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makkiz [27]
3 years ago
5

Someone please help with 9th question !!!!

Mathematics
1 answer:
meriva3 years ago
5 0

LIMITS  \:  \: AND \:  \:  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  DERIVATIVES


Heya !

Check the attachment.
Hope it helps you :)

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mrs_skeptik [129]
Is the associative

your very much welcome
5 0
3 years ago
The standard deviation of a normal random variable x is $20. given that p(x ≤ $10) = 0.1841. from this we can determine that the
attashe74 [19]
Let \mu be the mean of X. Then

\mathbb P(X\le10)=0.1841\iff\mathbb P\left(\underbrace{\dfrac{X-\mu}{20}}_Z\le\dfrac{10-\mu}{20}\right)=0.1841

For a random variable Z following the standard normal distribution, we have

\mathbb P(Z\le z)=0.1841\implies z\approx-0.8999

Transform the random variable to get this critical value in terms of \mu:

\dfrac{10-\mu}{20}=-0.8999\implies\mu\approx27.997
3 0
3 years ago
Lan is standing 20 feet above sea level. Mary is swimming 10 feet below sea level. How much higher is lan's
Savatey [412]

Answer:

30 feet

Step-by-step explanation:

sea level = 0

up 20 feet

below 10 feet

total 30 feet

4 0
3 years ago
A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 447 gram setting. It is
lord [1]

Answer:

A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 447 gram setting.

H_0:\mu = 447\\H_a:\mu\neq 447

Mean = \mu = 447\\s = 21\\x = 443

n = 19

Since n < 30 , so we will use t test

t=\frac{x-\mu}{\frac{s}{\sqrt{n}}}

Substitute the values :

t=\frac{443-447}{\frac{21}{\sqrt{19}}}

t=−0.8302

t calculated = -0.830

degree of freedom = n-1 = 19-1 = 18

A level of significance=α=0.025

t_{(df,\frac{\alpha}{2})}=2.093

t critical = 2.093

t calculated < t critical

So, We failed to reject null hypothesis

Decision rule  -0.830< 2.093 So, We failed to reject null hypothesis

4 0
3 years ago
Help solve for “q”<br> —————————————
VMariaS [17]

Digram:-

\\

\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\put(5,1){\vector(1,0){4}}\put(5,1){\vector(-1,0){4}}\put(5,1){\vector(1,1){3}}\put(2,2){$\underline{\boxed{\large\sf a + b = 180^{\circ}}$}}\put(4.5,1.3){$\sf a^{\circ}$}\put(5.7,1.3){$\sf b^{\circ}$}\end{picture}

\\

STEP :-

\dashrightarrow \tt(4q - 1) {}^{ \circ}  +  {117}^{ \circ}  = 18 {0}^{ \circ}

{Linear pair}

\\  \\

\dashrightarrow \tt(4q - 1) {}^{ \circ}= 18 {0}^{ \circ}   - {117}^{ \circ}

\\

\dashrightarrow \tt(4q - 1) {}^{ \circ}=63^{ \circ}

\\

\dashrightarrow \tt4q - 1{}^{ \circ}=63^{ \circ}

\\

\dashrightarrow \tt4q =63^{ \circ} + 1{}^{ \circ}

\\

\dashrightarrow \tt4q =64{}^{ \circ}

\\

\dashrightarrow \tt \: q =  \dfrac{64}{4}^{ \circ}

\\

\dashrightarrow \tt \: q =  \dfrac{16 \times 4}{4}^{ \circ}

\\

\dashrightarrow \tt \: q =  \dfrac{16 \times \cancel4}{\cancel4}^{ \circ}

\\

\dashrightarrow \tt \: q =  \dfrac{16}{1}

\\

\dashrightarrow \bf q = 16 \degree

\\  \\

Verification:

\\

\dashrightarrow \tt(4 \times 16- 1) {}^{ \circ}  +  {117}^{ \circ}  = 18 {0}^{ \circ}

\\

\dashrightarrow \tt(64- 1) {}^{ \circ}  +  {117}^{ \circ}  = 18 {0}^{ \circ}

\\

\dashrightarrow \tt63^{ \circ}  +  {117}^{ \circ}  = 18 {0}^{ \circ}

\\

\dashrightarrow \tt180^{ \circ}  = 18 {0}^{ \circ}

\\

LHS = RHS

HENCE VERIFIED!

8 0
3 years ago
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