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2 years ago

There are 1,000 cans of cat food produced daily at a factory. in the meeting last week, the

1 answer:

4 0

Using the binomial distribution, it is found that the true statement related to the number of cans of cat food the company expects to be dented is given by:

B. The factory can expect to have an average of 50 cans dented in a day.

<h3>What is the binomial probability distribution?</h3>

It is the probability of exactly x successes on n repeated trials, with p probability of a success on each trial.

The expected value of the binomial distribution is:

E(X) = np

In this problem, we have that there are 1000 cans, each with a probability of 1/20 of being dented, hence:

n = 1000, p = 1/20 = 0.05.

The expected value is then given by:

E(X) = np = 1000 x 0.05 = 50.

Researching the problem on the internet, the correct option is given by:

B. The factory can expect to have an average of 50 cans dented in a day.

More can be learned about the binomial distribution at brainly.com/question/24863377

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Use the Law of Sines to find the measure of angle R to the nearest degree.

IrinaK [193]

$\frac{\sin(R)}{PQ} = \frac{\sin(Q)}{PR}$

$\frac{\sin(R)}{24.7} = \frac{\sin(43^{\circ})}{16.9}$

$\sin(R) = 24.7*\frac{\sin(43^{\circ})}{16.9}$

$\sin(R) \approx 0.9967668$

$R \approx \arcsin(0.9967668)$

$R \approx 85.39137896^{\circ}$

$R \approx 85^{\circ}$

The final answer is 85 degrees

4 0

3 years ago

Use the method in the Example to find the square roots of 1 + √3i

kiruha [24]

Answer:

$\sqrt{1 + \sqrt{3} i}=\pm (1.58+i 0.548)$

Step-by-step explanation:

Given

z = 1 + √3 i

Let $\sqrt{1+\sqrt(3) i}=p+iq$

Squaring both sides

$1+\sqrt(3) i=p^2-q^2+2ipq$

Comparing real and imaginary part

Re(LHS)=Re(RHS)

$1=p^2-q^2$ ...........................(1)

comparing Im(LHS)=Im(RHS)

√3=2pq

$q=\frac{\sqrt{3}}{2p}$

Substitute q in equation (1)

$1=p^2-(\frac{\sqrt{3}}{2p})^2$

$p^4-p^2-0.75=0$

Let $x=p^2$

$x^2-x-0.75=0$

$x=\dfrac{1\pm \sqrt{1^2+4\times 0.75}}{2}$

$x=\frac{1\pm 4}{2}$

we take only Positive value because $p^2=x$

x=2.5

$p^2=2.5$

thus $p=\pm 1.58$

$q=\pm 0.548$

thus,

$\sqrt{1 + \sqrt{3} i}=\pm (1.58+i 0.548)$

5 0

3 years ago

Please answer this question, i request

Jet001 [13]

${\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}$

$\star \: \tt \cot \theta = \dfrac{7}{8}$

${\large{\textsf{\textbf{\underline{\underline{To \: Evaluate :}}}}}}$

$\star \: \tt \dfrac{(1 + \sin \theta)(1 - \sin \theta) }{(1 + \cos \theta) (1 - \cos \theta) }$

${\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}$

Consider a $\triangle$ ABC right angled at C and $\sf \angle \: B = \theta$

Then,

‣ Base [B] = BC

‣ Perpendicular [P] = AC

‣ Hypotenuse [H] = AB

$\therefore \tt \cot \theta = \dfrac{Base}{ Perpendicular} = \dfrac{BC}{AC} = \dfrac{7}{8}$

Let,

Base = 7k and Perpendicular = 8k, where k is any positive integer

In $\triangle$ ABC, H² = B² + P² by Pythagoras theorem

$\longrightarrow \tt {AB}^{2} = {BC}^{2} + {AC}^{2}$

$\longrightarrow \tt {AB}^{2} = {(7k)}^{2} + {(8k)}^{2}$

$\longrightarrow \tt {AB}^{2} = 49{k}^{2} + 64{k}^{2}$

$\longrightarrow \tt {AB}^{2} = 113{k}^{2}$

$\longrightarrow \tt AB = \sqrt{113 {k}^{2} }$

$\longrightarrow \tt AB = \red{ \sqrt{113} \: k}$

Calculating Sin $\sf \theta$

$\longrightarrow \tt \sin \theta = \dfrac{Perpendicular}{Hypotenuse}$

$\longrightarrow \tt \sin \theta = \dfrac{AC}{AB}$

$\longrightarrow \tt \sin \theta = \dfrac{8 \cancel{k}}{ \sqrt{113} \: \cancel{ k } }$

