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JulsSmile [24]
3 years ago
15

6 2/3 divided by 2 6/7

Mathematics
2 answers:
Umnica [9.8K]3 years ago
8 0
6 2/3 divided by 2 6/7 is 2.333 continued
Arturiano [62]3 years ago
5 0

Answer:

7/3 in decimal form it's 2.3 repeated

Step-by-step explanation:

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Musya8 [376]
He gave away 14 bouncy balls
3 0
3 years ago
Question 2 (1 point)<br> How much does the prefix milli multiply the value of a base unit?
I am Lyosha [343]

Answer:1000

Step-by-step explanation:

                                     

7 0
4 years ago
If log, 7 -3 what is the value of x?<br> A. 21<br> B. 3V7<br> C. 37<br> D. 343
gregori [183]

Answer:

B

Step-by-step explanation:

logₓ 7 = 3

x³ = 7

Take cube root both sides

x= \sqrt[3]{7}

4 0
3 years ago
A recent survey of 51 students reported that the average amount of time they spent listening to music was 11.5 hours per week, w
charle [14.2K]

Answer:

90% confidence interval for the mean time per week spent listening to the radio is  11.5 \pm 1.676 \times \frac{9.2}{\sqrt{51} } .

Step-by-step explanation:

We are given that a recent survey of 51 students reported that the average amount of time they spent listening to music was 11.5 hours per week, with a sample standard deviation of 9.2 hours.

Firstly, the pivotal quantity for 90% 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 average amount of time spent listening to music = 11.5

             s = sample standard deviation = 9.2 hours

             n = sample of students = 51

             \mu = population mean per week spent listening to the radio

<em>Here for constructing 90% confidence interval we have used One-sample t test statistics as we know don't about population standard deviation.</em>

<em />

So, 90% confidence interval for the population mean, \mu is ;

P(-1.676 < t_5_0 < 1.676) = 0.90  {As the critical value of t at 50 degree of

                                        freedom are -1.676 & 1.676 with P = 5%}  

P(-1.676 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 1.676) = 0.90

P( -1.676 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 1.676 \times {\frac{s}{\sqrt{n} } } ) = 0.90

P( \bar X-1.676 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+1.676 \times {\frac{s}{\sqrt{n} } } ) = 0.90

<u><em>90% confidence interval for</em></u> \mu = [ \bar X-1.676 \times {\frac{s}{\sqrt{n} } } , \bar X+1.676 \times {\frac{s}{\sqrt{n} } } ]

                  = [ 11.5-1.676 \times {\frac{9.2}{\sqrt{51} } } , 11.5+1.676 \times {\frac{9.2}{\sqrt{51} } } ]

                  = [9.34 hours , 13.66 hours]

Therefore, 90% confidence interval for the mean time per week spent listening to the radio is [9.34 hours , 13.66 hours].

4 0
4 years ago
Helppp pls yall guys​
gizmo_the_mogwai [7]

Answer:

4/10

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
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