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3 years ago

2) Write the equation of a line lin point-slope form) with a slope of -2 passing through the

Mathematics

1 answer:

shepuryov [24]3 years ago

8 0

Answer:

y+4=-2(x-4)

Step-by-step explanation:

y-y1=m(x-x1)

y-(-4)=-2(x-4)

y+4=-2(x-4)

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is the diameter of a circle with center p. the coordinates of the endpoints of the diameter are (1, 3) and (6, 6). what are the

Rashid [163]

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3 years ago

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3 years ago

Find x and y please

Alexandra [31]

Answer:

Answer:

x → 4.24
y → 4.24

Step-by-step explanation:

» x and y can only be got through using trigonometric ratios:

✍ For x :

${ \tt{ \sin( \theta) = \frac{opposite}{hypotenuse} }} \\ \\ { \tt{ \sin(45 \degree) = \frac{x}{6} }} \\ \\ { \tt{x = 6 \times \sin(45 \degree) }} \\ { \tt{x = 4.24}}$

✍ For y :

${ \tt{ \cos( \theta) = \frac{adjacent}{hypotenuse} }} \\ \\ { \tt{ \cos( 45 \degree) = \frac{y}{6} }} \\ \\ { \tt{y = 6 \times \cos(45 \degree) }} \\ { \tt{y = 4.24}}$

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2 years ago

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An evolutionary biologist examined the relative fitness of Escherichia coli bacteria grown for 2000 generations, about 300 days,

Varvara68 [4.7K]

Answer:

Step-by-step explanation:

Hello!

The variable of interest is:

X: fitness of a line of E. coli grown on an acidic environment.

n= 6 E. coli lines

Recorded fitness for each line: 1.24, 1.22, 1.23, 1.24, 1.18, 1.09

The relative fitness of 1 indicates that both bacteria types are equally fit.

A relative fitness larger than 1 indicates that the acid-evolved line is more fit than the parental line kept at neutral pH when both are grown in acidic conditions.

Meaning that if the average fitness of the E. coli lines grown on an acidic environment is greater than 1 then they are better adjusted to live in acidic conditions, symbolically: μ > 1

The statistic hypotheses are:

H₀: μ ≤ 1

H₁: μ > 1

α: 0.05

Assuming that the variable has a normal distribution you have to apply a one-sample t-test:

$t= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } } ~~t_{n-1}$

X[bar]= 1.20

S= 0.06

$t_{H_0}= \frac{1.20-1}{\frac{0.06}{\sqrt{6} } } = 8.40$

The p-value for this test is 0.0002

Since the p-value= 0.0002 is less than α:0.05 the decision is to reject the null hypothesis.

Then at a 5% significance level, there is significant evidence to conclude that the bacteria evolved in acidic pH are better adapted to acidic conditions.

I hope you have a SUPER day!

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3 years ago

Find the common ratio of the sequence. 2, -10, 50, -250, ...

ludmilkaskok [199]

The common ratio is -5, note that each value is the previous value multiplied by -5.

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2 years ago

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