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3 years ago

What TWO expressions equal 12−(−33) ?

Mathematics

1 answer:

erastova [34]3 years ago

8 0

Answer:

12+33 and 33 - (-12)

Step-by-step explanation:

12 - (-33) = 45

12 + 33 gives us 45

33 - (-12) gives us 45

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When conducting a two-tailed z test with a = 0.01, the test value was computed to be 2.07. The decision would be to reject or to

vekshin1

You reject the Null Hypothesis only if the p-value is less than alpha.
p < 0.01

To find the p-value, you need to look up test value 2.07 in a standard normal table. The p-value is probability that Z > 2.07. For a two-tailed test, you include both positive and negative cases. |Z| > 2.07.

When you look up 2.07 you get about 0.98.
This means there is about 2% chance Z > 2.07 or 4% chance |Z| > 2.07.

For the two-tailed test we use p = 0.04
.04 > .01
Therefore we do Not reject the Null Hypothesis.

3 0

3 years ago

What is the simplest form of ^4 sqrt 81x^8y^5

lara31 [8.8K]

ANSWER

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

EXPLANATION

We want to simplify:

$\sqrt[4]{81 {x}^{8} {y}^{5} }$

We can split the radical sign to obtain:

$\sqrt[4]{81} \times \sqrt[4]{ {x}^{8} } \times \sqrt[4]{ {y}^{5} }$

$\sqrt[4]{81} \times \sqrt[4]{ {x}^{8} } \times \sqrt[4]{ {y}^{4} \times y}$

$\sqrt[4]{81} \times \sqrt[4]{ {x}^{8} } \times \sqrt[4]{ {y}^{4}} \times \sqrt[4]{y}$

$\sqrt[4]{ {3}^{4} } \times \sqrt[4]{ {x}^{8} } \times \sqrt[4]{ {y}^{4}} \times \sqrt[4]{y}$

Recall that:

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

${3}^{4 \times \frac{1}{4} } \times {x}^{8 \times \frac{1}{4} } \times {y}^{4 \times \frac{1}{4} }\times \sqrt[4]{y}$

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

6 0

4 years ago

Read 2 more answers

find a curve that passes through the point (1,-2 ) and has an arc length on the interval 2 6 given by 1 144 x^-6

taurus [48]

Answer:

$f(x) = \frac{6}{x^2} -8$ or $f(x) = -\frac{6}{x^2} + 4$

Step-by-step explanation:

Given

$(x,y) = (1,-2)$ --- Point

$\int\limits^6_2 {(1 + 144x^{-6})} \, dx$

The arc length of a function on interval [a,b]: $\int\limits^b_a {(1 + f'(x^2))} \, dx$

By comparison:

$f'(x)^2 = 144x^{-6}$

$f'(x)^2 = \frac{144}{x^6}$

Take square root of both sides

$f'(x) =\± \sqrt{\frac{144}{x^6}}$

$f'(x) = \±\frac{12}{x^3}$

Split:

$f'(x) = \frac{12}{x^3}$ or $f'(x) = -\frac{12}{x^3}$

To solve fo f(x), we make use of:

$f(x) = \int {f'(x) } \, dx$

For: $f'(x) = \frac{12}{x^3}$

$f(x) = \int {\frac{12}{x^3} } \, dx$

Integrate:

$f(x) = \frac{12}{2x^2} + c$

$f(x) = \frac{6}{x^2} + c$

We understand that it passes through $(x,y) = (1,-2)$ .

So, we have:

$-2 = \frac{6}{1^2} + c$

$-2 = \frac{6}{1} + c$

$-2 = 6 + c$

Make c the subject

$c = -2-6$

$c = -8$

$f(x) = \frac{6}{x^2} + c$ becomes

$f(x) = \frac{6}{x^2} -8$

For: $f'(x) = -\frac{12}{x^3}$

$f(x) = \int {-\frac{12}{x^3} } \, dx$

Integrate:

$f(x) = -\frac{12}{2x^2} + c$

$f(x) = -\frac{6}{x^2} + c$

We understand that it passes through $(x,y) = (1,-2)$ .

So, we have:

$-2 = -\frac{6}{1^2} + c$

$-2 = -\frac{6}{1} + c$

$-2 = -6 + c$

Make c the subject

$c = -2+6$

$c = 4$

$f(x) = -\frac{6}{x^2} + c$ becomes

$f(x) = -\frac{6}{x^2} + 4$

3 0

3 years ago

how can something that could be a need become a want? A) if it is highly desired B) if the need is already fulfilled C) if it is

sergij07 [2.7K]

The answer is B) if the need is already fulfilled

8 0

4 years ago

Find the percent decrease. Round to the nearest percent.<br> From 145 pounds to 138 pounds

alina1380 [7]

Answer:

-4.83% however rounded should be05.00

Step-by-step explanation:

5 0

3 years ago