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3 years ago

An angle measures 72 degrees more than the measure of its complementary angle what is the measure of each angle

Mathematics

1 answer:

labwork [276]3 years ago

8 0

Good evening

Answer:

<h2>measure of Angle 1 = 81°</h2><h2>Measure of angle 2 = 9°</h2>

Step-by-step explanation:

we need to solve this system

x + y = 90 (1)

x = y + 72 (2)

If we substitute x by y+72 into the equation (1) we will get

(y+72) + y = 90 ⇔ 2y + 72 = 90 ⇔ 2y = 90 - 72 = 18 ⇔ y = 9

then x = 90 - 9 = 81

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Question 4 options: Find the mean and standard deviation for a binomial distribution with 680 trials and a probability of succes

lys-0071 [83]

Answer:

Mean for a binomial distribution = 374

Standard deviation for a binomial distribution = 12.97

Step-by-step explanation:

We are given a binomial distribution with 680 trials and a probability of success of 0.55.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 680 trials

r = number of success

p = probability of success which in our question is 0.55

So, it means X <em>~ </em> $Binom(n=680, p=0.55)$

<em><u>Now, we have to find the mean and standard deviation of the given binomial distribution.</u></em>

Mean of Binomial Distribution is given by;

E(X) = n $\times$ p

So, E(X) = 680 $\times$ 0.55 = 374

Standard deviation of Binomial Distribution is given by;

S.D.(X) = $\sqrt{n \times p \times (1-p)}$

= $\sqrt{680 \times 0.55 \times (1-0.55)}$

= $\sqrt{680 \times 0.55 \times 0.45}$ = 12.97

Therefore, Mean and standard deviation for binomial distribution is 374 and 12.97 respectively.

3 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

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