$\longrightarrow \tt \sin \theta = \purple{ \dfrac{8}{ \sqrt{113} } }$

Calculating Cos $\sf \theta$

$\longrightarrow \tt \cos \theta = \dfrac{Base}{Hypotenuse}$

$\longrightarrow \tt \cos \theta = \dfrac{BC}{ AB}$

$\longrightarrow \tt \cos \theta = \dfrac{7 \cancel{k}}{ \sqrt{113} \: \cancel{k } }$

$\longrightarrow \tt \cos \theta = \purple{ \dfrac{7}{ \sqrt{113} } }$

Solving the given expression :- 

$\longrightarrow \: \tt \dfrac{(1 + \sin \theta)(1 - \sin \theta) }{(1 + \cos \theta) (1 - \cos \theta) }$

Putting,

• Sin $\sf \theta$ = $\dfrac{8}{ \sqrt{113} }$

• Cos $\sf \theta$ = $\dfrac{7}{ \sqrt{113} }$

$\longrightarrow \: \tt \dfrac{ \bigg(1 + \dfrac{8}{ \sqrt{133}} \bigg) \bigg(1 - \dfrac{8}{ \sqrt{133}} \bigg) }{\bigg(1 + \dfrac{7}{ \sqrt{133}} \bigg) \bigg(1 - \dfrac{7}{ \sqrt{133}} \bigg)}$

Using (a + b ) (a - b ) = a² - b²

$\longrightarrow \: \tt \dfrac{ { \bigg(1 \bigg)}^{2} - { \bigg( \dfrac{8}{ \sqrt{133} } \bigg)}^{2} }{ { \bigg(1 \bigg)}^{2} - { \bigg( \dfrac{7}{ \sqrt{133} } \bigg)}^{2} }$

$\longrightarrow \: \tt \dfrac{1 - \dfrac{64}{113} }{ 1 - \dfrac{49}{113} }$

$\longrightarrow \: \tt \dfrac{ \dfrac{113 - 64}{113} }{ \dfrac{113 - 49}{113} }$

$\longrightarrow \: \tt { \dfrac { \dfrac{49}{113} }{ \dfrac{64}{113} } }$

$\longrightarrow \: \tt { \dfrac{49}{113} }÷{ \dfrac{64}{113} }$

$\longrightarrow \: \tt \dfrac{49}{ \cancel{113}} \times \dfrac{ \cancel{113}}{64}$

$\longrightarrow \: \tt \dfrac{49}{64}$

$\qquad \: \therefore \: \tt \dfrac{(1 + \sin \theta)(1 - \sin \theta) }{(1 + \cos \theta) (1 - \cos \theta) } = \pink{\dfrac{49}{64} }$

$\begin{gathered} {\underline{\rule{300pt}{4pt}}} \end{gathered}$

${\large{\textsf{\textbf{\underline{\underline{We \: know :}}}}}}$

✧ Basic Formulas of Trigonometry is given by :-

$\begin{gathered}\begin{gathered}\boxed { \begin{array}{c c} \\ \bigstar \: \sf{ In \:a \:Right \:Angled \: Triangle :} \\ \\ \sf {\star Sin \theta = \dfrac{Perpendicular}{Hypotenuse}} \\\\ \sf{ \star \cos \theta = \dfrac{ Base }{Hypotenuse}}\\\\ \sf{\star \tan \theta = \dfrac{Perpendicular}{Base}}\\\\ \sf{\star \cosec \theta = \dfrac{Hypotenuse}{Perpendicular}} \\\\ \sf{\star \sec \theta = \dfrac{Hypotenuse}{Base}}\\\\ \sf{\star \cot \theta = \dfrac{Base}{Perpendicular}} \end{array}}\\\end{gathered} \end{gathered}$