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3 years ago

Help a girl out with geometry

Mathematics

1 answer:

Darya [45]3 years ago

7 0

Follow the steps

106=x+79+x+45
106=2x+124
-124= -124
-18=2x
( -18)/2=(2x)/2 <----- solving for x
-9=x

then
angle LMF= -9+79
=70

angle FMN= -9+45
= 36
(when you add 70 and 36 together you get 106 which is the measurement of angle LMN or in other word the whole thing)

In conclusion the answer to the problem is:
Angle LMF is equal to 70.

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Given f(x)=x^2+5 and g(x)=2x^3-1 find (f/g)(-3)

AlekseyPX

For this case we have the following functions:

$f (x) = x ^ 2 + 5\\g (x) = 2x ^ 3-1$

By definition of composition of functions we have:

$(f / g) (x) = \frac {f (x)} {g (x)}$

Then substituting:

$f (-3) = (- 3) ^ 2 + 5 = 9 + 5 = 14\\g (-3) = 2 (-3) ^ 3-1 = 2 (-27) -1 = -54-1 = -55$

So:

$(f / g) (- 3) = \frac {14} {- 55} = - 0.25$

Answer:

$(f / g) (- 3) = \frac {14} {- 55} = - 0.25$

7 0

3 years ago

julsineya [31]

Answer:

In order to ride on the ride an individual must be at least 4 feet tall.

This means that the individual must be equal to 4 ¾ feet tall.

h = 4 ⅓

and the individual could be greater than 4¾ feet tall.

h > 4 ¾

So the individual could be greater than and equal to 4¾ feet tall.

h ≥ 4¾

4 0

3 years ago

Assume that women's heights are normally distributed with a mean given by mu equals 62.5 in,and a standard deviation given by

Misha Larkins [42]

Answer:

(a) 0.5899

(b) 0.9166

Step-by-step explanation:

Let X be the random variable that represents the height of a woman. Then, X is normally distributed with

$\mu$ = 62.5 in

$\sigma$ = 2.2 in

the normal probability density function is given by

$f(x) = \frac{1}{\sqrt{2\pi}2.2}\exp{-\frac{(x-62.5)^{2}}{2(2.2)^{2}}}$ , then

(a) $P(X < 63) = \int\limits_{-\infty}^{63}f(x) dx$ = 0.5899

(in the R statistical programming language) pnorm(63, mean = 62.5, sd = 2.2)

(b) We are seeking $P(\bar{X} < 63)$ where n = 37. $\bar{X}$ is normally distributed with mean 62.5 in and standard deviation $2.2/\sqrt{37}$ . So, the probability density function is given by