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2 years ago

You are playing a board game with a spinner that has numbers 1, 2, 3, and 4 on

Mathematics

1 answer:

Whitepunk [10]2 years ago

6 0

Answer:

1/4

Step-by-step explanation:

Let x = number the spinner lands on

P(X = 1) = 1/4

P(X = 2) = 1/4

P(X = 3) = 1/4

P(X = 4) = 1/4

The spins are independent events as their probabilities are not affected by whether or not the other event happens

Therefore, the probability that you land on a number 2 given that your first spin landed on number 1 is still 1/4

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0.000007 in standform

Murljashka [212]

Answer:

7.0 x 10 to the 6th

Step-by-step explanation:

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2 years ago

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Round to the nearest hundredth if necessary. <br> 2 qt =______ mL<br> help me pleaseeeee

Rasek [7]

It would be 1892.71

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3 years ago

If a rectangular region has perimeter p inches and area a square inches, is the region square?

Flura [38]

If this is all the information you have, it is not enough to conclude the region is square. we need more information

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2 years ago

The graph shows the velocity f(t) of a runner during a certain time interval:

omeli [17]

Answer:

Option B

Step-by-step explanation:

(0,2)
(6,8)
(8,0)

Slope

m=8-2/6
m=6/6=1

And

m=0-8/8-6
m=-8/2
m=-4

These are acceleration

Y intercept is (0,2) as x coordinate is 0

This is initial velocity 2m/s

The runner stopped at 8s

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2 years ago

The attention span of children (ages 3 to 5) is claimed to be Normally distributed with a mean of 15 minutes and a standard devi

Agata [3.3K]

Answer:

If the observed sample mean is greater than 17.07 minutes, then we would reject the null hypothesis.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 15 minutes

Sample size, n = 10

Alpha, α = 0.05

Population standard deviation, σ = 4 minutes

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 15\text{ minutes}\\H_A: \mu > 15\text{ minutes}$

Since the population standard deviation is given, we use one-tailed z test to perform this hypothesis.

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Now, $z_{critical} \text{ at 0.05 level of significance for one tail} = 1.64$

Thus, we would reject the null hypothesis if the z-statistic is greater than this critical value.

Thus, we can write:

$z_{stat} = \displaystyle\frac{\bar{x} - 15}{\frac{4}{\sqrt{10}} } > 1.64\\\\\bar{x} - 15>1.64\times \frac{4}{\sqrt{10}}\\\bar{x} - 15>2.07\\\bar{x} > 15+2.07\\\bar{x} > 17.07$

Thus, the decision rule would be if the observed sample mean is greater than 17.07 minutes, then we would reject the null hypothesis.

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3 years ago