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mamaluj [8]
3 years ago
14

Why is the end behavior of a quadratic function different from a linear function?

Mathematics
1 answer:
inna [77]3 years ago
4 0

A linear function can’t repeat but a quadratic function can

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A parallelogram is shown below:
nordsb [41]

9514 1404 393

Answer:

  A.  2 square feet

  B.  draw a diagonal; 1 square foot

Step-by-step explanation:

A. The area of a parallelogram is given by the formula ...

  A = bh

Here the base (b) is 3 feet, and the height (h) is 2/3 ft. The area is ...

  A = (3 ft)(2/3 ft) = 2 ft^2 . . . . area of the parallelogram

__

B. <em>Drawing a diagonal</em> (AC or BD) will divide the figure into two congruent triangles. Each triangle has the same base and height as the original parallelogram. The area of one of the triangles is given by ...

  A = 1/2bh

Using the given dimensions, ...

  A = 1/2(3 ft)(2/3 ft) = 1 ft^2 . . . . area of one triangle

7 0
3 years ago
Evaluate using P E M D A S 2 x 3 + 4
Juli2301 [7.4K]

Step-by-step explanation:

We would do multiplication first, because of order of operations.

2x3=6

Now we add.

6+4=10

hope it helps! :3

8 0
3 years ago
Read 2 more answers
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!

7 0
3 years ago
Find the value of x.
Ira Lisetskai [31]

Answer:

The value of x is 0

Step-by-step explanation:

5 0
3 years ago
Which set of ordered pairs does not represent a function?
Leokris [45]
The last one does not represent a function because for it to be a function none of the x’s can repeat and in the last one the 0 repeats.
4 0
3 years ago
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