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3 years ago

7y-23 plus 23x-16 plus 8x -21 find x and Y show steps please

1 answer:

6 0

7y - 23 + 23x -16 + 8x - 21
31x + 7y - 60
What is the original problem equal to? You can't go any further unless you know.

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Two different types of polishing solutions are being evaluated for possible use in a tumble-polish operation for manufacturing i

murzikaleks [220]

Answer:

p-value: 1.000

There is enough evidence at the 1% level of significance to suggest that the proportions are not equal.

Step-by-step explanation:

We will be conducting a difference of 2 proportions hypothesis test

The hypothesis for this test is:

H_0: p1 - p2=0

H_a: p1 - p2 ≠0

(p1 ) = 252/300 = 0.84

(p2) = 195/300 = 0.65

This is a 2 tailed test with a significance level of 1%. So our critical values are: z > 2.575 and z < -2.575

See the attached photo for the calculations for this test

8 0

2 years ago

What is the factor of 28/88

Llana [10]

The factors are:
4, 2, 1

6 0

2 years ago

what is the area of a triangle if the base is 7 less than 5 times a number and the height is 3 times the number and eight

Flauer [41]

Area of a triangle is half multiplied by the height multiplied by the base.So multiplying the two together and halfing is shown below

6 0

2 years ago

Proportions happen because two events or scenarios are ___

Anni [7]

<span>Proportions happen because two events or scenarios are related or associated with one another </span>

6 0

2 years ago

Can someone help me find the equivalent expressions to the picture below? I’m having trouble

miss Akunina [59]

Answer:

Options (1), (2), (3) and (7)

Step-by-step explanation:

Given expression is $\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}$ .

Now we will solve this expression with the help of law of exponents.

$\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}=\frac{\sqrt[3]{(2^3)^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}$

$=\frac{\sqrt[3]{2\times 3} }{3\times2^{\frac{1}{9}}}$

$=\frac{2^{\frac{1}{3}}\times 3^{\frac{1}{3}}}{3\times 2^{\frac{1}{9}}}$

$=2^{\frac{1}{3}}\times 3^{\frac{1}{3}}\times 2^{-\frac{1}{9}}\times 3^{-1}$

$=2^{\frac{1}{3}-\frac{1}{9}}\times 3^{\frac{1}{3}-1}$

$=2^{\frac{3-1}{9}}\times 3^{\frac{1-3}{3}}$

$=2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }$ [Option 2]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2$ [Option 1]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2$

$=(2^2)^{\frac{1}{9}}\times (3^2)^{-\frac{1}{3} }$

$=\sqrt[9]{4}\times \sqrt[3]{\frac{1}{9} }$ [Option 3]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(2^2)^{\frac{1}{9}}\times (3^{-2})^{\frac{1}{3} }$

$=\sqrt[9]{2^2}\times \sqrt[3]{3^{-2}}$ [Option 7]

Therefore, Options (1), (2), (3) and (7) are the correct options.

6 0

2 years ago