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3 years ago

What is the confidence level (written as a percentage) for the interval ? (round your answer to 2 decimal places.)?

Mathematics

2 answers:

valentinak56 [21]3 years ago

7 0

You don't provide any info other than the question.

Whitepunk [10]3 years ago

6 0

I think the question is incorrect !

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2x + y = -3<br> 5x - y = -4

Georgia [21]

What do u need help on?

5 0

3 years ago

The total price for 7 movie tickets is 92.40 . If each ticket costs the same amount, how much does one movie ticket cost?

horsena [70]

Answer:

$13.20

Step-by-step explanation:

total cost of tickets = $92.40

total number of tickets = 7

cost of each ticket

= total cost of tickets ÷ total number of tickets

= 92.40 ÷ 7

= $13.20

hope this helps

8 0

3 years ago

Read 2 more answers

Can somebody please help me out ?

Sever21 [200]

Answer:

Step-by-step explanation:

238,900/7,917=30.1755

7 0

3 years ago

Read 2 more answers

Please anyone help please ASAP

denis23 [38]

Answer:

Angle WTS will be 180 degree minus 15x. You can write this as an expression:

180-15x

Now you have an expression for each angle in the triangle. Triangles will always have 180 degrees in them. You can add the expressions together and set it equal to 180. Then you can solve the equation for the x variable.

Step-by-step explanation:

Angle SWT + angle WTS + angle TSW = 180 degrees

Replace each angle with the expression then solve.

3 0

3 years ago

Read 2 more answers

A study sampled 350 upperclassmen (Group 1) and 250 underclassmen (Group 2) at high schools around the city of Portland. The stu

iVinArrow [24]

Answer:

$-0.061 < P_1 -P_2< 0.025$

Step-by-step explanation:

Give data:

$n_1 = 350$

$n_2 =250$

$x_1 = 23$

$x_2 = 21$

$\hat{P} 1 = \frac{x_1}{n_1} = \frac{23}{350} = 0.066$

$\hat{P} 2 = \frac{x_2}{n_2} = \frac{21}{250} = 0.084$

for 95% confidence interval

$\alpha = 1 - 0.95 = 0.05 and \alpha/2 = 0.025$

$z_{\alpha/2} = 1.96$ from standard z- table

confidence interval for P_1 and P_2 is

$\hat{P} 1 - \hat{P} 2 \pm z_{\alpja/2} \sqrt{\frac{\hat{P} 1(1-\hat{P} 1)}{n_1} +\frac{\hat{P} 2(1-\hat{P} 2)}{n_2}}$

$(0.066 - 0.084) \pm 1.96 \sqrt{\frac{0.066(1-0.066)}{350} +\frac{0.084(1-0.084)}{250}}$

$-0.018 \pm 0.043$

confidence interval is

$-0.018 - 0.043 < P_1 -P_2$

$-0.061 < P_1 -P_2< 0.025$

7 0

3 years ago