All categories

1 year ago

{(-2/3)^-2}^2 can someone pls help me in this

Mathematics

1 answer:

anzhelika [568]1 year ago

8 0

The value of the expression given in the task content is; 81/16.

<h3>What is the value of the expression?</h3>

According to the task content;

We are required to determine the value of the expression given in the task content.

{(-2/3)^-²}^²

= {(-2/3)^-⁴

= (-3/2)-⁴

= 81/16

Therefore, the value of the expression given in the task content is; 81/16.

You might be interested in

C=59(F−32)

neonofarm [45]

I says every time you INCREASE by 1 degree Farenheit you go up 59 degrees C, so it is false. Same for III, if you go up or down 1 degree C that's not a change of 59 in degrees F.

That leaves B, which is right if you experiment with the math.

8 0

2 years ago

MA = (4x - 2) and mzB = (11x + 17)°. Find x if the angles are

charle [14.2K]

The answer is a.) 165 sorry if that’s wrong but i hope it’s right

3 0

2 years ago

Read 2 more answers

If middle school and senior school are winning football games in the ratio 3:9 . If the Middle school won 17 games, calculate ho

Helen [10]

The Middle school won 17 games so the senior school won 51 games

4 0

2 years ago

Solve for x: 3|x - 3| + 2 = 14

astra-53 [7]

D) is the correct one

3 0

2 years ago

Read 2 more answers

The amount of time Ricardo spends brushing his teeth follows a Normal distribution with unknown mean and standard deviation. Ric

Papessa [141]

Let $X$ denote the number of minutes spent brushing his teeth, and let $\mu$ be the mean and $\sigma$ the standard deviation for this distribution.

$\mathbb P(X$

The z-score corresponding to this probability is approximately $z=-0.2533$ , which means

$\dfrac{1-\mu}\sigma=-0.2533\iff\mu-0.2533\sigma=1$

Next, (note the sign change)

$\mathbb P(X>2)=0.02\implies\mathbb P(X\le2)=\mathbb P\left(\dfrac{X-\mu}\sigma\le\dfrac{2-\mu}\sigma\right)=0.98$

The corresponding z-score is approximately $z=2.0538$ , so you have

$\dfrac{2-\mu}\sigma=2.0538\iff\mu+2.0538\sigma=2$

Solving the two equations for $\mu$ and $\sigma$ , you'll find that the mean is approximately $\mu=1.1098$ and the standard deviation is approximately $\sigma=0.4334$ .

7 0

2 years ago