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3 years ago

Please answer the followingplease explain how you got your answer

Mathematics

1 answer:

Liula [17]3 years ago

4 0

Answer:

P(1) = 4/20 or 1/5

P(2) = 3/20

P(3) = 7/20

P(4) = 6/20 or 3/10

Step-by-step explanation:

Crystal spun the spinner 20 times. 1 was the outcome 4 times out of 20 spins, so the experimental probability would be 4/20 or 1/5. Same goes for all the other ones.

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A jar contains 0.75 liter of blackberry juice and 0.60 liter of blueberry juice. Patrick poured 0.25 liter of guava juice into t

mixas84 [53]

Answer:

1.40 litres; order of operations

Step-by-step explanation:

Part A:

First, write it out in the full form.

#1 (.75 litres of blackberry juice + .60 litres of blueberry juice + .25 litres of guava juice) - .20 litres drank = 1.4 litres of juice remaining in the jar

Part B:

Removing the wordy descriptions from the expression, we get the following:

#2 (.75 + .60 +.25) - .20

or, by adding the numbers in the parentheses first:

#3 (1.60) - .20 = 1.40

#4 1.40 litres of juice left in the jar

The order of operations was used in moving from step #2 to step #3, as operations within the parentheses were conducted before subtracting the .20 litres of juice which were drank.

8 0

3 years ago

You are trying to catch a train, which will be leaving in 2 minutes. There are 2 kilometers between you and the train station. I

noname [10]

K, you have 2 minutes, one minute for one kilometer.
If you're in 60km/h, you're on 1km/m, so, if you cover 1 km in one minute, you need to stay in that speed to get to the train station.

4 0

3 years ago

What is the value of z?

Amiraneli [1.4K]

Answer:

if a z-score is equal to +1, it is 1 standard deviation above the mean.

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

The mean number of words per minute (WPM) read by sixth graders is 8888 with a standard deviation of 1414 WPM. If 137137 sixth g

Bingel [31]

Noticing that there is a pattern of repetition in the question (the numbers are repeated twice), we are assuming that the mean number of words per minute is 88, the standard deviation is of 14 WPM, as well as the number of sixth graders is 137, and that there is a need to estimate the probability that the sample mean would be greater than 89.87.

Answer:

"The probability that the sample mean would be greater than 89.87 WPM" is about $\\ P(z>1.56) = 0.0594$ .

Step-by-step explanation:

This is a problem of the distribution of sample means. Roughly speaking, we have the probability distribution of samples obtained from the same population. Each sample mean is an estimation of the population mean, and we know that this distribution behaves normally for samples sizes equal or greater than 30 $\\ n \geq 30$ . Mathematically

$\\ \overline{X} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$ [1]

In words, the latter distribution has a mean that equals the population mean, and a standard deviation that also equals the population standard deviation divided by the square root of the sample size.

Moreover, we know that the variable Z follows a normal standard distribution, i.e., a normal distribution that has a population mean $\\ \mu = 0$ and a population standard deviation $\\ \sigma = 1$ .

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [2]

From the question, we know that

The population mean is $\\ \mu = 88$ WPM
The population standard deviation is $\\ \sigma = 14$ WPM

We also know the size of the sample for this case: $\\ n = 137$ sixth graders.

We need to estimate the probability that a sample mean being greater than $\\ \overline{X} = 89.87$ WPM in the distribution of sample means. We can use the formula [2] to find this question.

The probability that the sample mean would be greater than 89.87 WPM

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ Z = \frac{89.87 - 88}{\frac{14}{\sqrt{137}}}$

$\\ Z = \frac{1.87}{\frac{14}{\sqrt{137}}}$

$\\ Z = 1.5634 \approx 1.56$

This is a standardized value and it tells us that the sample with mean 89.87 is 1.56 standard deviations above the mean of the sampling distribution.

We can consult the probability of P(z<1.56) in any cumulative standard normal table available in Statistics books or on the Internet. Of course, this probability is the same that $\\ P(\overline{X} < 89.87)$ . Then