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3 years ago

9

The debate team is showing a video of their recent debate. The first showing begins at 3:15 p.m. The second showing starts at 4:

50 p.m with a 1/2 break between the showings. The second showing started 20 minutes late. Will the second showing be over by 6:30 p.m? EXPLAIN you answer.

1 answer:

Ghella [55]3 years ago

7 0

Given:
3:15 pm - start of first showing.
30 minutes break;
20 minutes late - 2nd showing.
4:50 - start of second showing.

4:50 - 0:20 = 4:30 should be the start of the second showing.
4:00 - 4:30 = 1/2 hour break
3:15 - 4:00 = duration of the 1st showing. 45 minutes in all.

4:50 + 0:45 = 5:35 pm end of the 2nd showing.

Yes, the 2nd showing will be over by 6:30 pm.

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Nastasia [14]

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3 years ago

ryzh [129]

A the answer is A....

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3 years ago

I NEED HELP PLEASE

nadezda [96]

Answer:

(3,2)

Step-by-step explanation:

The solution that satisfies both equations is where the two lines intersects.

The two graphs intersect at x=3 and y =2

3 0

3 years ago

Read 2 more answers

In a survey of 1,003 adults concerning complaints about restaurants, 732 complained about dirty or ill-equipped bathrooms and 38

Ganezh [65]

Answer:

a

$0.716 < p < 0.744$

b

$0.3498 < p < 0.4089$

c

With the result obtained from a and b the manager can be 95 % confidence that the proportion of the population that complained about dirty or ill-equipped bathrooms are within the interval obtained at a

and that

the proportion of the population that complained about loud or distracting diners at other tables are within the interval obtained at b

Step-by-step explanation:

From the question we are told that

The sample size is $n = 1003$

The number that complained about dirty or ill-equipped bathrooms is $e = 732$

The number that complained about loud or distracting diners at other tables is $q = 381$

Given that the the confidence level is 95% then the level of significance is mathematically represented as

$\alpha = (100- 95)\%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table , the value is

$Z_{\frac{\alpha }{2} } = 1.96$

Considering question a

The sample proportion is mathematically represented as

$\r p = \frac{e}{n}$

=> $\r p = \frac{732}{1003}$

=> $\r p = 0.73$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{ \r p (1- \r p)}{n} }$

$E = 1.96* \sqrt{ \frac{ 0.73 (1- 0.73)}{1003} }$

$E = 0.01402$

The 95% confidence interval is

$\r p - E < p < \r p +E$

$0.73 - 0.01402 < p < 0.73 + 0.01402$

$0.716 < p < 0.744$

Considering question b

The sample proportion is mathematically represented as

$\r p = \frac{q}{n}$

=> $\r p = \frac{381}{1003}$

=> $\r p = 0.3799$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{ \r p (1- \r p)}{n} }$

$E = 1.96* \sqrt{ \frac{ 0.3799 (1- 0.3799)}{1003} }$

$E = 0.0300$

The 95% confidence interval is

$\r p - E < p < \r p +E$

$0.3798 - 0.0300 < p < 0.3798 + 0.0300$

$0.3498 < p < 0.4089$

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3 years ago

Please help!!<br><br><br> What is the value of x??

alexira [117]

Answer:

x = 58°

Step-by-step explanation:

angle formed by Tangent and secant: (far arc - near arc) / 2

(160 - x) / 2 = 51

160 -x = 102

x = 160 - 102 = 58°

4 0

3 years ago