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3 years ago

Find the value of h in the triangle

Mathematics

1 answer:

sasho [114]3 years ago

7 0

The triangles ABC and ADC are similar.
Therefore
$\dfrac{h}{12}=\dfrac{16}{20}\\\\\dfrac{h}{12}=\dfrac{4}{5}\ \ \ |\cdot12\\\\h=\dfrac{48}{5}\\\\h=9.6\ m$

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(8*10^7)(2*10^6) simplify the expression and write in scientific notation

lesya [120]

Grouping similar factors together makes this easier:

(8)(2)(10^7)(10^6) = 16(10^13) = 1.6(10^14)

8 0

4 years ago

Find the Derivative y’ implicitly.

Brums [2.3K]

$e^{x^2y} - e^y = e^y(e^{x^2} - 1) = x$

We should ISOLATE x

$e^y= \frac{x}{e^{x^2} - 1}$

Find the Natural Log of Both Sides to Make the Left Side "y"

$y = ln(\frac{x}{e^{x^2}-1})$

Now, FIND THE DERIVATIVE Using Chain Rule!!!

$y' = \frac{1}{\frac{x}{e^{x^2}-1}} * \frac{(1(e^{x^2}-1) - x(2x*e^{x^2})}{(e^{x^2}-1)^2}= {\frac{e^{x^2}-1}{x}}* \frac{(1(e^{x^2}-1) - x(2x*e^{x^2})}{(e^{x^2}-1)^2} = {\frac{1}{x}}* \frac{(1(e^{x^2}-1) - x(2x*e^{x^2})}{(e^{x^2}-1)} = {\frac{1}{x}}* \frac{(e^{x^2}-1 - 2x^2e^{x^2})}{(e^{x^2}-1)}$

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3 years ago

he used a map of bus routes to get from the airport to his cousin's house. The distance from the airport to his cousin's house i

Anarel [89]

The scale factor for the problem is 14

7 0

3 years ago

When the population distribution is normal, the statistic median {|X1 − X tilde|, . . . , |Xn − X tilde|}/0.6745 can be used to

Andreyy89

Answer:

The corresponding point estimate is 0.882.

The sample standard deviation is 1.373.

Step-by-step explanation:

The data set is:

S = {25.01, 25.87, 26.34, 26.51, 26.75, 27.24, 27.40, 27.63, 27.83, 27.90, 28.08, 28.13, 28.37, 28.58, 28.59, 28.96, 29.20, 29.22, 29.38, 30.88}

Compute the mean as follows:

$\bar X=\frac{1}{n}\sum X\\\\=\frac{1}{20}\times [25.01+25.87+...+30.88]\\\\=\frac{1}{20}\times 557.87\\\\=27.8935$

Subtract the mean from each value and take the modulus of those values.

The new data set is:

S₁ = {2.8835, 2.0235, 1.5535, 1.3835, 1.1435, 0.6535, 0.4935, 0.2635, 0.0635, 0.0065, 0.1865, 0.2365, 0.4765, 0.6865, 0.6965, 1.0665, 1.3065, 1.3265, 1.4865, 2.9865}

Arrange these values in ascending order as follows:

S₂ = {0.0065 , 0.0635 , 0.1865 , 0.2365 , 0.2635 , 0.4765 , 0.4935 , 0.6535 , 0.6865 , 0.6965 , 1.0665 , 1.1435 , 1.3065 , 1.3265 , 1.3835 , 1.4865 , 1.5535 , 2.0235 , 2.8835 , 2.9865}

There are 20 observations in the data set.

The median value for an even set of values is the mean of the middle two values.

In this case the median will be the mean of the 10th and 11th observations.

$\text{Median}=\frac{10^{th}obs.+11^{th}obs.}{2}=\frac{0.6965+1.0665}{2}=0.8815\approx 0.882$

Thus, the corresponding point estimate is 0.882.

Compute the standard deviation as follows:

In set S₁ we computed the absolute mean deviations.

Now take the square of these values and divide by (n - 1) to compute the sample variance:

$\sigma^{2}=\frac{1}{n-1}\sum (|X_{i}-\bar X|)^{2}$

$=\frac{1}{20-1}\times [(2.8835)^{2}+(2.0235)^{2}+...+(2.9865)^{2}]\\\\=\frac{1}{19}\times 35.7953\\\\=1.88396$

Compute the sample standard deviation as follows:

$\sigma=\sqrt{\sigma^{2}}=\sqrt{1.88396}=1.373$

Thus, the sample standard deviation is 1.373.

6 0

3 years ago

The mean weight of frozen yogurt cups in an ice cream parlor is 8 oz.Suppose the weight of each cup served is normally distribut

vaieri [72.5K]

Answer:

10.03% probability of getting a cup weighing more than 8.64oz

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

$\mu = 8, \sigma = 0.5$