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3 years ago

Using the graph below, what is f(5)?

Mathematics

1 answer:

Olin [163]3 years ago

5 0

F(5) would be the y value of where the line is when it is at x 5

The line crosses at -6.5

F(5) = -6.5

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If the rectangle below are similar what Is the length of X

tino4ka555 [31]

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9/4=2.25*2=4.5
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3 years ago

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A weather forecast says that there is a 40% chance of rain today find the odds against rain

san4es73 [151]

Answer:

60%

Step-by-step explanation:

100%-40%-60%

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2 years ago

Convert 4/3 to an mixed number

wariber [46]

1 1/3. One whole number is 3/3 so 3/3=1 then u put 1/3

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3 years ago

An English professor assigns letter grades on a test according to the following scheme. A: Top 15% of scores B: Scores below the

lisabon 2012 [21]

Answer:

59 to 66

Step-by-step explanation:

Mean test scores = u = 74.2

Standard Deviation = $\sigma$ = 9.6

According to the given data, following is the range of grades:

Grade A: 85% to 100%

Grade B: 55% to 85%

Grade C: 19% to 55%

Grade D: 6% to 19%

Grade F: 0% to 6%

So, the grade D will be given to the students from 6% to 19% scores. We can convert these percentages to numerical limits using the z scores. First we need to to identify the corresponding z scores of these limits.

6% to 19% in decimal form would be 0.06 to 0.19. Corresponding z score for 0.06 is -1.56 and that for 0.19 is -0.88 (From the z table)

The formula for z score is:

$z=\frac{x-u}{\sigma}$

For z = -1.56, we get:

$-1.56=\frac{x-74.2}{9.6}\\\\ x = 59$

For z = -0.88, we get:

$-0.88=\frac{x-74.2}{9.6}\\\\ x = 66$

Therefore, a numerical limits for a D grade would be from 59 to 66 (rounded to nearest whole numbers)

7 0

3 years ago

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago