All categories

3 years ago

What's the answer to (-6a^-2b^4c^0)^3

Mathematics

2 answers:

omeli [17]3 years ago

5 0

I hope this helps you

(-2)^3.3^3.a^(-2).3. b^4.3 .c0^0.3

-2^3. 3^3. a^-6. b^12 .1

Korvikt [17]3 years ago

4 0

-6a -2b^4

^if you simplify, that would be your answer.

You might be interested in

A random sample of 10 parking meters in a resort community showed the following incomes for a day. Assume the incomes are normal

GenaCL600 [577]

Answer:

A 95% confidence interval for the true mean is [$3.39, $6.01].

Step-by-step explanation:

We are given that a random sample of 10 parking meters in a resort community showed the following incomes for a day;

Incomes (X): $3.60, $4.50, $2.80, $6.30, $2.60, $5.20, $6.75, $4.25, $8.00, $3.00.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean income = $\frac{\sum X}{n}$ = $4.70

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = $1.83

n = sample of parking meters = 10

$\mu$ = population mean

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.262 < $t_9$ < 2.262) = 0.95 {As the critical value of t at 9 degrees of

freedom are -2.262 & 2.262 with P = 2.5%}

P(-2.262 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.262) = 0.95

P( $-2.262 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu$ < $2.262 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.262 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.262 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.262 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.262 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $4.70-2.262 \times {\frac{1.83}{\sqrt{10} } }$ , $4.70+ 2.262 \times {\frac{1.83}{\sqrt{10} } }$ ]

= [$3.39, $6.01]

Therefore, a 95% confidence interval for the true mean is [$3.39, $6.01].

The interpretation of the above result is that we are 95% confident that the true mean will lie between incomes of $3.39 and $6.01.

Also, the margin of error = $2.262 \times {\frac{s}{\sqrt{n} } }$

= $2.262 \times {\frac{1.83}{\sqrt{10} } }$ = 1.31

4 0

3 years ago

The map at the right has an Euler Path in which you can pass through each doorway exactly once. Which 2 rooms must be a starting

bogdanovich [222]

Answer:I think the answer is d

Step-by-step explanation:

5 0

3 years ago

A car has mass 1500 kg and is traveling at a speed of 35 miles/hour. what is its kinetic energy in joules? (Be sure to convert m

photoshop1234 [79]

Answer:

Factor by which kinetic energy increase = 4 times

Step-by-step explanation:

Given,

Mass of the car, v1 = 1500 kg

initial speed of car = 35 miles/h

$=\dfrac{35\times 1609.34}{3600}\ m/s$

= 15.64 m/s

Initial kinetic energy of the car is given by,

$k_1\ =\ \dfrac{1}{2}.m.v_1^2$

$=\ \dfrac{1}{2}\times 1500\times (15.64)^2\ joule$

= 183606.46 J

Final velocity of car v2 = 70 miles/hour

$=\dfrac{70\times 1609.34}{3600}$

= 31.29 m/s

So, final kinetic energy of car is given by

$k_2\ =\ \dfrac{1}{2}.m.v_2^2$

$=\ \dfrac{1}{2}\times 1500\times (31.29)^2$

= 734425.84 J

Now, the ratio of final to initial kinetic energy can be given by,

$\dfrac{k_2}{k_1}=\ \dfrac{734425.84}{183606.46}$

$=>\ k_2\ =\ 4k_1$

Hence, the kinetic energy will increase by 4 times.

7 0

3 years ago

Factor the following problems -13=-4a-2b+c 3=4a+2b+c 5=16a+4b+c Please Help!

Pie

1) -13 = -4a - 2b + c
-13 = -2(2a + b) + c

2) 3 = 4a + 2b +c
3 = 2(2a + b) + c

3) 5 = 16a + 4b + c
5 = 4(4a + b) + c

[ Final Answers are in bold ]

Hope this helps!

8 0

3 years ago

Match the measureable quantity giving by its formula with the appropriate unit(s)

Anarel [89]

Can you give the things you have to match?

8 0

3 years ago

Read 2 more answers