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3 years ago

15

A wheel has ten chairs and people are seated two per chair. Every minute a chair passes the exit platform. The wheel opens at 11

am for thirty minutes – how many people took a ride in that time?

1 answer:

Anit [1.1K]3 years ago

6 0

I’m not really sure,but I got 60

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Help please ASAP. Correct answer will get brainly

djverab [1.8K]

Answer:

I think it's 3,6.....................

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2 years ago

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The Parkview middle school football team will be playing in the state championship the cochairs free tickets that he would be di

ser-zykov [4K]

Answer: 156

Step-by-step explanation: 52*3=156 tickets

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3 years ago

Find the interquartile range of these data.<br><br> 5, 12, 9, 8, 6, 2, 6

Kryger [21]

Answer: 7

Step-by-step explanation:

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3 years ago

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HELPPPPPP ASAP What is the length of the third side of the window frame below? (5 points) (Figure is not drawn to scale.) 58a 1)

ollegr [7]

Answer:

The answer is 3) 40.

Step-by-step explanation:

since you're working with the Pythagorean theorem, you should know that a^2+b^2=c^2, therefore, we know that we only have two sides. we have a and c. we know that a = 42 and c = 58, giving us the equation of 42^2+b^2=58^2,

we want to find b!

so we would simply subtract the two.

58^2 - 42^2 = b

b = 1600

now we want to square root,

b = sqrt1600

which in the end, gives us the answer of 40.

4 0

2 years ago

In a simple random sample of 300 boards from this shipment, 12 fall outside these specifications. Calculate the lower confidence

Lyrx [107]

Answer:

The 95% confidence interval for the percentage of all boards in this shipment that fall outside the specification is (1.8%, 6.2%).

Step-by-step explanation:

In a random sample of 300 boards the number of boards that fall outside the specification is 12.

Compute the sample proportion of boards that fall outside the specification in this sample as follows:

$\hat p =\frac{12}{300}=0.04$

The (1 - <em>α</em>)% confidence interval for population proportion <em>p</em> is:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

The critical value of <em>z</em> for 95% confidence level is,

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use a <em>z</em>-table.

Compute the 95% confidence interval for the proportion of all boards in this shipment that fall outside the specification as follows:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.04\pm1.96\sqrt{\frac{0.04(1-0.04)}{300}}\\=0.04\pm0.022\\=(0.018, 0.062)\\\approx(1.8\%, 6.2\%)$

Thus, the 95% confidence interval for the proportion of all boards in this shipment that fall outside the specification is (1.8%, 6.2%).

6 0

3 years ago