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neonofarm [45]
3 years ago
5

Tyler ate x fruit snacks, and Han ate 3/4 less that. write an expression for the number of fruit snacks Han ate​

Mathematics
1 answer:
Alika [10]3 years ago
3 0

x = ×-3/4

because you hace the variable x and you do not know the number of fruit snacks he ate and you need to have the same variables to properly find the answer.

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Please help me and show your work!!
11Alexandr11 [23.1K]
\bf \cfrac{\sqrt{250x^{16}}}{\sqrt{2x}}\qquad 
\begin{cases}
250=2\cdot 5\cdot 5\cdot 5\\
\qquad 2\cdot 5\cdot 5^2\\
x^{16}=x^{8\cdot 2}\\
\qquad (x^8)^2
\end{cases}\implies \cfrac{\sqrt{2\cdot 5\cdot 5^2\cdot (x^8)^2}}{\sqrt{2\cdot  x}}

\bf \cfrac{5x^8\sqrt{2\cdot 5}}{\sqrt{2}\cdot \sqrt{x}}\implies \cfrac{5x^8\underline{\sqrt{2}}\cdot \sqrt{5}}{\underline{\sqrt{2}}\cdot \sqrt{x}}\implies \cfrac{5x^8\sqrt{5}}{\sqrt{x}}\impliedby 
\begin{array}{llll}
\textit{and rationalizing the}\\
\textit{denominator}
\end{array}

\bf \cfrac{5x^8\sqrt{5}}{\sqrt{x}}\cdot \cfrac{\sqrt{x}}{\sqrt{x}}\implies \cfrac{5x^8\sqrt{5}\cdot \sqrt{x}}{(\sqrt{x})^2}\implies \cfrac{5x^8\sqrt{5x}}{x}\implies 5x^8x^{-1}\sqrt{5x}
\\\\\\
5x^{8-1}\sqrt{5x}\implies 5x^7\sqrt{5x}
4 0
3 years ago
Jaylen bed is 6 feet long and his desk is 3 feet long. he wants to leave a two foot space between his bed and desk. if there is
Montano1993 [528]
No, he would need 132 inches
6 0
2 years ago
Read 2 more answers
The average production cost for major movies is 57 million dollars and the standard deviation is 22 million dollars. Assume the
Degger [83]

Using the normal distribution, we have that:

  • The distribution of X is X \approx (57,22).
  • The distribution of \mathbf{\bar{X}} is \bar{X} \approx (57, 5.3358).
  • 0.0597 = 5.97% probability that a single movie production cost is between 55 and 58 million dollars.
  • 0.2233 = 22.33% probability that the average production cost of 17 movies is between 55 and 58 million dollars. Since the sample size is less than 30, assumption of normality is necessary.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

Z = \frac{X - \mu}{\sigma}

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
  • By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation s = \frac{\sigma}{\sqrt{n}}.

In this problem, the parameters are given as follows:

\mu = 57, \sigma = 22, n = 17, s = \frac{22}{\sqrt{17}} = 5.3358

Hence:

  • The distribution of X is X \approx (57,22).
  • The distribution of \mathbf{\bar{X}} is \bar{X} \approx (57, 5.3358).

The probabilities are the <u>p-value of Z when X = 58 subtracted by the p-value of Z when X = 55</u>, hence, for a single movie:

X = 58:

Z = \frac{X - \mu}{\sigma}

Z = \frac{58 - 57}{22}

Z = 0.05.

Z = 0.05 has a p-value of 0.5199.

X = 55:

Z = \frac{X - \mu}{\sigma}

Z = \frac{55 - 57}{22}

Z = -0.1.

Z = -0.1 has a p-value of 0.4602.

0.5199 - 0.4602 = 0.0597 = 5.97% probability that a single movie production cost is between 55 and 58 million dollars.

For the sample of 17 movies, we have that:

X = 58:

Z = \frac{X - \mu}{s}

Z = \frac{58 - 57}{5.3358}

Z = 0.19.

Z = 0.19 has a p-value of 0.5753.

X = 55:

Z = \frac{X - \mu}{s}

Z = \frac{55 - 57}{5.3358}

Z = -0.38.

Z = -0.38 has a p-value of 0.3520.

0.5753 - 0.3520 = 0.2233 = 22.33% probability that the average production cost of 17 movies is between 55 and 58 million dollars. Since the sample size is less than 30, assumption of normality is necessary.

More can be learned about the normal distribution at brainly.com/question/4079902

#SPJ1

8 0
1 year ago
What is 1/20 as a decimal
Alexandra [31]

Your answer would be 0.05

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Emma invested $750 in an account paying an interest rate of 1.9% compounded quarterly. Assuming no deposits or withdrawals are m
storchak [24]

Answer:

10.2

Step-by-step explanation:

3 0
2 years ago
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