What is a shape that has no sides the same length that is a quadrilateral

Mathematics

1 answer:

Orlov [11]2 years ago

3 0

Answer:

Step-by-st

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The owner of a fish market has an assistant who has determined that the weights of catfish are normally distributed, with mean

LiRa [457]

Answer:

$P(X\geq 3.4)=0.0228$

Step-by-step explanation:

Given the mean is 3.2, standard deviation is 0.8 and the sample size is 64.

-We calculate the probability of a mean of 3.4 as follows:

#First determine the z-value:

$z=\frac{\bar X -\mu}{\sigma/\sqrt{n}}\\\\=\frac{3.4-3.2}{0.8/\sqrt{64}}\\\\=2.000$

#We then determine the corresponding probability on the z tables:

$Z(X\geq 3.4)=1-P(X$

Hence, the probability of obtaining a sample mean this large or larger is 0.0228

8 0

2 years ago

Find x when f(x)=1 . (These means find the input that gives an output of 1.)<br><br> f(x)=1/x-2

Andreyy89

Answer:

x = 1/3

Step-by-step explanation:

1 = 1/x -2

add 2 to both sides

3 = 1/x

x = 1/3

5 0

2 years ago

2x² - 5x+1 has roots alpha and beta. Find alpha⁴+beta⁴ without solving the equation.

shepuryov [24]

Answer:

Step-by-step explanation:

$\alpha+\beta=\dfrac{5}{2} \\\\\alpha*\beta=\dfrac{1}{2} \\\\\alpha^2+\beta^2=\dfrac{21}{4} \ (see\ previous\ post)\\\\(\alpha+\beta)^4=\dfrac{625}{16} \\\\=\alpha^4+\beta^4+4*\alpha^3*\beta+6*\alpha^2*\beta^2+4*\alpha*\beta^3\\\\=\alpha^4+\beta^4+4*(\alpha*\beta)(\alpha^2+\beta^2)+6*\alpha^2*\beta^2\\\\\alpha^4+\beta^4=(\alpha+\beta)^4-4*(\alpha*\beta)(\alpha^2+\beta^2)-6*\alpha^2*\beta^2\\\\= \dfrac{625}{16} -4*\dfrac{1}{2} *\dfrac{21}{4} -6*(\dfrac{1}{2})^2 \\\\$