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butalik [34]
2 years ago
13

Using the WASA, USA task if the measure of angle 32 is 118 degrees. Find

Mathematics
1 answer:
Karo-lina-s [1.5K]2 years ago
6 0

Step-by-step explanation:

The first step is to locate and define the variables.

We are working with complementary angles which  are "either of two angles whose sum is 90°".  Therefore, our job is to find the degrees of 2 angles. We want to determine the algebraic equation for the sum of these two angles, we need a variable representing the second, complementary angle, let's say x. Next, let's use the information provided to write only the expression.

It states "an angle measures 32 degrees MORE THAN the measure of its complementary angle". Whenever we see the words "more than" we know we will be adding. If the complementary, second angle is represented as the variable x, then 32 more than x (x + 32) represents the first angle measure.  

To summarize what we have so far:

x+32 = (is) the value of the FIRST angle measure

x = (is) the value of the SECOND angle measure

We know that when these two angles are added TOGETHER they equal 90°

Therefore, x + 32 + x = 90° . But we aren't done yet! We have to solve for x in order to find the measure of BOTH angles:  

x + 32 + x = 90°  Our algebraic equation for this word problem

2x + 32 = 90°   Combine like terms  

 2x = 58   Subtract 32 from both sides of the equation, so that variables are on one side and              constants on the other side.

   x = 29°   Divide by 2 on both sides of the equation (we want to solve for x)

Now that we have solved for x, the measure of the SECOND, complementary angle, we can solve for the first angle.  

x+32 = (is) the value of the FIRST angle measure

29 + 32 = 61°   substitute x for its value, 29

Lastly, Check your work:  

29 + 61 = 90°  

90° = 90° ✓

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Let X denote the random variable for the weight of a swan. Then each swan in the sample of 36 selected by the farmer can be assigned a weight denoted by X_1,\ldots,X_{36}, each independently and identically distributed with distribution X_i\sim\mathcal N(26,7.2).

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\mathbb P(X_1+\cdots+X_{36}>1000)=\mathbb P\left(\displaystyle\sum_{i=1}^{36}X_i>1000\right)

Note that the left side is 36 times the average of the weights of the swans in the sample, i.e. the probability above is equivalent to

\mathbb P\left(36\displaystyle\sum_{i=1}^{36}\frac{X_i}{36}>1000\right)=\mathbb P\left(\overline X>\dfrac{1000}{36}\right)

Recall that if X\sim\mathcal N(\mu,\sigma), then the sampling distribution \overline X=\displaystyle\sum_{i=1}^n\frac{X_i}n\sim\mathcal N\left(\mu,\dfrac\sigma{\sqrt n}\right) with n being the size of the sample.

Transforming to the standard normal distribution, you have

Z=\dfrac{\overline X-\mu_{\overline X}}{\sigma_{\overline X}}=\sqrt n\dfrac{\overline X-\mu}{\sigma}

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