Help:( i’ll give brainliest

Supply the missing information for each of the right triangles described below. In each example, θ = angle, O = opposite side, A

blagie [28]

Answer:

$\therefore O = 1.9096$

Step-by-step explanation:

Six trigonometric ratio:

$sin\theta=\frac{opposite}{hypoteuse}$

$cos\theta=\frac{adjacent}{hypotenuse}$

$tan \theta=\frac{opposite}{adjacent}$
$cosec\theta=\frac{hypotenuse}{opposite}$
$sec\theta=\frac{hypotenuse}{adjacent}$
$tan \theta=\frac{adjacent}{opposite}$

Given that,

θ = angle, O= opposite side, A= adjacent side and H = hypotenuse.

H=11, sin θ=0.1736

We know that,

$sin\theta=\frac{opposite}{hypoteuse}$

$\Rightarrow 0.1736=\frac{O}{11}$

$\Rightarrow O = 11 \times 0.1736$

$\Rightarrow O = 1.9096$

$\therefore O = 1.9096$

3 0

4 years ago

Which expression is equal to 3√54−4√24 ? 11√6 √6 −11√6 −√6

romanna [79]

3√54-4√24
9√6-4√24
9√6-8√6
√6

6 0

4 years ago

Help me with this!!!!!!!

WINSTONCH [101]

Oou I know this!!!! It 0.5 because the point F is at -1 and point G is at 2.
Soooo,
-1, 0, 1, 2
-1, -0.5, 0, 0.5, 1, 1.5, 2
Sooo the answer is 0.5 !!!

7 0

3 years ago

Graph a line with a Y-intercept of 3 containing the point (2,5). help this has to be done before 11:59

Sindrei [870]

The answer:

y = x + 3

5 0

3 years ago

The weight of an adult swan is normally distributed with a mean of 26 pounds and a standard deviation of 7.2 pounds. A farmer ra

Snezhnost [94]

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 $X_i\sim\mathcal N(26,7.2)$ .

You want to find

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

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

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

so that in this case,

$Z=6\dfrac{\overline X-26}{7.2}$

and the probability is equivalent to

$\mathbb P\left(\overline X>\dfrac{1000}{36}\right)=\mathbb P\left(6\dfrac{\overline X-26}{7.2}>6\dfrac{\frac{1000}{36}-26}{7.2}\right)$
$=\mathbb P(Z>1.481)\approx0.0693$

5 0

3 years ago