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3 years ago

A different pattern is made using 15 straight lines and 12 arcs

Mathematics

1 answer:

Volgvan3 years ago

8 0

Answer:

Super Idol的笑容都没你的甜八月正午的阳光都没你耀眼热爱105 °C的你滴滴清纯的蒸馏水.

Step-by-step explanation:

Super Idol的笑容都没你的甜八月正午的阳光都没你耀眼热爱105 °C的你滴滴清纯的蒸馏水. Super Idol的笑容都没你的甜八月正午的阳光都没你耀眼热爱105 °C的你滴滴清纯的蒸馏水. Super Idol的笑容都没你的甜八月正午的阳光都没你耀眼热爱105 °C的你滴滴清纯的蒸馏水. Super Idol的笑容都没你的甜八月正午的阳光都没你耀眼热爱105 °C的你滴滴清纯的蒸馏水. Super Idol的笑容都没你的甜八月正午的阳光都没你耀眼热爱105 °C的你滴滴清纯的蒸馏水. Super Idol的笑容都没你的甜八月正午的阳光都没你耀眼热爱105 °C的你滴滴清纯的蒸馏水. Super Idol的笑容都没你的甜八月正午的阳光都没你耀眼热爱105 °C的你滴滴清纯的蒸馏水. Super Idol的笑容都没你的甜八月正午的阳光都没你耀眼热爱105 °C的你滴滴清纯的蒸馏水. Super Idol的笑容都没你的甜八月正午的阳光都没你耀眼热爱105 °C的你滴滴清纯的蒸馏水. Super Idol的笑容都没你的甜八月正午的阳光都没你耀眼热爱105 °C的你滴滴清纯的蒸馏水. Super Idol的笑容都没你的甜八月正午的阳光都没你耀眼热爱105 °C的你滴滴清纯的蒸馏水. Super Idol的笑容都没你的甜八月正午的阳光都没你耀眼热爱105 °C的你滴滴清纯的蒸馏水. Super Idol的笑容都没你的甜八月正午的阳光都没你耀眼热爱105 °C的你滴滴清纯的蒸馏水. Super Idol的笑容都没你的甜八月正午的阳光都没你耀眼热爱105 °C的你滴滴清纯的蒸馏水. Super Idol的笑容都没你的甜八月正午的阳光都没你耀眼热爱105 °C的你滴滴清纯的蒸馏水. Super Idol的笑容都没你的甜八月正午的阳光都没你耀眼热爱105 °C的你滴滴清纯的蒸馏水. Super Idol的笑容都没你的甜八月正午的阳光都没你耀眼热爱105 °C的你滴滴清纯的蒸馏水. Super Idol的笑容都没你的甜八月正午的阳光都没你耀眼热爱105 °C的你滴滴清纯的蒸馏水.

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A student is getting ready to take an important oral examination and is concerned about the possibility of having an "on" day or

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The students should request an examination with 5 examiners.

Step-by-step explanation:

Let X denote the event that the student has an “on” day, and let Y denote the

denote the event that he passes the examination. Then,

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

The events ( $Y|X$ ) follows a Binomial distribution with probability of success 0.80 and the events ( $Y|X^{c}$ ) follows a Binomial distribution with probability of success 0.40.

It is provided that the student believes that he is twice as likely to have an off day as he is to have an on day. Then,

$P(X)=2\cdot P(X^{c})$

Then,

$P(X)+P(X^{c})=1$

⇒

$2P(X^{c})+P(X^{c})=1\\\\3P(X^{c})=1\\\\P(X^{c})=\frac{1}{3}$

Then,

$P(X)=1-P(X^{c})\\=1-\frac{1}{3}\\=\frac{2}{3}$

Compute the probability that the students passes if request an examination with 3 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{3}_{x=2}{{3\choose x}(0.80)^{x}(1-0.80)^{3-x}}]\times\frac{2}{3}+[\sum\limits^{3}_{x=2}{{3\choose x}(0.40)^{3}(1-0.40)^{3-x}}]\times\frac{1}{3}$

$=0.715$

The probability that the students passes if request an examination with 3 examiners is 0.715.

Compute the probability that the students passes if request an examination with 5 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{5}_{x=3}{{5\choose x}(0.80)^{x}(1-0.80)^{5-x}}]\times\frac{2}{3}+[\sum\limits^{5}_{x=3}{{5\choose x}(0.40)^{x}(1-0.40)^{5-x}}]\times\frac{1}{3}$