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3 years ago

Simplify completely x2 + 4x - 45/x2 +10x+9 and find the restrictions on the variable.

Mathematics

1 answer:

murzikaleks [220]3 years ago

4 0

Answer:

The correct answer is,

${x}^{2} + 14x - \frac{45}{ {x}^{2} } + 9$

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Winona’s bowling scores for the past nine weeks are 195, 180, 195, 212, 208, 231, 179, 246, and 195. Find the mean, median, and

belka [17]

Answer: Mean = 204.6 , Median = 195 and mode =195

Step-by-step explanation:

The given data for Winona’s bowling scores for the past nine weeks : 195, 180, 195, 212, 208, 231, 179, 246, 195.

The formula to find Mean is given by :-

$\text{Mean}=\dfrac{\text{Sum of all observations}}{\text{Number of observations}}\\\\=\dfrac{1841}{9}=204.555555556\approx204.6$

For median , first we arrange data values in order as

179, 180, 195, 195, 195, 208, 212, 231, 246

Since the number of data values is 9 (odd).

∴ Median = Middle most data value = 195

Also, Mode = Most occurring data value = 195

3 0

3 years ago

Plzzzzzzzzzz help me

Ierofanga [76]

The answer would be the top right graph where none of the lines intersect since it shows that there is no possible real solutions for the system if none of the lines intersect on a certain point.

Remember: If the lines intersect, there is a solution; if the lines do not intersect, there is no solution; if the lines fall on the same line, or the equations are equivalent, then the solution is all real numbers.

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3 years ago

A study of the effects of color on easing anxiety compared anxiety test scores of participants who completed the test printed on

NikAS [45]

Answer:

Step-by-step explanation:

To solve this, we would follow these simple steps. We have

unvrs :

The arithmetic mean, x-bar for the yellow paper group (Y) = 20.6

The arithmetic mean, x-bar for the green paper group (G) = 21.75

Recall that, H0: µY = µG

And from the data we have, we can see that

H0: µY< µG

We proceed to say that the

T-Test-statistic = -0.404

Also, the p-value: 0.349

From our calculations, we can see that the p-value > 0.05, and as such, we conclude that we will not reject H0. This is because there is not enough evidence to show that test that is printed on the yellow paper decreases anxiety at a 0.05 significance level.

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3 years ago

Plz read and answerrrrrrrrrrr

natulia [17]

Answer:

16.7%.

Step-by-step explanation:

There are initially $24 + 15 + 8 = 47$ pencils in the bag.

Take a pencil out of this bag of 47 pencils. 15 out of the 47 pencils blue. Let $A$ represent the event of getting a blue pencil on the first pick. The probability of getting a blue pencil is:

$\displaystyle P(A) = \frac{15}{47}$ .

There are now $47 - 1 = 46$ pencils left in the bag. However, given that the first pencil removed from the bag is blue, the number of red pencils in the bag will still be 24. Take another pencil out of this bag of 46 pencils. Let $B$ represent the event of getting a red pencil on the second pick. The possibility that the second pencil is red given that the first pencil is blue will be: