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3 years ago

What's the answer to this question

Mathematics

1 answer:

Licemer1 [7]3 years ago

4 0

To create a similar line, you have to make a line of any length, open up the compass to the length of the original line, then draw an arc with that length on the created line. Choice 2 follows these set of steps.

:)

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Santa's Wonderland is an extravagant holiday light display that is open from early November through January each year. The entry

Fofino [41]

Ok,
f(0.35)= 7f/20

f(-5.2)=-26f/5

f(10)= 10f

f(-0.5)= -f/2

as for the last question I am not quite sure, sorry....hope I helped a little :)

5 0

4 years ago

Rounded to the nearest hundredth, what is the positive solution to the quadratic equation 0 = 2x2 + 3x – 8? Quadratic formula: x

Lynna [10]

ANSWER

1.39

EXPLANATION

The given quadratic equation is

$0 = 2 {x}^{2} + 3x - 8$

This is the same as,

$2 {x}^{2} + 3x - 8 = 0$

Comparing to

$a {x}^{2} + bx + c = 0$

We have

a=2, b=3,c=-8

Using the quadratic formula, the solution is given by:

$x = \frac{ - b \pm \: \sqrt{ {b}^{2} - 4ac} }{2a}$

We substitute the values to get,

$x = \frac{ - 3\pm \: \sqrt{ {3}^{2} - 4(2)( - 8)} }{2(2)}$

$x = \frac{ - 3\pm \: \sqrt{ 73} }{4}$

The positive root is

$x = \frac{ - 3 + \: \sqrt{ 73} }{4} = 1.39$

to the nearest hundredth.

4 0

3 years ago

Read 2 more answers

Evaluate: |18| A:1/18 B:-1/-18 C:-18 D18

masya89 [10]

Answer will be D. Absolute value can never be negative.

4 0

3 years ago

Average precipitation for the first 7 months of the year, the average precipitation in toledo, ohio, is 19.32 inches. if the ave

Colt1911 [192]

Part A:

The probability that a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\sigma}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that a randomly selected year will have precipitation greater than 18 inches for the first 7 months is given by:

$P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{2.44}\right) \\ \\ =1-P(z\ \textless \ -0.5410) \\ \\ =1-0.29426=\bold{0.7057}$

Part B:

The probability that an n randomly selected samples of a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\frac{\sigma}{\sqrt{n}}}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that 5 randomly selected years will have precipitation greater than 18 inches for the first 7 months is given by:

 $P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{\frac{2.44}{\sqrt{5}}}\right) \\ \\ =1-P(z\ \textless \ -1.210) \\ \\ =1-0.1132=\bold{0.8868}$

7 0

3 years ago

The formula for estimating the number, N, of a certain product sold is N = 8800ln(7t + 9), where t is the number of years after

arsen [322]

To determine or predict the expected number of sales after 2 years, we substitute 2 to the t of the givne equation.
N = (8800)xln(7(2) + 9)
N = 27,592.34
Thus, it is expected that the number of sales after 2 years is 27,592 units.

6 0

3 years ago