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3 years ago

Which choice does NOT represent the sample space for rolling a six-sided die one time?

Mathematics

2 answers:

Setler79 [48]3 years ago

4 0

Answer:

for imagine math it is the top left corner choice that says 1 with two lines coming from it and then a 2 and 3 from each line and so on until 6

Step-by-step explanation:

Karolina [17]3 years ago

3 0

On Imagine math, It is the very first choice, with the 1 at the top and then the 2 and the 3, then the 4 and 5 and 6. Hope this helped!

Step-by-step explanation:

You might be interested in

A circular pizza box logo has a sector with a central angle of 120 degrees and a diameter of 9cm. Find the area.

Mumz [18]

Answer:

A circular pizza box logo has a sector with a central angle of 40 degrees and a diameter of 25 inches. Find the area of this sector.

-------

r = d/2 = 12.5 inches

Step-by-step explanation:

4 0

3 years ago

Chen and Dave divide £350 in the ratio 3:4 ed Work out how much money Dave gets

kodGreya [7K]

Answer:

£200

Step-by-step explanation:

Sum the parts of the ratio, 3 + 4 = 7 parts

Divide the amount by 7 to find the value of one part of the ratio

£350 ÷ 7 = £50 ← value of 1 part of the ratio , then

4 parts = 4 × £50 = £200 ← amount Dave gets

5 0

3 years ago

Jada and Diego are practicing the piano for an upcoming rehearsal. The number of minutes Jada practiced is listed below:

katen-ka-za [31]

Answer:

Step-by-step explanation:

First, rewrite the numbers in numerical order:

8, 10, 10, 15, 15, 20, 20, 20, 20, 25, 25, 25, 35

Then find the middle number: (note there's 6 numbers on each side of 20)

8 10 10 15 15 20 20 20 20 25 25 25 35

Hope this helps!

7 0

3 years ago

How many years will it take $39,450 to double at an interest rate of 3.2%, continuously compounded? *

taurus [48]

Years = natural log (total / principal) / rate
Years = ln (78,900 / 39,450) / .032
Years = ln (2) / .032
Years = 0.69314718056 / .032
Years = 21.6608493925 
And here's a calculator to check that answer:
http://www.1728.org/compint.htm

4 0

3 years ago

A college student is interested in investigating the TV-watching habits of her classmates and surveys 20 people on the number of

Alla [95]

Answer:

80% confidence interval of the true average number of hours of TV watched per week is [8.28 hours, 11.02 hours].

Step-by-step explanation:

We are given that a college student is interested in investigating the TV-watching habits of her classmates and surveys 20 people on the number of hours they watch per week. The results are provided below;

Hours of TV per week (X): 6, 14, 13, 6, 16, 10, 19, 4, 5, 5, 18, 8, 7, 14, 8, 8, 9, 12, 6, 5.

Firstly, the Pivotal quantity for 80% confidence interval for the true average is given by;

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

where, $\bar X$ = sample mean number of hours of TV watched per week = $\frac{\sum X}{n}$ = 9.65

s = sample standard deviation = $\sqrt{\frac{\sum (X -\bar X)^{2} }{n-1} }$ = 4.61

n = sample of people = 20

$\mu$ = true average number of hours of TV watched per week

Here for constructing 80% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.

So, 80% confidence interval for the true average, $\mu$ is ;

P(-1.33 < $t_1_9$ < 1.33) = 0.80 {As the critical value of t at 19 degrees of

freedom are -1.33 & 1.33 with P = 10%}

P(-1.33 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.33) = 0.80

P( $-1.33 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.33 \times {\frac{s}{\sqrt{n} } }$ ) = 0.80

P( $\bar X-1.33 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.33 \times {\frac{s}{\sqrt{n} } }$ ) = 0.80

80% confidence interval for $\mu$ = [ $\bar X-1.33 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.33 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $9.65-1.33 \times {\frac{4.61}{\sqrt{20} } }$ , $9.65+1.33 \times {\frac{4.61}{\sqrt{20} } }$ ]

= [8.28 hours, 11.02 hours]

Therefore, 80% confidence interval of the true average number of hours of TV watched per week is [8.28 hours, 11.02 hours].

7 0

3 years ago