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Delicious77 [7]
2 years ago
7

Please Help ASAP!! Due today!!

Mathematics
1 answer:
emmasim [6.3K]2 years ago
3 0

Answer:

-4 equals p=2 and a and R are at point Y downwards.

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The table below describes a sequence of transformations that justify that two figures are similar. Which sentence below shows th
yuradex [85]

Answer:

dilation

Step-by-step explanation:

whenever the shape is dilated, or scaled by a factor of any number, the shape will change size and will become similar rather than congruent. congruent shapes have the same dimensions as the parent or original shape.

7 0
3 years ago
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The values of the 2s in 2,201,538 are.
Ronch [10]

Answer:

First Two: 2 million,

Second Two: 2  hundred thousand

8 0
3 years ago
5x - 4y = 19 <br> x + 2y = 8
klemol [59]

Answer:

(5x-4y)=19

5×19-4×19

95-76

19

Step-by-step explanation:

x+2y=8

x=8-2

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7 0
3 years ago
A spinner has three unusual sections: red, yellow, and blue.. The table shows the results of Nolan’s spins. Find the probability
Anika [276]

Answers:

  • Red = 1/3
  • Yellow = 7/15
  • Blue = 1/5

====================================================

Explanation:

Add up all the frequencies in the table to see that there are 120+168+72 = 360 spins total.

  • Of those 360 spins, 120 of them land on red. The probability of landing on red is 120/360 = (120*1)/(120*3) = 1/3
  • Since 168 land on yellow, this means the probability of landing on yellow is 168/360 = (24*7)/(24*15) = 7/15
  • Lastly, we have 72 occurrences of blue out of 360 total spins. The probability of landing on blue is 72/360 = (72*1)/(72*5) = 1/5
4 0
3 years ago
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 <em>distribution of sample means</em>. 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 <em>normally</em> 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 <em>normal standard distribution</em>, 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 <em>distribution of sample means</em>. 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 <em>standardized value </em> and it tells us that the sample with mean 89.87 is 1.56<em> standard deviations</em> <em>above</em> the mean of the sampling distribution.

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

\\ P(z

However, we are looking for P(z>1.56), which is the <em>complement probability</em> of the previous probability. Therefore

\\ P(z>1.56) = 1 - P(z

\\ P(z>1.56) = P(\overline{X}>89.87) = 0.0594

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

5 0
3 years ago
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