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3 years ago

Can someone explain how to do problems like (-2+4) or (4+ -6)

Mathematics

2 answers:

sladkih [1.3K]3 years ago

8 0

Answer:

Step-by-step explaination

-2 + 4 is 2 4+ -6 is -2

KiRa [710]3 years ago

4 0

Answer:

All you have to do is just add the numbers inside the parenthesis. (4+-2) is the same as 4-2 which is equal to 2. (4+-6) is the same as 4-6 which is equal to -2.

Step-by-step explanation:

If this has helped please mark as brainliest

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N = 9

g100num [7]

Answer:

The p-value is 0.809.

Step-by-step explanation:

In this case we need to perform a significance test for the standard deviation.

The hypothesis is defined as follows:

H₀: σ₀ = 4 vs. Hₐ: σ₀ ≤ 4

The information provided is:

n = 9

s = 3

Compute the Chi-square test statistic as follows:

$\chi^{2}=\frac{(n-1)s^{2}}{\sigma_{0}^{2}}$

$=\frac{(9-1)\cdot (3)^{2}}{(4)^{2}}\\\\=\frac{8\times 9}{16}\\\\=4.5$

The test statistic value is 4.5.

The degrees of freedom is:

df = n - 1

= 9 - 1

= 8

Compute the p-value as follows:

$p-value=P(\chi^{2}_{9}>4.5)=0.809$

*Use a Chi-square table.

Thus, the p-value is 0.809.

7 0

3 years ago

Tems11 [23]

Answer:

one board game cost 15 dollars

75/5=15

15y

y = number of board games and 15= the price of each board game

so to calculate any number of board games bought just input the number of board games bought for y in the expression 15y and you will get the price.

Step-by-step explanation:

6 0

2 years ago

Find the length of the curve. R(t) = cos(8t) i + sin(8t) j + 8 ln cos t k, 0 ≤ t ≤ π/4

arsen [322]

we are given

$R(t)=cos(8t)i+sin(8t)j+8ln(cos(t))k$

now, we can find x , y and z components

$x=cos(8t),y=sin(8t),z=8ln(cos(t))$

Arc length calculation:

we can use formula

$L=\int\limits^a_b {\sqrt{(x')^2+(y')^2+(z')^2} } \, dt$

$x'=-8sin(8t),y=8cos(8t),z=-8tan(t)$

now, we can plug these values

$L=\int _0^{\frac{\pi }{4}}\sqrt{(-8sin(8t))^2+(8cos(8t))^2+(-8tan(t))^2} dt$

now, we can simplify it

$L=\int _0^{\frac{\pi }{4}}\sqrt{64+64tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{1+tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{sec^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8sec(t) dt$

now, we can solve integral

$\int \:8\sec \left(t\right)dt$

$=8\ln \left|\tan \left(t\right)+\sec \left(t\right)\right|$

now, we can plug bounds

and we get

$=8\ln \left(\sqrt{2}+1\right)-0$

so,

$L=8\ln \left(1+\sqrt{2}\right)$ ..............Answer

5 0

3 years ago

2^8-x+16=15 HELLO! How is everyone lately?

Lera25 [3.4K]

Answer:

I’m fine ig,

how are you doin?

Step-by-step explanation : )

6 0

3 years ago

Read 2 more answers

Refer to the table below if needed.

elena-14-01-66 [18.8K]

A quadrant is the area that is divided into the x and y axes

The quadrants in which tan $\theta$ and cot $\theta$ are positive are I and III

<h3>How to determine the quadrants</h3>

The tangent of an angle is calculated as:

$\tan(\theta) = \frac{sin(\theta)}{\cos(\theta)}$

While the cotangent of the angle is calculated as:

$\cot(\theta) = \frac{cos(\theta)}{\sin(\theta)}$

The above equations mean that:

For the tangent and cotangent of an angle to be positive, then the sine and the cosine of the angle must have the same sign.

In the first quadrant, the sine and the cosine angles are positive.
In the third quadrant, the sine and the cosine angles are negative.

Hence, the quadrants in which tan $\theta$ and cot $\theta$ are positive are I and III

Read more about trigonometry ratios at:

brainly.com/question/8120556

7 0

2 years ago