prefer Crest toothpaste.What is the ratio of dentist whi prefer Crest to those who do

The average of a set of 18 consecutive integers is 22.5. what is the smallest integer in the set?

quester [9]

The sum of the 18 numbers is 22.5 x 18 = 405.
Let the numbers be x, x + 1, x + 2, . . ., x + 17

Sum of n term of an arithmetic sequence = n/2(a + l)
18/2(x + x + 17) = 405
9(2x + 17) = 405
2x + 17 = 405/9 = 45
2x = 45 - 17 = 28
x = 28/2 = 14

Therefore, the smallest integer is 14.

6 0

2 years ago

the average number of spectators at a football competition for the first five days was 3144.the attendance on the sixth day was

Ksivusya [100]

Answer:

3285

Step-by-step explanation:

5x3144=15720

3990

15720+3990=19710

19710/6=3285

8 0

2 years ago

A random sample of 250 students at a university finds that these students take a mean of 15.2 credit hours per quarter with a st

Orlov [11]

Answer:

95% confidence interval for the mean credit hours taken by a student each quarter is [14.915 hours , 15.485 hours].

Step-by-step explanation:

We are given that a random sample of 250 students at a university finds that these students take a mean of 15.2 credit hours per quarter with a standard deviation of 2.3 credit hours.

Firstly, the pivotal quantity for 95% 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 credit hours per quarter = 15.2 credit hours

s = sample standard deviation = 2.3 credit hours

n = sample of students = 250

$\mu$ = population mean credit hours per quarter

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

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

P(-1.96 < $t_2_4_9$ < 1.96) = 0.95 {As the critical value of t at 249 degree of

freedom are -1.96 & 1.96 with P = 2.5%}

P(-1.96 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $15.2-1.96 \times {\frac{2.3}{\sqrt{250} } }$ , $15.2+1.96 \times {\frac{2.3}{\sqrt{250} } }$ ]

= [14.915 hours , 15.485 hours]

Therefore, 95% confidence interval for the mean credit hours taken by a student each quarter is [14.915 hours , 15.485 hours].

The interpretation of the above confidence interval is that we are 95% confident that the true mean credit hours taken by a student each quarter will be between 14.915 credit hours and 15.485 credit hours.

8 0

1 year ago

francesca wants to give one of her photographs to a friend as a gift. help her choose the photograph and determine the style and

emmainna [20.7K]

It's a photo of a bridge & it's 3x5" (inches)

3 0

2 years ago

The means and mean absolute deviations of the individual times of members of two relay swim teams are shown in the table below.

shepuryov [24]

See the attached picture for the solution:

7 0

2 years ago

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