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EleoNora [17]
2 years ago
7

Help .................................................

Mathematics
2 answers:
kati45 [8]2 years ago
5 0
AC-AB=BC
53-4x-6=2x+5
42=6x
x=7
BC=2x+5=2(7)+5=19
Ksenya-84 [330]2 years ago
4 0
The correct answer is 19 because x=7
Hope this helped :)
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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.

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3 years ago
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2 years ago
5x^2 + 19x + 12=0<br> how do i figure this out
zlopas [31]

Answer:

x=−4/5 or x=−3

Step-by-step explanation:

Step 1: Factor left side of equation.

(5x+4)(x+3)=0

Step 2: Set factors equal to 0.

5x+4=0 or x+3=0

x=−4/5 or x=−3

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2 years ago
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Vera_Pavlovna [14]

Answer:

13.7795 inches

Step-by-step explanation:

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In a cookie mix, we find spice cookies, snickerdoodle cookies, and no-bake cookies in a ratio of 3:3:5. If a bag of the mix cont
mixas84 [53]

Answer:

see the explaination

Step-by-step explanation:

let x be 3:3:5

3x:3x:5x

3x is splice cookies

3x is snickerdoodle cookies

5x is no bake cookies

where 5x is equal to 45

x is equal to 45÷5=9

so x is equal to 9

now,

3x = 3×9=27

3x = 3×9=27

5x = 45 (given)

now add all, total cookies =27+27+45

=99 cookies

so there are total 99 cookies in a cookie mix

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