PLEASE HELP ASAP!!!!!!!!!!!!!!!

Please help me with this question.

ra1l [238]

Answer:

Area = 43.26

Step-by-step explanation:

There is a short way to do this.

Area = 1/2 s*t * Sin(R)

s = 14

t = 7

Area = 1/2 * 14 * 7 * Sin(62)

Area = 7 * 7 * sin(62)

Area = 49 * sin(62)

Area = 43.26

==================

There is another way you could do this.

Use the Cos law to find r

Then use Heron's formula to find the area. Trust me, you don't want to do that.

Doing the way I suggested gives 43.19. The difference is in what I did in the rounding. But I wouldn't want to type all the numbers.

4 0

3 years ago

You place a 16 foot ladder against a building. The angle that the ladder forms with

gulaghasi [49]

Answer:The base of the ladder should be placed so that it is one foot away from the building for every four feet of hight to where the ladder rests against the building. This is known as the 4 to 1 rule.

Step-by-step explanation:

7 0

3 years ago

Which of the following choices lists the values of the side lengths of a triangle with 45-45-90 degree angles and a hypotenuse =

Gnoma [55]

Answer:

<u>The correct answer is D. 12√2, 12√2, 24.</u>

Step-by-step explanation:

Let's recall that a property of the right triangles 45 - 45 - 90 is that the lengths of the sides of this type of triangles are in the ratio 1 : 1: √2.

Thus, if the hypotenuse is 24 units, the length of the legs is 12√2. We calculated it this way:

24/√2 = x/1

x√2 = 24

x = 24/√2

x = (24 * √2) / (√2 * √2) (For not having a square root in the denominator)

x = 24√2/2

x = 12√2 (Most simplified expression)

<u>The correct answer is D. 12√2, 12√2, 24.</u>

5 0

3 years ago

Let X denote the length of human pregnancies from conception to birth, where X has a normal distribution with mean of 264 days a

Kaylis [27]

Answer:

Step-by-step explanation:

Hello!

X: length of human pregnancies from conception to birth.

X~N(μ;σ²)

μ= 264 day

σ= 16 day

If the variable of interest has a normal distribution, it's the sample mean, that it is also a variable on its own, has a normal distribution with parameters:

X[bar] ~N(μ;σ²/n)

When calculating a probability of a value of "X" happening it corresponds to use the standard normal: Z= (X[bar]-μ)/σ

When calculating the probability of the sample mean taking a given value, the variance is divided by the sample size. The standard normal distribution to use is Z= (X[bar]-μ)/(σ/√n)

a. You need to calculate the probability that the sample mean will be less than 260 for a random sample of 15 women.

P(X[bar]<260)= P(Z<(260-264)/(16/√15))= P(Z<-0.97)= 0.16602

b. P(X[bar]>b)= 0.05

You need to find the value of X[bar] that has above it 5% of the distribution and 95% below.

P(X[bar]≤b)= 0.95

P(Z≤(b-μ)/(σ/√n))= 0.95

The value of Z that accumulates 0.95 of probability is Z= 1.648

Now we reverse the standardization to reach the value of pregnancy length:

1.648= (b-264)/(16/√15)

1.648*(16/√15)= b-264

b= [1.648*(16/√15)]+264

b= 270.81 days

c. Now the sample taken is of 7 women and you need to calculate the probability of the sample mean of the length of pregnancy lies between 1800 and 1900 days.

Symbolically:

P(1800≤X[bar]≤1900) = P(X[bar]≤1900) - P(X[bar]≤1800)

P(Z≤(1900-264)/(16/√7)) - P(Z≤(1800-264)/(16/√7))

P(Z≤270.53) - P(Z≤253.99)= 1 - 1 = 0

d. P(X[bar]>270)= 0.1151

P(Z>(270-264)/(16/√n))= 0.1151

P(Z≤(270-264)/(16/√n))= 1 - 0.1151

P(Z≤6/(16/√n))= 0.8849

With the information of the cumulated probability you can reach the value of Z and clear the sample size needed:

P(Z≤1.200)= 0.8849

$Z= \frac{X[bar]-Mu}{Sigma/\sqrt{n} }$