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3 years ago

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There are 23 people in this class. what is the probability that at least 2 of the people in the class share the same birthday? s

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1 answer:

BlackZzzverrR [31]3 years ago

4 0

23/2 = 0.08
0.08 = 8/100 or 8%

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What is the equation of the line parallel to x = 7, passing through the point (-4, 1)? A. X = -4 B. X = 1 C. Y = -4 D. Y = 1

77julia77 [94]

Answer:

x=-4

Step-by-step explanation:

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3 years ago

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Find the inverse of ????−7 −2

Masja [62]

Answer:

$A^{-1}=\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}$

Step-by-step explanation:

$A=\begin{bmatrix}-7 & -2\\ 4 & 1\end{bmatrix}$

$det\ A=-7\times 1-(-2)\times 4\\\Rightarrow det\ A=1$

$A^{-1}=\frac{1}{det\ A}\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}\\\Rightarrow A^{-1}=\frac{1}{1}\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}\\\Rightarrow A^{-1}=\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}$

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$A.A^{-1}=\begin{bmatrix}-7 & -2\\ 4 & 1\end{bmatrix}\times \begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}\\\Rightarrow A.A^{-1}=\begin{pmatrix}\left(-7\right)\cdot \:1+\left(-2\right)\left(-4\right)&\left(-7\right)\cdot \:2+\left(-2\right)\left(-7\right)\\ 4\cdot \:1+1\cdot \left(-4\right)&4\cdot \:2+1\cdot \left(-7\right)\end{pmatrix}\\\Rightarrow A.A^{-1}=\begin{pmatrix}1&0\\ 0&1\end{pmatrix}$

Hence, confirmed

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4 years ago

Please help me thanks if u do

aniked [119]

Answer: a

Step-by-step explanation: look at the chart

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3 years ago

The graph of the equation y=7x-3 has what slope?

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The slope of the line would be 7

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4 years ago

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Someone answer. Will give brainlest

Leto [7]

Area inside the semi-circle and outside the triangle is (91.125π - 120) in²

Solution:

Base of the triangle = 10 in

Height of the triangle = 24 in

Area of the triangle = $\frac{1}{2} bh$

$$=\frac{1}{2} \times 10\times24$

Area of the triangle = 120 in²

Using Pythagoras theorem,

$\text{Hypotenuse}^2=\text{base}^2+\text{height}^2$

$\text{Hypotenuse}^2=10^2+24^2$

$\text{Hypotenuse}^2=100+576$

$\text{Hypotenuse}^2=676$

Taking square root on both sides, we get

Hypotenuse = 23 inch = diameter

Radius = 23 ÷ 2 = 11.5 in

Area of the semi-circle = $\frac{1}{2}\pi r^2$

$$=\frac{1}{2} \pi \times (13.5)^2$

Area of the semi-circle = 91.125π in²

Area of the shaded portion = (91.125π - 120) in²

Area inside the semi-circle and outside the triangle is (91.125π - 120) in².

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3 years ago