Ask question

Ask question

All categories

3 years ago

10

Find the value of the expression. what's the Value

2 answers:

lbvjy [14]3 years ago

8 0

Answer:

1

Step-by-step explanation:

(⅗)⁰ = 1

1^-2 = 1/1² = 1/1 = 1

Contact [7]3 years ago

4 0

Answer:

1

Step-by-step explanation:

(3/5) ^0 ^-2

We know that a^b^c = a^(b*c)

(3/5) ^(0*-2) = (3/5)^0

We know a^0 = 1 as long as a does not equal1

(3/5)^0 =1

You might be interested in

Determine whether the formula describes y as a function of x. Explain your reasoning.. . y = -x

Alex17521 [72]

The given formula above, y = -x, is a sample of a function. A function is an equation that would give only one value of y for every value of x. This is the same as saying that exactly one output is set by each input. If we take x = -1, the value of y is 1 and nothing else.

7 0

3 years ago

Help me with this plz.

kondaur [170]

The answer is F because my brother is in high school and he knows the's things

7 0

3 years ago

Read 2 more answers

What is x in 3.6x=1.6x+24

Ipatiy [6.2K]

Answer:

x=12

Step-by-step explanation:

3.6x=1.6x+24

We simplify the equation to the form, which is simple to understand

3.6x=1.6x+24

We move all terms containing x to the left and all other terms to the right.

+3.6x-1.6x=+24

We simplify left and right side of the equation.

+2x=+24

We divide both sides of the equation by 2 to get x.

x=12

7 0

3 years ago

How to I find this? I tried using Pythagorean theorem, but it didn't work

jarptica [38.1K]

The Pythagorean theorem works.

$12^2+9^2=c^2$
$144+81=c^2$
$225=c^2$
$15=c$

Therefore the perimeter is 12 + 9 + 15 = 36 in. (B)

4 0

3 years ago

Read 2 more answers

The amount of lateral expansion (mils) was determined for a sample of n = 8 pulsed-power gas metal arc welds used in LNG ship co

Alona [7]

Answer:

95% Confidence interval for the variance:

$3.6511\leq \sigma^2\leq 34.5972$

95% Confidence interval for the standard deviation:

$1.9108\leq \sigma \leq 5.8819$

Step-by-step explanation:

We have to calculate a 95% confidence interval for the standard deviation σ and the variance σ².

The sample, of size n=8, has a standard deviation of s=2.89 miles.

Then, the variance of the sample is

$s^2=2.89^2=8.3521$

The confidence interval for the variance is:

$\dfrac{ (n - 1) s^2}{ \chi_{\alpha/2}^2} \leq \sigma^2 \leq \dfrac{ (n - 1) s^2}{\chi_{1-\alpha/2}^2}$

The critical values for the Chi-square distribution for a 95% confidence (α=0.05) interval are:

$\chi_{0.025}=1.6899\\\\\chi_{0.975}=16.0128$

Then, the confidence interval can be calculated as:

$\dfrac{ (8 - 1) 8.3521}{ 16.0128} \leq \sigma^2 \leq \dfrac{ (8 - 1) 8.3521}{1.6899}\\\\\\3.6511\leq \sigma^2\leq 34.5972$

If we calculate the square root for each bound we will have the confidence interval for the standard deviation:

$\sqrt{3.6511}\leq \sigma\leq \sqrt{34.5972}\\\\\\1.9108\leq \sigma \leq 5.8819$

6 0

3 years ago