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3 years ago

Find the missing angle..? x=?

Mathematics

2 answers:

Sedaia [141]3 years ago

7 0

Answer:

x = 95°

Step-by-step explanation:

x and 95 are vertical angles and congruent, thus

x = 95°

larisa [96]3 years ago

7 0

Answer:

x = 95°

Step-by-step explanation:

opposite angles

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Use the following information for the question. Male players at the high school, college and professional ranks use a regulation

Virty [35]

Answer:

13.567%

Step-by-step explanation:

We solve the above question, using z score formula

z = (x-μ)/σ, where

x is the raw score = 23.1 ounces

μ is the population mean = 22.0 ounces

σ is the population standard deviation = 1.0 ounce

More than = Greater than with the sign = >

Hence, for x > 23.1 ounces

z = 23.1 - 22.0/1.0

= 1.1

Probability value from Z-Table:

P(x<23.1) = 0.86433

P(x>23.1) = 1 - P(x<23.1)

P(x>23.1) = 1 - 0.86433

P(x>23.1) = 0.13567

Converting to percentage

= 0.13567 × 100

= 13.567%

Therefore, the percentage of regulation basketballs that weigh more than 23.1 ounces is 13.567%

8 0

2 years ago

*PLS HELP!!*<br>{The Picture Is In The Attachment}

mash [69]

Answer:

B is 45 A is 135

Step-by-step explanation:

45 + 135 = 180

45 x 3 = 135

135 is three times as large as 45

135 and 45 are supplementary angles

6 0

2 years ago

Find the Fourier series of f on the given interval. f(x) = 1, ?7 < x < 0 1 + x, 0 ? x < 7

Zolol [24]

$f(x)=\begin{cases}1&\text{for }-7$

The Fourier series expansion of $f(x)$ is given by

$\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi x}7+\sum_{n\ge1}b_n\sin\frac{n\pi x}7$

where we have

$a_0=\displaystyle\frac17\int_{-7}^7f(x)\,\mathrm dx$
$a_0=\displaystyle\frac17\left(\int_{-7}^0\mathrm dx+\int_0^7(1+x)\,\mathrm dx\right)$
$a_0=\dfrac{7+\frac{63}2}7=\dfrac{11}2$

The coefficients of the cosine series are

$a_n=\displaystyle\frac17\int_{-7}^7f(x)\cos\dfrac{n\pi x}7\,\mathrm dx$
$a_n=\displaystyle\frac17\left(\int_{-7}^0\cos\frac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\cos\frac{n\pi x}7\,\mathrm dx\right)$
$a_n=\dfrac{9\sin n\pi}{n\pi}+\dfrac{7\cos n\pi-7}{n^2\pi^2}$
$a_n=\dfrac{7(-1)^n-7}{n^2\pi^2}$

When $n$ is even, the numerator vanishes, so we consider odd $n$ , i.e. $n=2k-1$ for $k\in\mathbb N$ , leaving us with

$a_n=a_{2k-1}=\dfrac{7(-1)-7}{(2k-1)^2\pi^2}=-\dfrac{14}{(2k-1)^2\pi^2}$

Meanwhile, the coefficients of the sine series are given by

$b_n=\displaystyle\frac17\int_{-7}^7f(x)\sin\dfrac{n\pi x}7\,\mathrm dx$
$b_n=\displaystyle\frac17\left(\int_{-7}^0\sin\dfrac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\sin\dfrac{n\pi x}7\,\mathrm dx\right)$
$b_n=-\dfrac{7\cos n\pi}{n\pi}+\dfrac{7\sin n\pi}{n^2\pi^2}$
$b_n=\dfrac{7(-1)^{n+1}}{n\pi}$

So the Fourier series expansion for $f(x)$ is

$f(x)\sim\dfrac{11}4-\dfrac{14}{\pi^2}\displaystyle\sum_{n\ge1}\frac1{(2n-1)^2}\cos\frac{(2n-1)\pi x}7+\frac7\pi\sum_{n\ge1}\frac{(-1)^{n+1}}n\sin\frac{n\pi x}7$