All categories

3 years ago

Plz help me this is hard for me

Mathematics

2 answers:

MaRussiya [10]3 years ago

8 0

Answer:

it is so simple lol

Step-by-step explanation:

just know that that angle is 90 degrees which means that you should do 90-41

Stels [109]3 years ago

3 0

Answer:

Step-by-step explanation:

90*-41*=49

You might be interested in

Determine the area under the standard normal curve that lies to the left of (a) Z=081. (b) Z=0.43. (c) Z= -0.33. and (d) Z= 1.04

nasty-shy [4]

Using the normal distribution, the areas to the left are given as follows:

a) 0.7910.

b) 0.6664.

c) 0.3707.

d) 0.8508.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X, and is also the area to the left of Z.

Hence:

The area to the left of Z = 0.81 is of 0.7910.

The area to the left of Z = 0.43 is of 0.6664.

The area to the left of Z = -0.33 is of 0.3707.

The area to the left of Z = 1.04 is of 0.8508.

More can be learned about the normal distribution at brainly.com/question/4079902

#SPJ1

4 0

1 year ago

Consider the following equations. f(x) = − 4/ x3, y = 0, x = −2, x = −1. Sketch the region bounded by the graphs of the equation

Sindrei [870]

Answer:

1.5 unit^2

Step-by-step explanation:

Solution:-

- A graphing utility was used to plot the following equations:

$f ( x ) = - \frac{4}{x^3}\\\\y = 0 , x = -1 , x = -2$

- The plot is given in the document attached.

- We are to determine the area bounded by the above function f ( x ) subjected boundary equations ( y = 0 , x = -1 , x = - 2 ).

- We will utilize the double integral formulations to determine the area bounded by f ( x ) and boundary equations.

We will first perform integration in the y-direction ( dy ) which has a lower bounded of ( a = y = 0 ) and an upper bound of the function ( b = f ( x ) ) itself. Next we will proceed by integrating with respect to ( dx ) with lower limit defined by the boundary equation ( c = x = -2 ) and upper bound ( d = x = - 1 ).

The double integration formulation can be written as:

$A= \int\limits_c^d \int\limits_a^b {} \, dy.dx \\\\A = \int\limits_c^d { - \frac{4}{x^3} } . dx\\\\A = \frac{2}{x^2} |\limits_-_2^-^1\\\\A = \frac{2}{1} - \frac{2}{4} \\\\A = \frac{3}{2} unit^2$