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3 years ago

In the function y = (23)x + 11, the asymptote located at y = ___?

1 answer:

3 0

It is hard to graph but here is a website that me and my former crew put togather https://www.freemathhelp.com/asymptotes.html

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(2×²y²)⁰ I need help with exponents

balandron [24]

Answer:

Step-by-step explanation:

Any real value to the power of 0 is 1.

7 0

3 years ago

Read 2 more answers

I need help finding the area

aniked [119]

<h3>Given :</h3>

Base of triangle = 7 yd
Height of triangle = 10 yd

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Area of triangle

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We know:-

When base and height of triangle is given we use this formula:

$\bigstar \boxed{ \rm Area \: of \: triangle = \frac{base \times height}{2} }$

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So:-

$\\ \\$

$\dashrightarrow \sf \: Area \: of \: triangle = \dfrac{base \times height}{2} \\$

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$\dashrightarrow \sf \: Area \: of \: triangle = \dfrac{7 \times 10}{2} \\$

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$\dashrightarrow \sf \: Area \: of \: triangle = \dfrac{7 \times 5 \times2 }{2} \\$

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$\dashrightarrow \sf \: Area \: of \: triangle = \dfrac{7 \times 5 \times\cancel2 }{\cancel2} \\$

$\\ \\$

$\dashrightarrow \sf \: Area \: of \: triangle = \dfrac{7 \times 5 \times1 }{1} \\$

$\\ \\$

$\dashrightarrow \sf \: Area \: of \: triangle =7 \times 5$

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$\dashrightarrow \bf \: Area \: of \: triangle =35 {yd}^{2} \\$

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$\therefore \underline{\textsf{ \textbf {\: Area \: of \: triangle = \red{35}}} { \red{\bf{yd} }^{ \red2} }}$

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$\small\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf \small{Formulas\:of\:Areas:-}}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}\end{gathered}$ ]

7 0

1 year ago

At a manufacturing plant where switches are made, it is a known fact that 2% of all switches are defective. If two switches are

olchik [2.2K]

Answer:

The probability that exactly one switch is good is

$P(x) =0.0392$

Step-by-step explanation:

The probability that a switch is defective is:

$P(D) = \frac{2}{100} =0.02$

The probability that a switch is not defective is

$P(D') = 1-P(D)=0.98$

Therefore, if two switches are selected, the probability that exactly 1 is good is:

$P(1=1)=P (D) P (D ') + P (D') P (D)$

$P(x)=(0.02)(0.98) + (0.98)(0.02)$

$P(x) =0.0392$

6 0

3 years ago

Read 2 more answers

Please help me with my homework idk if it is right or wrong

jeka57 [31]

Q1: 7×6 = 42
Q2: 42×1/8 = 42/8 = 5.25 (5 1/4) in^3
Q3: (0.5)^3 = 0.5×0.5×0.5 = 0.125 m^3
Q4: 4(5/2)^2 = 4×5/2×5/2 = 4×25/4 = 25 in^3
Q5: 3(13/2)(3/2) = 117/4 = 29.25 (29 1/4) in^3
Q6: (0.9)^3 = 0.729 cm^3

6 0

2 years ago

The n candidates for a job have been ranked 1, 2, 3,…, n. Let X 5 the rank of a randomly selected candidate, so that X has pmf p

Anon25 [30]

Answer:

A. E(x) = 1/n×n(n+1)/2

B. E(x²) = 1/n

Step-by-step explanation:

The n candidates for a job have been ranked 1,2,3....n. Let x be the rank of a randomly selected candidate. Therefore, the PMF of X is given as

P(x) = {1/n, x = 1,2...n}

Therefore,

Expectation of X

E(x) = summation {xP(×)}

= summation {X×1/n}

= 1/n summation{x}

= 1/n×n(n+1)/2

= n+1/2

Thus, E(x) = 1/n×n(n+1)/2

Value of E(x²)

E(x²) = summation {x²P(×)}

= summation{x²×1/n}

= 1/n

3 0

2 years ago